Programs and Programming

	Learning Objectives
	Understand the terms source code and executable
	Be able to use an editor to enter a source program into a file
	Be able to compile the source code to produce the executable
	Be able to run the program
	Understand the basic layout of a simple program - including the keyword main, the use of the semi-colon and the cout statement
	Combine the above to be able to write simple programs to output text to the screen

	Source Code from Notes
	Additional Examples
	Quick Check Exercises
	Additional Exercises to try

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Basic Programming

Source Code from Notes

Example

   // Program written October 1997
   // by jmb

#include <iostream.h>

main()
{
        cout <<< "Hello World!\n";
        // \n causes new line 
}

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Basic Programming

Case Study

PROBLEM:

Write a program to display the text, This is my first program, on the screen.

SOLUTION:

The program must start with #include as the the program will output text to the screen.

The main body of the program is contained within curly braces which follow the keyword main().

The text is displayed on the screen using the cout statement. Note that this statement requires a semi--colon at the end to indicate the end of the statement.

Comments have also been included to make the programme easier to understand.

// My first program
// Written by jmb
// October 1998

#include <iostream.h>

main()
{
	cout <<  "This is my first program \n";
	// \n is required to produce newline 
}

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Basic Programming Additional Exercises

1. Write a program to display your name, address and telephone number on the screen.

2. Correct the mistakes in the following program, type it in and compile the program.

   
   // Example Program
      Week 1
      April 1997 
   //
   mAin(
   
           / display data to screen 
           cout << "Example Program\n";
           cout << 'for Introduction to Programming Class\n 
           cout >> " Autumn Term 1999/n";
           cout >> " Illustrates the basic components of C++\n;
   }
   

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Answers

1. It includes the header file iostream.h into the source code. This file is required for input and output to the program. Therefore, if the cout statement is used to output text/data to the screen it is required.

2. Source code is the actual C++ program which is typed into the file. This is then compiled into machine code which can be executed by the computer.

3. No - a comment is not a program statement, so does not require to be terminated by a semi-colon.

4. The text to be printed must be within double quotes.

   
   cout << "Hello World\n";
   

5. Keywords in C++ must be lowercase COUT must be written as cout

6. The output operator is << and not >>

7. The \n causes the a new line. It is required at the end of Hello WOrld to ensure the cursor returns to the next line.

8. The \n (newline) is text to be printed and must be within the double quotes.

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Back to Main Page for Lecture 1

 Variables,  Data Types, Constants and Operators 
 
 






Learning Objectives


 
Be able to declare variables


Be able to write simple arithmetic expressions


Be able to input values typed at the keyboard into a program using cin



Be able to output values to the screen using cout 


Be  familiar with the concept of constants 


Be able to combine above to write simple programs which require the use to
input values, make calculations and output the results








Source Code from Notes





  Additional Examples



 
Quick Check Exercises 



 
Additional Exercises to try



 
Test Cases for Tutorial










 Previous Lecture
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 Variables, Data Types, Constants and Operators 

 Source Code from Notes 
 
 




Example 1



//  program to add two integer values 
//  and display result                 

#include <iostream.h>

main()
{
        // declare variables 
        int     value1, value2, sum;

        // assign values and compute the result 
        value1 = 32;
        value2 = 27;
        sum = value1 + value2;

        // display the result 
        cout << "The sum of " << value1 << " and  << value2  << " is " <<
		sum << "\n" ;
}






Example 2

This is the example program to calculate the area of a rectangle


#include <iostream.h>

main()
{
        // type definitions 
        // 3 variables - length, breadth and area 

        float   length, breadth, area;

        // input phase - read length and breadth from user 

        cout << "Enter the length and breadth\n";
         cin >> length >> breadth;

        // computation - calculate area 

        area = length * breadth;

        // output phase 

        cout << "The area of the rectangle with length  " << length << " and breadth "
	     << breadth  << " is " << area << "\n"; 

}





Example 3

This example  calculates the average value of three numbers entered by
the user.



// Program to calculate the average of three numbers
#include <iostream.h>

main()
{
        float   number1,number2,number3;
        float   average;

        // Enter values
        cout << "Enter three numbers: ";
        cin >> number1 >> number2 >> number3;

        // calculate average
        average = (number1 + number2 + number3)/3.0;

        // Output result
        cout << "The average value of " << number1 << " , " << number2 <<
             " and " << number3 << " is "  << average << "\n";
}






Example 4


This example prompts the user to input a number and then calculates the
reciprocal of this number.



// Program to calculate the reciprocal of a number
#include <iostream.h>

main()
{
        float   number;
        float   reciprocal;

        // Enter Value
        cout <<"Enter a number: ";
        cin >> number;

        // Calculate reciprocal
        reciprocal = 1.0 / number;

        // output result
        cout << "The reciprocal of " << number << " is " << reciprocal << "\n";
}






Example 5

This example illustrates the use of constants. It calculates the area and
circumference of a circle, when given the radius.


// program to calculate circumference and area of circle 

#include <iostream.h>

main()
{
	const float PI = 3.14156;
        // define variables
        float   radius,circumference,area;

        // Input radius
	cout << "Enter the radius: ";
        cin >> radius;

        // calculate radius and circumference
        circumference = PI * 2 * radius;
        area = PI * radius * radius;

        // display results
        cout << "The circumference is " << circumference << " and the rea is " << area 
	     << "\n"; 
                                           
}








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Variables,  Data Types, Constants and Operators

 Case Study 





PROBLEM:

Write a program to calculate an employees wages. They are paid at a  basic
rate of £6.65
hour, but are taxed at a rate of 23%
and
pay a pension contribution of 4% and 
pay a union
subscription of  1.25 per week. 

Write a program to read in the hours worked and calculate the take home
weekly wages of the employee.




SOLUTION:

The structure of the program will take the following form:-

#include <iostream.h>

// Declare constants 

main()
{
	// declare variables 

	// prompt user to enter number of hours worked and read in value 

	// Calculate wages 

	// output weekly wages 
}



Constants

We are told that the employee is paid 6.65 per hour is taxed at a rate of
23%, pays 4% pension contributions and a union fee of 1.25. We should make
these values be constants - since they have been given to use. In addition,
this makes it easier to change these values at a later date and if we use
meaningful names will make the program easier to read. Therefore we will
declare:-


const float hourly_rate = 6.65;
const float tax_rate    = 0.23;
const float pension_rate = 0.04;
const float union_fee   = 1.25;


I have converted the percentages to floating point values, since for example
23% = 23/100 * value.


Variables 

We are obviously going to require a variable to store the number of hours
worked, which we will assume can be a fractional number of hours and
therefore use a floating point value. We will also use variables for the
gross_pay, pension, tax, deductions and total_pay. All of which will be
floating point values.


        float   hours, gross_pay, tax, pension, deductions, total_pay;




Input

We require to prompt the user to enter the number of hours worked and then
read this into the variable hours .

        cout << "Enter the number of hours worked \n";
        cin >> hours;



 Calculation

There are a number of different ways this could be implemented. I have chosen
here to do it in a number of different stages. Calculating initially the
gross wages and then the deductions.


        gross_pay = hours * hourly_rate;
        tax = gross_pay * tax_rate;
        pension = gross_pay * pension_rate;
        deductions = tax + pension + union_fee;
        total_pay = gross_pay - deductions;




Output

Finally we have to output the result. This can be achieved by the statement

        cout << "The total weekly pay is " << total_pay << "\n";




Total Program 


#include <iostream.h>


main()
{
	// Declare constants 
	const float hourly_rate = 6.65;
	const float tax_rate    = 0.23;
	const float pension_rate = 0.04;
	const float union_fee   = 1.25;
        //declare variables 
        float   hours, gross_pay, tax, pension, deductions, total_pay;

        //prompt user to enter number of hours worked and read in value 
        cout << "Enter the number of hours worked \n";
        cin >> hours;

        // Calculate wages 
        gross_pay = hours * hourly_rate;
        tax = gross_pay * tax_rate;
        pension = gross_pay * pension_rate;
        deductions = tax + pension + union_fee;
        total_pay = gross_pay - deductions;

        // output weekly wages 
        cout << "The total weekly pay is " << total_pay << "\n";
}












Back to Main Page for Lecture 2 









 
  
 

 Variables, Data Types, Constants and Operators 

 Additional Exercises 
 





 Write a program which prompts the user to enter 2 integer values. The
program will then print their sum, difference, product, quotient and
remainder.


Write a program which will prompt the user to enter a positive floating
point number. The program will then display the integral part and the
fractional part of the number.



For example:-

number = 2.33

integral = 2

fractional = 0.33



Write a program which prompts the user to enter a number of days, hours and
minutes. The program will then calculate and display this as a total number
of minutes.



For example

days = 1

hours = 12

minutes = 15

Total number of minutes = 2175









Back to Main Page for Lecture 2 








 


  

 

 Variables, Data Types, Constants and Operators 
 

 






What is wrong with this program?


main()
{
	p=47;
	cout << p;
}



What is wrong with this declaration?


int x = y = 95;



Write four different C++ statements that subtract 1 from the integer variable
x.


Which of the following are valid C++ statements? Evaluate the ones which are
valid.


x = 65/5*2

65 / 4 = k;

p = 65(4*7);

x = 34 - 4 * 7;

x = 34 - ( 4 * 7);
 
54 %  5 = h ;

x =  54 % 7;







 Answers

                                        
	














































The #include <iostream.h> is missing and the variable  p 
has not been declared.


