//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";
}

//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";

}

//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";

}

#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";
}

#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";
        }
}

#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";
        }
}

#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;
        }
}

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

#include &ltiostream.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";


}

Control Constructs

Additional Exercises

1. 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.

2. 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.

---

 



Control Constructs

---

3. What is wrong with the following?

int x;
if (x=1)
        x++;

4. What is wrong with the following?

int x, y, z;

if (x  y && < z)
        x = y;

5. Write an extract of code to test if the variable x is a positive number?
6. Write an extract to test if the variable x is a positive and even number?

---


 



---

7. 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.
8. 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)

 

---


 

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

---

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";
        }


}

---

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";

}

---

Iteration

Additional Exercises






---

9. 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.

10. 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.
11. 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.

---

 



Iteration

---

12. Change to following code to a while loop.

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

13. What is wrong with the following?

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

14. Write a while loop that calculates the sum of all numbers between 0 and 20 inclusive?
15. Convert the above loop to a do ... while loop.
16. Convert the above loop to a for loop.
17. What is wrong with the following?

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

---



---

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

18. 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;
}

19. 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.


 

 

---


 

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

---

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];
                }
        }
}

---

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";
}

---

Arrays and Structures

Additional Exercises






---

0. 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).

1. 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.

---

 



Arrays and Structures

---

2. Write a declaration for an array of 100 floats.
3. Initialise all elements in this array to have a value of 0
4. Assign a value of 10.5 to the 5th element in the array you have just declared.
5. What is wrong with the following?

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

6. What is wrong with the following?

float   index;
float   array[9];

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

---



---

7. float a[100];

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

8. a[4]=10.5;
9. 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.
10. The array index must always be an integer and hence the first line should have been

int     index;

#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);
}

// 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);
}

// 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);
}

#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));
}

#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;
}

#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];
        }
}

#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];
        }
}

#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;
}

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 ax&sup2;+bx+c=0, assuming that a is non-zero:-

• if b&sup2;-4ac is greater than 0, there are two root sgiven by (-b&plusmn;sqrt(b&sup2;-4ac))/2a
• if b&sup2;-4ac is equal than 0, there is one root given by -b/2a
• if b&sup2;-4ac is less than 0, there are no real roots
   

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

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;
}

#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";
}

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

        return;
}

#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;
}

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

m x n = m + (m x (n-1))

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

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

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

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

• 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.

0. If n is less than m and n divides m exactly (ie n%m ==0), then we can stop and the answer is n.
1. If m is less than n, then recall the function but swap m and n
2. 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

 
 
 

---

Functions

Additional Exercises






---

3. 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).

4. 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.
5. 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.
 
 

---

 



Functions

---

6. Explain the difference between a function declaration and a functions definition? Where is the function declaration placed?
7. Given the following function declarations, explain the meaning of each.
0. int function1(int a, float b);
1. void function2(float a, float b, float c);
2. int function3(void);
3. float function4(float x);
4. void function5(void);
8. Write an appropriate function call for each of the function declarations given above.
9. What does the following function do?

float function(float a, float b, float c)
{
        float tmp;

        tmp = (a + b + c )/ 3.0 ;

        return tmp;
}

10. Give an example function call for the above function.
11. Give the function declaration for the function given in question 4.



---

12. 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.

0. 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.
1. The function is sent 3 floating point arguments and returns no value to the calling program.
2. The function is sent no arguments but returns an integer.
3. The function is sent one floating point argument and returns a floating point argument.
4. The function is sent no variable and returns nothing.
13. 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();

14. It returns the average value of the three values sent to it.
15. x = function(a,b,c);
16. float function(float, float, float); or float function(float a, float b, float c);

---


 

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

---

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";

}

---

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.

---

Pointers

Additional Exercises






---

17. 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.

18. 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.


 

 

---

 



Pointers

---

0. How do you access the memory address of a variable?


 

 
 
 

1. How do you access the contents of a memory location whose address is stored in a pointer?
2. Given the following declarations:

int     *p, *q;
int n;

Which of the following is valid?

0. p+q
1. p+n
2. p-q
3. q-n
4. n-p
3. 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];

4. 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]?

---



---

5. The memory address of a variable is accessed using the address operator &. Given the variable x it's address is &
6. 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.
0. p+q - This is invalid. We cannot add two pointers.
1. p+n - This is valid. It increments the memory address p points to by n.

0. p-q - This is valid. The