The equals sign can only be used in a declaration to initialise a variable.
Therefore it should be

int	x=95, y =95;



x = x - 1;
x --;
--x;
x -= 1;




 missing ; 
 not valid - should be k = 64/4;
 not valid 
 valid  x = 6
 valid  
 not valid
 valid x = 5







 


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Back to Main Page for Lecture 2 



                                        
















 Control Constructs 
 
 








Learning Objectives


 
Know the relational and logical operators



Be able to use the if statement 


Be able to use the if....else statement



Be able to use the else ... if statement to implement multi-way
branching


Know when it is suitable to use the switch statement for
multi-way branching and be able to implement it













Source Code from Notes





  Additional Examples



 
Quick Check Exercises 



 
Additional Exercises to try



 
Test Cases for Tutorials






 




 Previous Lecture
 Home


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 Control Constructs 

 Source Code from Notes 
 





Example 1



//Program to calculate absolute value of an integer

#include <iostream.h>

main()
{
        int     number;

        //Enter number
        cout << "Enter a number: ";
        cin >> number;

        //test if negative
        if (number < 0)
                number = -number;

        //output results
        cout << "The absolute value is " << number << "\n";
}






Example 2



//Program to test if number is even or odd

#include <iostream.h>

main()
{
        int     number,remainder;

        cout << "Enter a number: ";
        cin >> number;
        
        remainder = number % 2;

        if (remainder ==0)
                cout << "The number is even\n";
        else
                cout << "The number is odd\n";

}







Example 3



//Program to illustrate the use of nested if statements

#include <iostream.h>

main()
{
        int number;

        cout << "Enter a number between 1 and 99: ";
        cin >> number;

        if (number > 0 && number < 100 ) 
        {
                if (number < 10)
                        cout <<  "One digit number\n";
                else
                        cout << "Two digit number\n";
        }
        else
                cout << "Number not in range\n";

}






Example 4




#include <iostream.h>

main()
{
        int number, n_digits = 0;

        cout << "Enter a number: ";
        cin >> number;

	if (number >= 100)
        	n_digits = 3;
	else if (number >=10)
        	n_digits = 2;
	else if (number >=0)
        	n_digits = 1;
	else
        	cout <<  "Value out of range\n";

	if (n_digits > 0)
		cout << "The number has " << n_digits << " digits\n";
}






Example 5




#include <iostream.h>

main()
{
        char    day;


        cout << "Which day?";
        cin >> day

        switch(day){
                case 's': cout << "Weekend\n";
                          break;
                case 'm': cout << "Week day\n";
                          break;
                case 't': cout << "Week day\n";
                          break;
                case 'w': cout << "Week day\n";
                          break;
                case 'f': cout << "Week day\n";
                          break;
                default:  cout << "Not a day\n";
	}
}






Example 6




#include <iostream.h>

main()
{
        char    day;

        cout << "Which day?";
        cin >> day;

        switch(day){
                case 's': case 'S': cout << "Weekend\n";
                          break;
                case 'm': case 'M': cout << "Week day\n";
                          break;
                case 't': case 'T': cout << "Week day\n";
                          break;
                case 'w': case 'W': cout << "Week day\n";
                          break;
                case 'f': case 'F': cout << "Week day\n";
                          break;

                default:  cout << "Not a day\n";
        }
}






Example 7




#include <iostream.h>

main()
{
        int 	input_value;

        cout << "Which day?";
	cin >> input_value;


        switch (input_value){
                case 1: cout << "The day is Monday\n";
                        break;
                case 2: cout << "The day is Tuesday\n";
                        break;
                case 3: cout << "The day is Wednesday\n";
                        break;
                case 4: cout << "The day is Thursday\n";
                        break;
                case 5: cout << "The day is Friday\n";
                        break;
                case 6: cout << "The day is Saturday\n";
                        break;
                case 7: cout << "The day is Sunday\n";
                        break;
                default : cout << "Invalid Input\n";
                        break;
        }
}







Back to Main Page for Lecture 3 








 
  

Control Constructs 

 Case Study 






PROBLEM:

Write a program to calculate the commission based wages of a computer
salesman. His basic wage is £50 per week and he is expected to sell at
least 10 computers. If he sells more than 10 computers, he receives an extra
£5.50 per computer he sells after the 10th computer. 




SOLUTION:

The basic algorithm could be drawn as:-





The structure of the program would then be:-

DECLARE CONSTANTS
DECLARE VARIABLES

PROMPT USER TO ENTER NUMBER OF COMPUTERS SOLD 
READ IN NUMBER



CALCULATE PAY
        IF SELLS MORE THAN 10 
                CALCULATE COMMISSION

OUTPUT PAYMENT




adding the actual code to the above:-

#include <iostream.h>


main()
{
	 //DECALRE CONSTANTS
        const float BASIC_PAY = 50;
        const int LEVEL = 10;
        const float COMMISSION = 5.50;

	 //DECLARE VARIABLES
        float pay, commission_pay;
        int num_computers;


        //PROMPT USER TO ENTER NUMBER OF COMPUTERS SOLD 
        cout << "Please enter number of computers sold: \en";
        //READ IN NUMBER
        cin >> num_computers;


        //CALCULATE PAY 

        if (num_computers > level)
        {
                commission_pay = (num_computers - LEVEL)* COMMISSION;
                pay = BASIC_PAY + commission_pay;
        }
        else
                pay = BASIC_PAY ;

        //OUTPUT PAYMENT
        cout << "Your wage for this week is " << pay << " pounds\en";


}



NOTE that this is only one possible solution to the problem. 


Try to think of some alternative methods of solving this problem.









Back to Main Page for Lecture 3 








 
  
 

 Control Constructs 

 Additional Exercises 
 









An employee in a company is  paid at a rate of £5.20 per hour. The
standard rate is paid for up to 37 hours worked in a week, after which 1.5
times the standard rate is paid. 22% tax is deducted from the gross pay as
well as 5% for national insurance  and 6% for the company pension scheme.
If the employee works less than 20 hours a week, the income tax rate is
only 20%.

Write a program to read in the hours worked and calculate the employees
pay, displaying the gross salary and all the deductions as well as the net
take home pay.

Remember to make proper use of constants.



A leap year is a year which is exactly divisible by 4, unless it is exactly
divisible by 100 in which case it is only a leap year if it is exactly
divisible by 400.  Write a program which prompts the user for a year and
then displays if it is a leap year or not.









Back to Main Page for Lecture 3 








 


  

 

 Control Constructs 
 

 





 What is wrong with the following?

int x;
if (x=1)
	x++;



What is wrong with the following?

int x, y, z;

if (x > y && < z)
	x = y;



Write an extract of code to test if the variable x is a positive
number?


Write an extract to test if the variable x is a positive and even
number?







 Answers
                                        











































































 The test should be  x== 1, with x=1 this will assign
the value of 1 to x rather than checking if x is a 1.


Assuming that it is intended to compare z to z, the test should be  x >
y && x < z 


 if (x > 0)



 if (x > 0 && x%2 == 0)






 


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 Iteration  
 
 








Learning Objectives


 
Be able to use a while loop



Be able to use a do..while loop 



Be able to use a for  loop



Know how to change from one type of loop to another


Know when one form of loop is more suitable than another i.e. would not use
for loop when checking if input to program is within desired range.













Source Code from Notes





  Additional Examples



 
Quick Check Exercises 



 
Additional Exercises to try











 Previous Lecture
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 Iteration 

 Source Code from Notes 
 
 




Example 1

This example reverses the digits of a number, and prints out the number in
the reverse order. Each digit is displayed as it is extracted by the
program. The final call to \fC\s8printf\fR\s0 returns the cursor to the
next line as the newline character has not been included within the
loop.


#include <iostream.h>

main()
{
        int     number, right_digit;

        cout << "Enter your number\n";
        cin >> number;

        while (number !=0)
        {
                right_digit = number % 10;
                cout >> right_digit;
                number /= 10;
        }

	cout << "\n";
}






Example 2

This example  checks that the input value is in the desired range



#include <iostream.h>

main()
{
        int     number;

        const int min = 1;
        const int max = 100;

        number = 0;

        while ( number < min || number > max)
        {
                cout << "Enter a number in the range 1 to 100\n";
                cin >> number;
        }
        cout << "the number is in range\n";
}







Example 3




#include <iostream.h>

main()
{
        char    letter;

        cout << "Press Y to continue\n";
        do
        {
                letter = getchar();
        } while (letter != 'y' && letter != 'Y');

        cout << "program execution will continue\n";

}






Example 4


The following program creates a conversion table for converting inches to
centimetre. The values in the table are calculated for inches in the range
1 to 10 in half inch increments.


#include <iostream.h>

main()
{
        float   inches,centimetres;

        // print table header
        cout << "inches\t centimetres\n";

        //calculate and print table entries
        for (inches = 1; inches <=10 ; inches += 0.5)
        {
                centimetres = 2.54 * inches;
                cout << inches << centimetres << "\n";
        }


}











Back to Main Page for Lecture 4 







 
  

Iteration 

 Case Study 





PROBLEM:

Write a program to calculate the average marks for a class of students. The
program will initially prompt for the number of students to be entered and
then for each student will ask for their marks for each of the 4 exams the
student sat. The program will then calculate  the average mark for each
student and the average mark for the whole class.




SOLUTION:

The solution to this will take the form:-




where the program repeats for each student the code to enter the student's
marks and calculate their average.

This could be implemented using a while loop to produce


//Using While Loop 

#include <iostream.h>

main()
{

        int     num_students, count;
        float   mark1, mark2, mark3, mark4;
        float   student_average, class_total = 0, class_average;


        // Enter no. of students
        cout << "Please enter the number of students in the class: ";
        cin >> num_students;


        // set count to zero
        count = 0;

        while (count < num_students)
        {
                // enter 4 marks for each student
                cout << "Please enter the marks for student "
                     << count << "\n";
                cin >> mark1 >> mark2 >> mark3 >> mark4;

                // Calculate average
                student_average = (mark1 + mark2 + mark3 + mark4) / 4;

                // Add average to total
                class_total += student_average;

                // Display average
                cout << "The average mark for student " << count 
                     << " is " << student_average << "\n\n";

                // Increment count for next student
                count ++;

        }


        // Calculate class average 
        class_average = class_total / num_students;

        // Output Class average
        cout << "The class average is " << class_average << "\n";

}


When this program is run it produces the following output ( with sample
figure inserted):-


Please enter the number of students in the class: 4
Please enter the marks for student 0
23 45 63 54
The average mark for student 0 is 46.25

Please enter the marks for student 1
67 87 67 87
The average mark for student 1 is 77

Please enter the marks for student 2
54 65 34 76
The average mark for student 2 is 57.25

Please enter the marks for student 3
34 35 36 25
The average mark for student 3 is 32.5

The class average is 53.25



	

The program can be drawn schematically to show how it fits in to the flow
structure given above. The code used for each part is shown alongside.





The same program could be converted to a for and only the
initialisation, loop statement and increment need to be changed. Here is the
same
program but written with the for loop.

	

//Using For Loop 

#include <iostream.h>

main()
{

        int     num_students, count;
        float   mark1, mark2, mark3, mark4;
        float   student_average, class_total = 0, class_average;

        // Enter no. of students
        cout << "Please enter the number of students in the class: ";
        cin >> num_students;

        // for statement includes initialisation, test and increment
        for (count = 0 ; count < num_students ; count ++)
        {
                // enter 4 marks for each student
                cout << "Please enter the marks for student "
                     << count << "\n";
                cin >> mark1 >> mark2 >> mark3 >> mark4;

                // Calculate average
                student_average = (mark1 + mark2 + mark3 + mark4) / 4;


                // Add average to total
                class_total += student_average;

                // Display average
                cout << "The average mark for student " << count
                     << " is " << student_average << "\n\n";

        }


        // Calculate class average 
        class_average = class_total / num_students;

        // Output Class average 
        cout << "The class average is " << class_average << "\n";

}







Back to Main Page for Lecture 4 








 
  
 

 Iteration 

 Additional Exercises 
 







In the train time table program, add loops to check that the times entered
are valid (ie times entered are between 0 and 24 hours for the start and stop
times and that the frequency and journey times are between 1 and 60
minutes).


Prevent the program from proceeding until suitable values are entered.



Write a program which reads in a  set of positive value real number values
terminated by a negative value and prints out the sum of all these values.




Write a program to find the largest, smallest and average values in a
collection of n numbers entered by the user where n is the
first data item entered.








Back to Main Page for Lecture 4 









 


  

 

 Iteration 
 

 






Change to following code to a while loop.

int	x, y;
for (x=1; x < 57; x+= 2)
	y = x;



What is wrong with the following?

int	n=0;
while (n < 100)
	value = n*n;


 Write a while loop that calculates the sum of all numbers
between 0 and 20 inclusive?


Convert the above loop to a  do ... while  loop.


Convert the above loop to a for loop.


What is wrong with the following?

i = 1;
while (i <= 10 )	
	cout << i;
	i++;

	





 Answers
                                        










































x = 1;
while (x < 57)
{
	y = x;
	x += 2;
}


The value of n within the loop is never incremented. It will therefore always
have a value of 0, which is less than 100, and hence this will produce an
infinite loop which will never terminate. The program will therefore
never end.


int	i, sum =0;
i = 0;
while (i <= 100)
{
	sum += i;
	i++;
}


int	i, sum =0;
i = 0;
do
{
	sum += i;
	i++;
}while (i <= 20);


int	i, sum =0;
for (i=0; i <= 20; i++)
{
	sum += i;
}


There are no braces around the cout and increment i statements, therefore
it will execute only the cout statement within the loop, and an infinite
loop will again occur.




 


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Back to Main Page for Lecture 4 



                                        
















 Arrays and Structures 
 
 








Learning Objectives


 
Be able to declare an array



 Be able to access elements of an array


 Be able to sequentially access arrays using loops



Be able to initialise arrays


Understand the use of structures


Be able to declare a structure and access the members of a structure
variable









Source Code from Notes





  Additional Examples



 
Quick Check Exercises


 
Additional Exercises to try











 Previous Lecture
  
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 Arrays and Structures 

Source Code from Notes 
 



Example

Program to input each of the temperatures over the 30 days to calculate the
average
temperature during the period.


// Program to input 30 temperature samples and calculate their average 
#include <iostream.h>

main()
{
        const int DURATION = 5;
        int     day;
        float   temperature[DURATION];
        float   average, total;

        //Enter temperature values 
        for (day =0; day < DURATION ;day++)
        {
                cout << "Enter temperature of day " << day << "\n";
                cin  >> temperature[day];
        }

        // initialise total variable 
        total = 0.0;

        // sum temperatures for each day 
        for (day=0; day < DURATION; day++)
        {
                total += temperature[day];
        }

        // calculate average 
        average = total/DURATION;


        //output result 
        cout  << "The average temperature is"  << average << "\n";

}





Example

The following example reads in a sentence which is terminated by a full
stop and calculates the number of lower case vowels in the sentence. An
array of char elements is defined. The length of this array is
limited to 100 elements.


// Program to count for number of vowels in a sentence 
// the sentence must be terminated by a full stop (.)  
#include <iostream.h>

main()
{
        const int max_length = 100; //max number of characters in sentence 
        char    sentence[max_length];
        int     i,count,length;

        i = 0;
        count = 0;

        // enter the sentence 
        cout << "Enter a sentence terminated by a full stop \n";
        do {
                cin >> sentence[i];
                i++;
        } while(sentence[i-1] != '.' && i < max_length);

        // number of characters in sentence 
        length = i;

        // count number of lower case vowels 

        for (i=0;i < length;i++) {
                if (sentence[i] == 'a' || sentence[i] == 'e' || sentence[i]
                   == 'i' || sentence[i] == 'o' || sentence[i] =='u') {
                        count++;
                }
        }

        // output number of vowels 
        cout << "The number of lower case vowels is " << count << "\n";

}



Example 


Two dimensional array example where the
average temperature each month for the last 20 years is declared and the user
is then prompted to enter the values


#include <iostream.h>

main()
{
        const int NO_MONTHS = 12;
        const int NO_YEARS = 20;
        float   temperature[NO_YEARS][NO_MONTHS];
        int     years,month;

        // input values to array 

        for (years = 0;years < NO_YEARS;years++)
        {
                for (month=0;month < NO_MONTHS;month++)
                {
                        cout << "Enter temperature ";
                        cin  >> temperature[years][month];
                }
        }
}







Back to Main Page for Lecture 5 









 

 

 Arrays and Structures  

 Additional Examples 
 




PROBLEM:

Write a program which will read in a sequence of positive
integers terminated by -1 to indicate the end of the input.
These values will be read into an array and the program should
then determine if the sequence is a palindrome. 

A palindrome
is a sequence of number/characters/words etc which is the same
when read from either direction. For example:

2 3 4 5 4 3 2

is a palindrome, since the sequence is the same when read from
left to right or from right to left.

2 3 4 5

Is NOT a palindrome.

SOLUTION:


The program does not know how big a sequence will be entered so
an array of characters larger than required is declared. The
program then reads in the sequence of integers one at a time
into the array. After it has read in each integer it checks if
it was a -1, if it was -1 it stops reading in. It also checks
to ensure that we can't read in any more integers than the
size of our array.

Once it has completed reading in the integers, the number of
values read in can be determined, as  i - 1 since we
are not interested in the -1.

We can then search through the array comparing the 1st element
with the last, the 2nd element with the 2nd last etc. As soon
as we find that they don't match we can terminate the search
as we know that the sequence is not a palindrome.

Note that we only have to loop for i < length/2 as we compare
the first half of the array to the second half.






#include <iostream.h>

main()
{
	const int SIZE = 100;
        int     array[SIZE];
        int     i=0, length, palindrome = 1;


        cout << "Enter A sequence of positive integer numbers\n";
        cout << "terminated by -1\n";

        do
        {
                cin >> array[i]);
                i++;
        }while(array[i-1] != -1 && i < SIZE);

        length = i -1; //don't need -1


        for (i=0; i < length/2 && palindrome == 1 ;i++)
        {
                if (array[i] != array[length - i - 1])
                        palindrome = 0;
        }


        if (palindrome == 1)
                cout << "array is a palindrome\n";
        else
                cout << "array is not a palindrome\n";
}










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 Arrays and Structures 

 Additional Exercises 
 









Write a program to read N integers into each of two arrays X and Y of size
20. The program will prompt the user to enter the values into each of the arrays. The
program will then compare each of the elements of X to the corresponding element in Y.
Then, in the corresponding element of a third array Z, store:-


 1

if X is larger than Y
 0

if X is equal to Y
 -1

if X is less than Y


Then print out a three column table displaying the contents of the arrays
X, Y  and Z, followed by a count of the number of elements of X that exceed
Y, and a count of the number of elements of X that are less than Y. Make up
your own test data.

Declare your arrays to be of a size greater than you will require, (larger
than N).





Define a structure to store the x and y coordinates of a point. The user
should then be prompted to enter the x and y coordinates for two points,
and the program should make use of the structure. The program will then
calculate the distance on the x axis between the two points and the
distance on the y axis between the two points.








Back to Main Page for Lecture 5 









 


  

 

 Arrays and Structures 
 

 






Write a declaration for an array of 100 floats.


Initialise all elements in this array to have a value of 0


Assign a value of 10.5 to the 5th element in the array you have just
declared.


What is wrong with the following?


int array[10],i;
for (i=0;i < 10;i++)
	array[i] = array[i+1];



What is wrong with the following?


float	index;
float	array[9];

for (index=0; index < 9; index++)
	array[index] = index;







 Answers
                                        





































 float a[100];

int i;
for (i=0;i<100;i++)
	a[i] = 0.0;



a[4]=10.5;


The array has 10 elements,with indexes from 0 to 9, but the last time round
the loop i = 9, and hence the code tries to do array[9] = array[10],
which would result in an error since a[10] does not exist.

 The array index must always be an integer and hence the first line
should have been

int	index;





 


 Back to Questions


Back to Main Page for Lecture 5 



                                        














 Functions 
 
 








Learning Objectives


 
Understand the use of the function definition and declaration



 Be able to write a function definition given a specification



Be able to call a function



Understand the use of void


 Be able to use standard functions


 Understand the scope of variables in functions



 Understand the use of recursive functions


 Understand the use of refernce variables and be able to use them to alter
values sent to a function


 Understand the concept of Default Arguments and be able to use them 


Understand the concept of function overloading 








Source Code from Notes





  Additional Examples



 
Quick Check Exercises 



 
Additional Exercises to try



 
Test Cases for Tutorial 6



 
Extra Examples on Recursion











 Previous Lecture
 Home


 Next Lecture





 
  
 

 Functions 

Source Code from Notes
 





Example 1

The following program uses a function to calculate the volume of the
cube with the largest volume from the dimensions
entered by the user for two cubes.



#include <iostream.h>

// function declaration 
float cube_volume(float length, float height, float width);

// main body of program 
main()
{
        float   length1,width1,height1,volume1;
        float   length2,width2,height2,volume2;

         // input cube dimensions 
         cout << "Enter dimensions of cube 1\n";
         cin  >> length1 >> height1 >> width1;

         cout << "Enter dimensions of cube 2\n";
         cin  >> length2 >> height2 >> width2;

        // calculate volume 
         volume1 = cube_volume(length1,height1,width1);
         volume2 = cube_volume(length2,height2,width2);

        // output results 
        if (volume1 > volume2)
                cout << "Cube 1 is largest, its volume is " 
                      << volume 1 << "\n";
        else if (volume2 > volume1)
                cout << "Cube 2 is largest, its volume is "
                     << volume2 << "\n";
        else
                cout << "both cubes have equal volumes of "
                     << volume1 << "\n";
}


// function definition 
float cube_volume(float length, float height, float width)
{
        float           volume;
        
        volume = length * width * height;
        return(volume);
}







Additional Example - convert Train Timetable from Iteration Tutorial to use
functions


// Calculate train time table - using functions

#include <isotream.h>

//Function declarations
int     to_24hr(int mins);
int     to_mins(int hr_24);

main()
{
        const int wait = 5;
        int     start, stop, time;
        int     start_min, stop_min;
        int     journey_time, freq_time;
        int     dep_red,dep_green,dep_yellow,arr_blue;
        int     dep_red_min,dep_green_min,dep_yellow_min,arr_blue_min;

        cout << "Enter the journey time (minutes)\n";
        cin >> journey_time;

        cout << "Enter the first train time\n";
        cin >> start;

        cout << "Enter the last train time\n";
        cin >> stop;

        cout << "Enter the frequency time\n";
        cin >> freq_time;

        // convert start and stop times to minutes
        start_min = to_mins(start);
        stop_min = to_mins(stop);

        // table header
        cout << "RED \t GREEN \t YELLOW \t BLUE\n";

        for (time=start_min; time<=stop_min; time+=freq_time)
        {
                // Calculate Times 
                dep_red_min = time;
                dep_green_min = (time + journey_time + wait);
                dep_yellow_min = (time + 2*(journey_time + wait));
                arr_blue_min = time + 3*journey_time +2*wait;

                // Convert times to 24 hour clock to display 
                dep_red = to_24hr(dep_red_min);
                dep_green = to_24hr(dep_green_min);
                dep_yellow = to_24hr(dep_yellow_min );
                arr_blue = to_24hr(arr_blue_min);

                cout << dep_red << "\t" << dep_green << "\t"
                     << dep_yellow << "\t" << arr_blue << "\n";
        }
} // end of main program 

// Definition of to_24hr function 
int     to_24hr(int mins)
{
        int     hr, min, total;
        hr = mins/60;
        min = mins%60;
        total = hr*100 + min;

        return (total);
}

// Definition of to_mins function 
int     to_mins(int hr_24)
{
        int hr, min, total_mins;

        hr = hr_24/100;
        min = hr_24%100;
        total_mins = hr*60 + min;

        return(total_mins);
}







Example - Recursion

 

// Program to calculate the factorials of positive integers 
// Program prints results for integers from 0 to 10        

#include <iostream.h>

long factorial (int n);

main()
{
        int j;

        for (j=0;j < 11;j++)
                cout << j << "! = " << factorial(j) << "\n";

}

long factorial (int n)
{
        long result;

        if (n==0)
                result = 1;
        else
                result = n * factorial(n-1);

        return(result);
}



Example

The following program prompts the user for the coordinates of 2 points. It
then calculates the length of the  line joining the two points.

#include <math.h>
#include <iostream.h>


struct coord {
        float    x,y,z;
};

// function declarations
float length (coord point1, coord point2);
float square (float x);

void main (void)
{
        coord   p1, p2; 
        float   line_length;


        cout << "Enter coords of point 1 : ";
        cin >> p1.x >> p1.y >> p1.z;

        cout << "Enter coords of point 2 : ";
        cin >> p2.x >> p2.y >> p2.z;

        line_length = length(p1, p2);
        cout << "Length = " << line_length << "\n";
}


float length (coord point1, coord point2)
{
        float   xlen,ylen,zlen;
        float   len_sqr;

        xlen = point1.x - point2.x;
        ylen = point1.y - point2.y;
        zlen = point1.z - point2.z;

        len_sqr = square(xlen) + square(ylen) + square(zlen) ;
        return (sqrt(len_sqr));
}


float square (float x)
{
        return ((x)*(x));
}



Example - Reference Variables


This example illustrates how reference variables can be used to permit
a function to change values in the calling program.

#include <iostream.h>
void swap(int &ra, int &rb);

main()
{
        int a,b;

        a = 10;
        b = 12;
        swap (a,b);
}
void swap(int &ra, int &rb)
{
        int temp;

        temp = ra;
        ra   = rb;
        rb   = temp;
}



Example


This example sends an array to the function. The function then
calculates both the largest and smallest value stored in the array.
These maximum and minimum values stored are then returned to the main calling
program as function arguments.

There is a msitake in the notes, the following program is correct and has
made MAX be global so the function also knows its value. An alternative
solution is given below where MAX is local and is sent to the function.


#include <iostream.h>
#include <math.h>

void max_min(float &max, float &min, float array[]);

const int MAX = 5;

main()
{
        float array1[MAX];
        float array2[MAX];
        float   max1, min1;

        float   max2, min2;
        int     i;

        for (i=0; i<MAX;i++)
        {
                cout << "Enter value  "<< i << " of array1\n";
                cin >> array1[i];
        }
        for (i=0; i<MAX;i++)
        {
                cout << "Enter value  "<< i << " of array2\n";
                cin >> array2[i];
        }

        max_min(max1,min1,array1);
        max_min(max2,min2,array2);

        cout << "Array 1 : max = " << max1 << " min = " << min1 << "\n";
        cout << "Array 2 : max = " << max2 << " min = " << min2 << "\n";
}

void max_min(float &max, float &min, float array[])
{
        int     i;

        min = array[0];
        max = array[0];

        for (i=1;i < MAX;i++)
        {
                if (array[i] > max)
                        max = array[i];
                if (array[i] < min)
                        min = array[i];
        }
}




Alternative Solution



#include <iostream.h>
#include <math.h>

void max_min(int size_array, float &max, float &min, float array[]);


main()
{
	const int MAX = 5;
        float array1[MAX];
        float array2[MAX];
        float   max1, min1;

        float   max2, min2;
        int     i;

        for (i=0; i<MAX;i++)
        {
                cout << "Enter value  "<< i << " of array1\n";
                cin >> array1[i];
        }
        for (i=0; i<MAX;i++)
        {
                cout << "Enter value  "<< i << " of array2\n";
                cin >> array2[i];
        }

        max_min(MAX, max1,min1,array1);
        max_min(MAX, max2,min2,array2);

        cout << "Array 1 : max = " << max1 << " min = " << min1 << "\n";
        cout << "Array 2 : max = " << max2 << " min = " << min2 << "\n";
}

void max_min(int size_array, float &max, float &min, float array[])
{
        int     i;

        min = array[0];
        max = array[0];

        for (i=1;i < size_array;i++)
        {
                if (array[i] > max)
                        max = array[i];
                if (array[i] < min)
                        min = array[i];
        }
}


Example - Sports Timer


A sports timer can time an event returning the elapsed time in seconds. Write
a
function to convert the time in seconds to hours, minutes and seconds,
returning
these values as function arguments. The function should be called from a main
calling program where the user is prompted to enter the elapsed time in
seconds.

Reference variable must be used in the function arguments for hours,
minutes and seconds.


#include <iostream.h>
#include <math.h>


// Function Declaration
void change_time(int elapsed_seconds, int &hours, int &mins, int &seconds);

main()
{
        int     elapsed_time;
        int     time_hours, time_minutes, time_seconds;


        cout << "Enter the time for the event\n";
        cin >> elapsed_time;

        //Function call - send address of variables to change 
        change_time(elapsed_time, time_hours, time_minutes, time_seconds);


        // Use variables as normal 
        cout << "The number of hours is " << time_hours << " \n";
        cout << "The number of minutes is "<< time_minutes << " \n";
        cout << "The number of seconds is " << time_seconds << " \n";
        cout << "The number of seconds is " << time_seconds << " \n";

}

// Function to convert total in seconds to hours, mins and seconds 
void change_time(int elapsed_seconds, int &hours, int &mins, int &seconds)
{
        int     tmp_minutes;

        seconds = elapsed_seconds%60;

        tmp_minutes = elapsed_seconds/60;

        mins = tmp_minutes%60;

        hours =  tmp_minutes /60;
}







Back to Main Page for Lecture 6 








 

 

 Functions  

 Additional Examples 
 





There are three additional worked examples:-




 Example 1




  Example 2 - using reference variables 




  Example 3 - Functions and arrays






 Example 1 


PROBLEM:

Write a function that will display a line of n asterisks on the
screen. n will be passed as a argument to the function. After
writing the line of n asterisks a new line should be printed.

Write a program to call this function, using it to display a grid of
m times n asterisks. The user should be prompted to enters the
values for m and n.


SOLUTION:

Considering initailly the function, the first thing to consider is 

What arguments are to be sent to the function?

Obviously, it need to be told how many asterisks to write, in this case
n which will be integer?
 The next thing to consider is what will be returned by the function?

Obviously, the function only prints to screen and therefore does not need to
return anything to the main calling program. The return type is therefore
void

We can now continue and write the function.

void line (int n)
{
        int i;

        for (i=0; i < n;i++)
                cout << "*";
        cout << "\n";
}





The complete program, has to call this function m times, and so a loop
will be used to call the program. The complete solution is then.
#include <iostream.h>

// function declaration */
void line( int n);

main()
{
        int     i;
        int     m,n;

        cout << "Enter  values for m and n \n";
	cin >> m >> n ;

	// Loop to call function m times 
        for (i=0; i < m; i++)
                line(n);
}


// function definition 
void line (int n)
{
        int i;

        for (i=0; i < n;i++)
                cout << "*";
        cout << "\n";
}



ADDITIONAL EXERCISE

Amend the main program written above, so that the following
type of pattern is displayed. This is shown for the case of m = 5.

*
**
***
****
***** 



SOLUTION

Note that no changes are required to the function, we simply need to amend
the main calling program.


#include <iostream.h>

// function declaration 
void line( int n);

main()
{
        int     i;
        int     m;

        cout << "Enter  a value for the number of rows\n";
	cin  >> m ;

	//Call to function in loop 
        for (i=1; i <= m; i++)
                line(i);
}









 Example 2 


PROBLEM:

Consider the calculation of the roots of the quadratic equation
ax²+bx+c=0, assuming that a is non-zero:-



if b²-4ac is greater than 0, there are two root sgiven by
(-b±sqrt(b²-4ac))/2a



if b²-4ac is equal than 0, there is one root given by
-b/2a


if b²-4ac is less than 0, there are no real roots



Write a function in C++ which will take a, b and c as arguments and returns
the roots as function arguments root1 and root2 (as appropriate). The number
of roots will be returned through the type specifier.


Write a main program which prompts the user to enter the values of a, b and
c. Then  after  using a call to the function written above, the program will
display the number of roots and their values as appropriate.




SOLUTION:


The function has to be sent the arguments a, b and c, it also requires two
arguments for the roots. These values will be altered by the program so must
be refernec variables. The function returns an integer number for the number
of 
roots.
The function declaration will therefore be:-


int root (float a, float b, flot c, float &root1, float &root2);



The rest of the function can now be written.


int root (float a, float b, float c, float &root1, float &root2)
{
        float   tmp;

        tmp = b*b - 4*a*c;

        if (tmp > 0)
        {
                root1 = (-b + sqrt(tmp))/(2*a);
                root2 = (-b - sqrt(tmp))/(2*a);
                return 2;
        }
        else if (tmp == 0)
        {
                root1 = -b/(2*a);
                return 1;
        }
        else
                return 0;
}


The main program can now be written. Remember to include the function
declaration and math.h as the sqrt function was used.


#include <iostream.h>
#include <math.h>

int root (float a, float b, float c, float &root1, float &root2);

main()
{
        float   a,b,c,root1,root2;
        int     num_roots;

        cout << "Enter values for a, b and c\n";

        cin >> a >> b >> c;

        num_roots = root(a,b,c,root1, root2);

        if (num_roots == 2)
                cout << "There are 2 real roots with values  " << root1 
                     << " and " << root2 << " \n";
        else if (num_roots == 1)
                 cout << "There is 1  real root with value " << root1 <<"
\n";
        else
                cout << "There are no real roots\n";
}








 Example 3 


Arrays can also be sent to functions. Arrays are also special in that the function
can change the contents of the array and the main calling program will know about the
changes after the function is called.

PROBLEM:
Write a function which when sent an array will fill each element of the array with
an integer number which is equal to the element number. The function will also be
sent the size of the array.

Write a main program to call this function, and then print out the contents of the
array.


SOLUTION:

The function will not return anything - hnece the type specifier is void.
The function will be sent an array of integers and an integer for the size. In the
function prototype, teh size of the array is not enetered in the square brackets for
the array - this is left blank. The empty square brackets simply indictate that an
array will be sent to the function.


void fill_array(int array[], int size)
{
	int i;
	
	for (i=0; i < size; i++)
		array[i] = i;

	return;
}


The main program will then take the form:-

#include <iostream.h>

// function declaration
void fill_array(int array[], int size);

main()
{
	int size = 100;
	int array[size];

	// call function
	fill_array(array, size);

	// write out contents of array
	for (i=0; i < size ; i++)
		cout << i << "\t" << array[i] << endl;
}


Note: That to call the function, only the name of the array to sent to the function
needs to be given in the function call. (In the notes on points it mentions how the
name of an array is a synonym for a pointer to the beginning of the array - hence
only the mname needs to be given.










Back to Main Page for Lecture 6 

















Recursion 




These pages contain some further examples of recursion and may help those
students who undertake the coursework for AI given by Dr Nick Taylor. The
minimax function required for this is best implemented using recursion.






Hints from Nick about the minimax function






What is recursion?


Recursion is a technique whereby functions can call themselves. Simple recursive
functions always have an if..else type of structure. The condition
of the if provides an exit from the recursion and is always tested 
before the recursive call.
If this condition is not true, the else
part calls the function again, but this time the value of one of the
arguments sent to the function will have changed (typically the value of the
variable tested in the condition). The value of the argument is changed in
such a manner that ultimately the base condition will be true.

This is explained more easily by considering the following examples. For each
problem both a recursive and an iterative solution to the problem are
presented - compare the structures of each!




Recursion Examples





 Example 1 
  Multiplication by repeated
addition




 Example 2 
  Euclids Algorithm for
Greatest Common Divisor





 Example 3
  Factorial Example





 Example 4
  Fibonacci Sequence








 Multiplication by repeated addition

PROBLEM:  
Write a recursive function to perform multiplication of two positive integers
(m and n)
 using only addition. The
function will take as its arguments two integers to multiply together (  m
x  n  ) and will return the product.

Hint: consider the following:


6 x 1  =  6
6 x 2  =  6 + (6 x 1)
6 x 3  =  6 + (6 x 2) = 6 + [6 + (6 x 1)] = 6 + 6 + 6





SOLUTION
This could be extended to the general case that
m x n = m + (m x (n-1))



Therefore, from the above we can repeat the multiplication process by
repeatedly calling the function, each time with n decreasing by 1.

The function stops calling itself once n reaches 1. This therefore gives the
condition to stop calling the function, since m x 1 is simply m and the
result m is returned.


int multiply(int m, int n)
{
	int result;

	if (n == 1)
		result =  m;
	else
		result =  m + multiply(m, n-1);
	return(result);
}


A recursive function always has a structure using an  if..else
statement. This allows the function to repeatedly call itself, but provides
the condition on which to stop.



EXAMPLE OF OPERATION

Consider what would happen if this function is called for m=6 and
n=3.
The function tests if n is equal to 1, it isn't so

 result = 6 + multiply(6,2);

so the function is recalled this time with m = 6  and n = 2. 
The function tests if n is equal to 1, it isn't so

 result = 6 + multiply(6,1);

so the function is recalled this time with   m = 6  and n = 1.
The function tests if n is equal to 1, it is so it sets


result = 6;


and returns this value. This then completes the statement 

 result = 6 + multiply(6,1); 

providing the value result  equal to 12. This  value is then
returned, and this then completes the statement

 result = 6 + multiply(6,2);

and finally the answer of 18 is returned.




ITERATIVE SOLUTION

This problem could also have been tackled using an iterative solution. An
iterative solution uses a loop.


int multiply (int m, int n)
{
	int tot =0;

	while ( n > 0O
	{
		tot += m;
		n--;
	}
	return tot;
}



 Back to top of page





 Euclid's Algorithm for Greatest Common Divisor

PROBLEM 2:
Euclids algorithm for calculating the greatest common divisor (GCD) of two
positive integers, GCD(m,n) is defined recursively below. The greatest common
divisor of two integers is the largest integer that divides them both. For
example, the GCD of 9 and 30 is 3.



GCD(m,n) is n if n<= m and n divides m

GCD(m,n) is GCD(n,m) if m < n

GCD(m,n) is GCD(n, remainder of m divided by n) otherwise.


Write a C++ function to implement this algorithm.



SOLUTION

In this case we require an else..if structure because there are 3 possible
routes depending on the values of m and n. 

If n is less than m and n divides
m exactly (ie n%m ==0), then we can stop and the answer is n.


If m is less than n, then recall the function but swap m and n


otherwise, recall function with arguments n and the remainder of m divided by
n (m%n)


This is implemented below.


int GCD (int m, int n)
{
	if (n<=m && m%n == 0)
		return n;
	else if (m < n)
		return GCD(n,m);
	else 
		return GCD (n, m%n);
}




 Back to top of page




 Factorials 

PROBLEM 3:
Write a recursive function to calculate the factorial of n (n!)
0! = 1
1! = 1
2! = 2 x 1 = 2 x 1!
3! = 3 x 2 x 1 = 3 x 2! 
4! = 4 x 3 x 2 x 1 = 4 x 3! 
5! = 5 x 4 x 3 x 2 x 1 = 5 x 4! 
etc.

n! = n x (n-1)!




SOLUTION:

From the above it is obvious to see a pattern emerging and

n! = n x (n-1)!


Therefore, we can calculate the factorial of n, by multiplying the factorial
of n-1 by n. We can calculate the factorial of n-1, by calling the function
again, but this time sending it n-1.

This is then repeated since,

(n-1)! = (n-1) x (n-2)!


We want to stop repeatedly calling the function, once n is 0. Hence the
condition on the if statement is that the recursion is terminated once n is
equal to 0. Otherwise, it recalls the function, but with n-1.


long factorial (int n)
{
        long result;

        if (n==0)
                result = 1;
        else
                result = n * factorial(n-1);
        return(result);
}




ITERATIVE SOLUTION

This could also be implemented iteratively using a loop.

long factorial (int n)
{
	long result;
	int i;

	result = 1;
	for (i=n; i >0 ; i--)
		result *= i;

	return (result);
}



 Back to top of page




Fibonacci Sequence

PROBLEM 4:
The Fibonacci numbers are generated by adding together the two previous
numbers in the sequence, initialising the 1st two numbers (F1 and F2) to 1.
The sequence

	1 1 2 3 5 8 13 21 .......

is then produced, where F3=F2+F1=2, F4 = F3+F2 =3, F5= F4+F3 = 5 etc.

Write a recursive function which returns the nth Fibonacci number (Fn), where
n is sent to the function as an argument.



SOLUTION:

The Fibonacci sequence can be generated by repeated calls to the function. We
want to keep on calling the function to calculate the 2 previous numbers in
the sequence. We want to stop calling the function if n is 1 or n is 2,
whereby we know the 1st Fibonacci number (F1) is 1 and the 2nd Fibonacci
number (F2) is 1.


int fib (int n)
{
	if (n == 1 || n == 2)
		return 1;
	else
		return (fib(n-1) + fib(n-2));
}




Iterative Solution

This could also have been implemented iteratively as:-

int fib_iterative (int n)
{
	int fib, fib_old1, fib_old2, i;

	if (n <= 2 )
		return 1;
	else
	{
		fib_old1 = 1;
		fib_old2 = 1;
		for (i=3; i <=n; i++)
		{
			fib = fib_old1 + fib_old2;
			fib_old2 = fib_old1;
			fib_old1 = fib;
		}
		return (fib);
	}
}	

			

Compare this to the Recursive Solution.


 Back to top of page







Back to Main Page for Lecture 6 










 
  

 

 Functions  

 Additional Exercises 
 







The following function raises an integer to a given power ( X^n )

int power (int x, int n)
{
        int p;
        for (p = 1; n > 0; n--)
                p = p * x;
        return p;
}


Add a main program to prompt the user to enter a integer and the desired
power then call the function, and finally print out the answer. Remember to
add a function prototype.


Revision

The power to raise the integer to,  n  must be a positive integer. Check
that
the
value entered is positive, if it is not do not let the user proceed until a
positive value is entered (see notes on Iteration).


The above is an iterative method of calculating the value of X^n, write a recursive function  to calculate the value of a number (base) raised
to 
a power (power). Assume the power is positive.



The Arrays Example  looked at writing a program to
determine if an array contained a palindrome. Rewrite this exercise, so that you now
have a function that when sent an array, will return a 1 if the array is
a palindrome and will return a 0 otherwise. The function will be sent the array of
integers.

Now re-write the main program to call this function and then display if the array
is a palindrome.








Back to Main Page for Lecture 6 









 


  

 

 Functions 
 

 





 Explain the difference between a function declaration and a functions
definition? Where is the function declaration placed?


Given the following function declarations, explain the meaning of each.


int function1(int a, float b);
   void function2(float a, float b, float c);
int function3(void);
float function4(float x);
void function5(void);



Write an appropriate function call for each of the function declarations given above.


What does the following function do?
float function(float a, float b, float c)
{
	float tmp;

	tmp = (a + b + c )/ 3.0 ;

	return tmp;
}



Give an example function call for the above function.



Give the function declaration for the function given in question 4.







 Answers
                                        
















































 The declaration is contained within the program before the function is called. It
is usually
outside any function, including the main program. It contains details of the function
name, the return type and the types of the function arguments.

The definition contains the actual code of what is to be implemented when the function
is called.

 

 The function is sent an integer argument which it will refer to as a
within the function and a floating point variable, which it will refer to as
b within the function. The function will return an integer quantity.

The function is sent 3 floating point arguments and returns no value to the calling
program.

The function is sent no arguments but returns an integer. 

The function is sent one floating point argument and returns a floating point
argument.
 The function is sent no variable and returns nothing.


 Example calls are given below, 

 
int x, z;
float y;

// x and y are assigned some values in the program 

z = function1(x, y);



float	x,y,z;

// x, y and z assigned values in program 

function2(x, y, z);


int a;

a = function3();


float a, b;

// b assigned a value in program 
a = function4(b);


 
function5();




It returns the average value of the three values sent to it.


x = function(a,b,c);


 float function(float, float, float); or float function(float a, float
b, float c); 






 


 Back to Questions


Back to Main Page for Lecture 6 



                                        
















 Pointers 
 
 








Learning Objectives


 
Understand the concept of pointers which point to memory locations



Know how * and & are used with pointers and what they do 


Be able to use command line arguments to input data to programs


Understand the concept of linked lists












Source Code from Notes





  Additional Examples



 
Quick Check Exercises 



 
Additional Exercises to try



 
Test Cases for Tutorial 7











 Previous Lecture
 Home


 Next Lecture





 
  
 

 Pointers 

Source Code from Notes
 





Example 

Example program to read in two numbers from the command line and add them
together.



#include <iostream.h>
#include <stdlib.h>

main(int argc,char *argv[])
{
        float           a,b,answer;

        // ensure correct number of arguments 
        if (argc != 3)
        {
                cout << "Wrong number of arguments for program\n";
                exit(1);
        }


        /* convert argument string to float */
        a = atof(argv[1]);
        b = atof(argv[2]);

        answer = a + b;
        cout << "Answer = " << answer << "\n";
}






Example

This example is more complex and is for a program which requires a
flag to indicate what operation should be performed on the two numbers. The
flags are indicated by the minus (-) sign.


#include <iostream.h>
#include <stdlib.h>

main(int argc,char *argv[])
{
        int     add_flag = 1;
        int     sub_flag = 0;
        int     mult_flag = 0;
        float   a,b;

        argc--;
        argv++;

        while ((argc > 0)&&(**argv == '-')){
                switch(*++*argv){
                        case 'a' : // add 
                                   add_flag = 1;
                                   break;
                        case 's' : // subtract 
                                   sub_flag = 1;
                                   add_flag = 0;
                                   break;
                        case 'm' : // multiply 
                                   mult_flag = 1;
                                   add_flag = 0;
                                   break;
                        default  : cout << "Unknown flag\n";
                                   exit(1);
                }
                argc--;
                argv++;
        }

        if (argc != 2)
        {
                cout << "Wrong number of arguments for program\n";
                exit(1);
        }

        a = atof(argv[0]);
        b = atof(argv[1]);

        if (add_flag == 1)
                cout << "answer = " << a+b << "\n";
        else if ( mult_flag == 1)
                cout << "answer = " << a*b << "\n";
        else if (sub_flag ==1)
                cout << "answer = " << a-b << "\n";

}










Back to Main Page for Lecture 7 









 

 

 Pointers  

 Additional Example 
 






PROBLEM:

Write a program which prompts the user to enter 10 integer numbers into an
array. The program will then search the array to find if a zero  is present
in the array. If a zero is present it will print out the number of elements
in the array before the zero occurs.

Write this program making use of pointers to search the array and compare it
to writing the solution using arrays only.




SOLUTION:

The program will still need to declare an array of 10 elements as normal. It
will also declare  a pointer to an integer.

        int     array[10];
        int     *array_ptr;

The program will then prompt the user to enter the elements into the array.
It uses the variable index to count when 10 integers have been entered.
The integers are read into successive elements. The name of the array is  a
pointer to the beginning of the array, this pointer is moved along the array
by adding the index to it.

        cout << "Enter 10 numbers \n";
        for (index =0; index < 10; index ++)
                cin >> *(array + index);


The pointer is then set to point to the beginning of the array.

        array_ptr = array;


The program then increments along the array looking for a zero and checking
that it has not reached the end of the array.

        while (*array_ptr != 0 && (array_ptr - array < 10))
                array_ptr ++;

This checks if the contents of the array poistion the array_ptr is pointing
to contains a zero and if it has yet looked at 10 elements. If both condtions
are false it increments the pointer to point to the next element in the
array.

Final Program:

#include <iostream.h>

main()
{
        int     array[10];
        int     *array_ptr;
        int     index;


        cout << "Enter 10 numbers \n";
        for (index =0; index < 10; index ++)
                cin >> *(array + index);

        array_ptr = array;

        while (*array_ptr != 0 && (array_ptr - array < 10))
                array_ptr ++;

        if (array_ptr - array < 10)
                cout << "There are " << array_ptr - array << 
                        " elements before the zero\n";
        else
                cout << "There is no zero in the array\n";
}


The same program could also have been written using the array and the integer
index. No pointers are required.

#include <iostream.h>

main()
{
        int     array[10];
        int     index;


        cout << "Enter 10 numbers \n";
        for (index =0; index < 10; index ++)
                cin >> array[index];

	// reset index back to 0
        index = 0;

        while (array[index] != 0 && index < 10)
                index ++;

        if (index < 10)
                cout << "There are " << index << 
                        " elements before the zero\n";
        else
                cout << "There is no zero in the array\n";
}


Both programs perform the same task.






Back to Main Page for Lecture 7 









  
 

 

 Pointers 

 Additional Exercises 
 





Write a function strn_comp to compare two strings. The
function will return a value TRUE if the strings are the same and FALSE
otherwise. Call this function from a main program.

The return value will be of type int with TRUE indicated by a
value of 1 and FALSE indicated by a value of 0.

Note: this can be achieved using array indices or pointers to manipulate the string.




Rewrite the main program for the example given under Additional Example  2 for Functions to take the values of a, b,
and c as command line arguments, instead of prompting the user to enter the
values.







Back to Main Page for Lecture 7 








 


  

 

 Pointers 
 

 





 
How do you access the memory address of a variable?


How do you access the contents of a memory location whose address is stored
in a pointer?


Given the following declarations:

int	*p, *q;
int n;

Which of the following is valid?


p+q

p+n

p-q

q-n

n-p



Given the following declaration

float	a[10];
float	*a_ptr;

What do the following statements do? Assume that they are executed in the
order given.


a_ptr = a;

a_ptr ++;
*a_ptr = 3.5;


a_ptr = a_ptr + 5;
*a_ptr = a[1];




When referring to command line arguments what is the argument count?

How many arguments are in the following command line?

prog -x -a file1 13

What is stored in argv[2]?






 Answers
                                        



















































The memory address of a variable is accessed using the address operator &.
Given the variable x it's address is &


The contents of a memory location pointed to by a pointer are accessed
using the de-referencing operator (*). The contents of the memory location
pointed to by p are accessed by *p.




p+q - This is invalid. We cannot add two pointers.

p+n - This is valid. It increments the memory address p points to by n.

p-q - This is valid. The subtraction of the two pointers gives the number of
integer variables stored between the two locations

q-n - This is valid. It decrements the memory address q points to by n

n-q - This is invalid. We cannot subtract a pointer from an integer.





This makes the pointer a_ptr point to the beginning of the array (element
number 0 of array a)

This increments the pointer so it now points to a[1].  The contents of a[1]
then becomes 3.5.

It increments the pointer by 5 places - so it now points to a[6]. It then
copies the value of a[1] (3.5) to a[6].



The argument count is the number of arguments specified on the command line
to run the program.

There are 5 arguments.

argv[2] contains -a






 


 Back to Questions


Back to Main Page for Lecture 7 



                                        
















 File Input and Output 
 
 








Learning Objectives


 
Be able to open a file to read from or write to




Understand the different modes in which files can be opened



Know how to check if the file has been opened correctly 


 Be able to read from or write to the file



Be able to close a file after use


Understand how to use functions such as eof()








Source Code from Notes





  Additional Examples



 
Quick Check Exercises 



 
Additional Exercises to try



 
Test Cases for Tutorial 8











 Previous Lecture
 Home


 Next Lecture







 
  
 

 File Input and Output 

Source Code from Notes 
 
 



Example 1


Example using command line arguments to provide the filenames of the files
to be opened.



#include <iostream.h>

main(int argc,char *argv[])
{
	istream	fin;
	ostream fout;

        if (argc != 3) {
                cerr  << "USAGE: program input_file output_file\n";
                exit(1);
        }
	
	fin.open(argv[1]);
	fout.open(argv[2]);

        if (!fin)
	{
                cerr << "Can't open file " << argv[1] << " for reading\n";
                exit(1);
        }
        if (!fout)
	{
                cerr << "Can't open file " << argv[2] << " for writing\n";
                exit(1);
        }

        // rest of program 

}




Example 


For example to open a file called Myfile and write some data to it

#include <fstream.h>
main()
{
        // open text file Myfile for writing
        // now
        ofstream my;
        my.open("Myfile");

        if (!my)           // if open didn't work..
        {
                cerr << "Error opening file \n";
                exit();         // output error and quit
        }

        // write to the file
        my << "My name is Judith Bell \n";
}







Example - reading from file and detecting feof 

This example opens a file called test. We will assume this file
contains a list of floats, which could for example represent
the temperature each day for several days. We will then read in these values
andcalculate the average value stored in the file.


Note: This program will read the last item in the file twice, since it does
not detect the end of file until it tries to read past it.



 // Program to read integers from  an input file
 // called test  and calculate average


#include<iostream.h>
#include<fstream.h>

main()
{

        ifstream fi;
        float input_value, sum = 0.0, average;
        int     no_values = 0;

        fi.open("test");

        if (!fi)
        {
                cout << "Can't open file test\n";
                exit(0);
        }

        while (!fi.eof())
        {
                fi >> input_value;
                sum += input_value;
                no_values ++;
        }

        average = sum / no_values;

        cout << "The average value in file test was "
             << average << "\n";

        fi.close();
}




Corrected Program 

#include<iostream.h>
#include<fstream.h>

main()
{

        ifstream fi;
        float input_value, sum = 0.0, average;
        int     no_values = 0;

        fi.open("test");

        if (!fi)
        {
                cout << "Can't open file test\n";
                exit(0);
        }

        fi >> input_value;
        while (!fi.eof())
        {
                sum += input_value;
                no_values ++;
                fi >> input_value;
        }
	average = sum / no_values;

        cout << "The average value in file test was "
	<< average << "\en";

	fi.close();
}



Corrected Program - using arrays

Note: this program also reads file name in as specified by the user.


#include <iostream.h>
#include <fstream.h>

main()
{
        ifstream fi;
        const int max = 100;   // declare a larger array than required
        float input_value[max], sum = 0.0, average;
        int     i, no_values = 0;
        char    file_name[20];

        cout << "enter filename";
        cin >> file_name;

        fi.open(file_name);

        if (!fi)
        {
                cout << "Can't open file test\n";
                exit(0);
        }

        i =0;
        while (!fi.eof())  // read data into array
        {
                fi >> input_value[i];
                i++;
        }

        no_values = i-1;  // discard last element in array

        for (i=0; i< no_values;i++)   // sum together all values in array
                sum += input_value[i];

        average = sum / no_values;

        cout << "The average file in file test was "
             << average << "\n";

        fi.close();
}




Example 

Here is a program that creates an output file, writes information to it,
closes the file and opens it again as an input file and reads in the
information.

#include <iostream.h>
#include <fstream.h>

main()
{
        ofstream        fo;

        fo.open("test_file");

        if (!fo)
        {
                cout << "Cannot open test_file\en";
                exit(0);
        }

        fo << "Hello\en";
        fo << 101.7;

        fo.close();     // close the file

        ifstream        fi;
        fi.open("test_file");

        if (!fi)
        {
                cout << "Cannot open test_file\en";
                exit(0);
        }

        char    str[80];
        float   x;

        fi >> str;
        fi >> x;

        cout << str << "\en" << x << "\en";

        fi.close();
}








Back to Main Page for Lecture 8 









 

 

 File Input and Output  

 Additional Example 
 







PROBLEM:

Write a program which reads in text from a file character by character and
then
writes out the text to another file, but this time in CAPITAL (or UPPER case)
letters.

The input and output filenames will be specified as command line arguments.

You will need to use the function toupper(). This function is
declared in ctype.h. It is sent a character and returns the
character in upper case through the type specifier.

You will need to use the fgetc() function to preserve the white
spaces.

Create your own input file - ie. in nedit create a file called
test_in and type in some text. For example

The cat sat on the mat

The output file 
obtained by running the program should then contain

THE CAT SAT ON THE MAT






SOLUTION:

The first thing the program should do, is check the number of command line
arguments and ensure that it is correct (In this case 3 - program name, input
filename and output filename). It can then open the input and output files
and check that they have been opened, if they are not correctly opened the
program will stop.

The program can then proceed and read character by character from the input
file. Each character read will then be converted to a capital letter and
written to the output file before the next character is read in. 


#include <iostream.h>
#include <fstream.h>
#include <ctype.h>

main(int argc, char *argv[])
{
        ifstream fi;
        ofstream fo;
        char    lower, upper;

	if (argc != 3)
	{
                cerr << "USAGE: program input_file output_file\n";
                exit(1);
	}
	
        fi.open(argv[1]);

        if (!fi)
        {
                cout << "Can't open input test file " << argv[1] << "\n";
                exit(0);
        }
        fo.open(argv[2]);
        if (!fo)
        {
                cout << "Can't open output test file " << argv[2] << "\n";
                exit(0);
        }

        lower = fi.get();

        while (!fi.eof())
        {
                upper = toupper(lower);
                fo << upper;
                lower = fi.get();
        }

        fi.close();
        fo.close();
}






Note carefully the loop for reading in and converting the characters. The
loop is written in this way to ensure that the End of File is correctly
detected. C programs will not detect the end of file until they have tried to
read the end of file. Therefore, you will not that the order ensures that the
next step afet reading a characetr, is checking if it is the end of file.
This is illustrated in the figure below.




If the code had been written as 


        while (!feof(fi))
        {
                lower = fgetc(fi);
                upper = toupper(lower);
                fo << upper;
        }


This would have produced an additional characetr on the end of the output
file, since it would have read the end of file character, tried to convert it
to upper case and written it to the output file before it checked if it was
the end of file. As shown below







If reading values from a file and checking for the end of
file always be careful of the End of File
Character and take care in the order in which you construct any
loops







Back to Main Page for Lecture 8 







  
 

 

 File Input and Output  

 Additional Exercises 
 






Write a program to read in the marks for a class. The file contains a number
to indicate how many students are in the class, and then the
name of a student followed by four marks, one for each of the subjects they
are studying. The format of the input file is shown below:-



3


bob
 10
 10
 10
 10 


fred
 2
 12
 14
 15 


john
 3
  4
  5
  6 



The program will then calculate the average mark for each student, it will
write the name of the student and their average mark to another file. For
the example above, the output file will contain



bob   
     10 


fred  
   10.75 


john   
    4.5 




In addition the program will print the overall average for the class to the
screen.



 NOTE:
 You can assume that the maximum length of the name for any student is 10
characters.









Back to Main Page for Lecture 8 







 


  

 

 File Input and Output 
 

 






Declare a file pointer, fi


Using this file pointer, open the text file file.txt for reading only.


Add code to check that the file has been opened correctly.


How would the code for question 2 be changed if you wished to oen the file for
writing rather than reading.


Assuming that the file file.txt opened for reading contains a column of
integers, give some example to code to read all the numbers into an array.








 Answers
                                        






































































ifstream fi;

  fi.open("file.txt");

 
if (!fi ) 
{
         cout << "Can't open file file.txt for reading\n";
         exit(1);
}

   oftream fi;

fi.open("file.txt");



fi >> array[i];
while (!fi.eof())
{
	i++;
	fi >> array[i];
}






 


 Back to Questions


Back to Main Page for Lecture 8 



                                        
















 Introduction to Classes 
 
 








Learning Objectives


 
Understand the basic concepts of classes and object orientated programming




Understand the concepts of member functions and member data




Be able to declare a class


 
Be able to access the members of class




Understand the terms private and public




Be able to create a constructor for a class









Source Code from Notes





  Additional Examples



 
Quick Check Exercises 



 
Additional Exercises to try



 
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 Introduction to Classes 

Source Code from Notes 
 
 


Example 

Example to illustrate accessing member data and member functions for a class


#include <iostream.h>

//Declaration of Class

class Savings
{
        public:
        unsigned      account_number;
        float         balance;

        unsigned deposit(unsigned amount)
        {
                balance += amount;
                return balance;
        }
};

//main program which uses class Savings
main()
{
        Savings a;   // declare an object a of data type Savings
        Savings b;   // declare an object b of data type Savings

        a.account_number = 1;  // set account number for a to 1
        a.balance = 0.0;       // set balance for a to 0
        a.deposit(10.0);       // deposits 10 pounds to a

        b.account_number = 2;  // set account number for b to 2
}



Example


#include <iostream.h>
// class declaration
class Rational
{
        public:
                void assign (int, int);
                double convert ();      // convert fraction to double
                void invert ();         // invert fraction
                void print ();          // print as a fraction
        private:
                int     num, den;
};
// main program
main()
{
        Rational x;

        x.assign(22,7);
        cout << " x = " ;
        x.print();
        cout << " = " << x.convert() << "\n";
        x.invert();
        cout << "1/x = ";
        x.print();
        cout << "\n";
}

//Rational function definitions
void Rational::assign(int n, int d)
{
        num = n;
        den = d;
}
double Rational::convert()
{
        return double(num)/den;
}
void Rational::invert()
{
        int temp = num;
        num = den;
        den = temp;
}
void Rational::print()
{
        cout << num << "/" << den;
}




Example - Constructor


Add Constructor to Student class example


#include >iostream.h<
class Student
{
        public:
        Student()
        {
                module_hours = 0;
                average_marks = 0.0;
        }
        //marks - returns average marks
        float marks()
        {
                return average_marks;
        }

        //hours_worked - returns number of study hours
        float hours_worked()
        {
                return module_hours;
        }


        // other public functions

        private:
        int     module_hours;
        float   average_marks;
};

int main()
{
        Student s; // create the object and initialise it

        // print out initialised values
        cout << "hours " << s.hours_worked()
             << " average_marks " << s.marks() << "\n";

        // rest of main 
}






Constrcutor Example - Rational Class


class Rational
{
        public:
                Rational(int n, int d)
                {
                        num = n;
                        den = d;
                }
                void print();
        private:
                int num,den;
};
main()
{
        Rational x(22,7), y(-2,5);
        cout << "x = ";
        x.print();
        cout << "y = ";
        y.print();
}






Back to Main Page for Lecture 9 










 

 

 Introduction to Classes 

 Additional Example 
 







PROBLEM:

Implement a Time class. Each object of this class will
represent a specific time of the day, storing the hours and minutes as
integers. Include a constructor, access functions, a function
advance(int h, int m) to advance the current time of day by
the specified amount and a function reset(int h, int m) to reset
the current time to the specified time. Also include a print
function.

Remember to include a normalise() function which checks that
 0 <= minutes < 60 and  0 <= hours < 24. If the
minutes are outside the specified range increment the hours and calculate
the minutes in the correct range. If the number of hours are greater than
24 then decrement by 24. This function will be called automatically each
time the time is changed \fIie.\fR by the constructor, the
reset and advance functions.

Write a short main program which declares objects of type Time
and manipulates them using the functions given.





SOLUTION:


#include <iostream.h>
#include <math.h>

class Time
{
        public:
                Time(int h, int m)
                {
                        hour = h;
                        min  = m;
                        normalise();
                }
                int  hours() { return hour;}
                int  mins()  { return min;}
                void advance(int h, int m);
                void reset(int h , int m);
                void print();
        private:
                int     hour, min;
                void normalise();
};

void Time::advance(int h, int m)
{
        hour += h;
        min += m;
        normalise();
}
void Time::reset(int h, int m)
{
        hour = h;
        min = m;
        normalise();
}
void Time::print()
{
        cout << hour << ":" << min << "\en";
}
void Time::normalise()
{
        while (min > 60)
        {
                min -= 60;
                hour++;
        } 
        while (hour > 24)
                hour -= 24;
}


main()
{
        Time time1(9,31), time2(10, 25);

        cout << "Time1 is ";
        time1.print();

        time1.advance(11,35);
        cout << "Time 1 is ";
        time1.print();

        time1.advance(11,35);
        cout << "Time 1 is ";
        time1.print();

        cout << "Time 2 is ";
        time2.print();
        time2.reset(12,11);
        cout << "Time 2 is ";
        time2.print();

}


The main program is simply to test the class and its member functions

If run the program will produce the following output


Time 1 is 9:31
Time 1 is 21:6
Time 1 is 8:41
Time 2 is 10:25
Time 2 is 12:11


On entering the main program, the contsructor is called to initialse Time1 to
containg the time 9:31 and it is called again to initialsise Time2 to be
10:25. The print() member function is then called to display the
value of Time1. This is then advanced by 11 hours and 35 minutes and
the new time is printed. This is repeated. Time2 is then prointed
and is the reset to the time of 12:11 and thus new time is displayed.




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 Introduction to Classes 

 Additional Exercises 
 










Back to Main Page for Lecture 9 







 


  

 

 Introduction to Classes 
 

 






Example the difference between a public and a private
member of a class.


Example the difference between a constructor and a destructor


What name must a constructor have?


What name must a destructor have?










 Answers
                                        






































































A public member is accessible from outside a class. A
private member can only be accessed within the class.

 
A constructor function is called automatically whenever we instantiate
(create) an object of that class and can be used to inialise the object. A
destructor is called automatically whenever the scope of the object
terminates (ie at the end of the program)

 
A constructor always has the same name as the class
  

A destructor has the same name as the class but prefixed with a ~ (tilde)







 


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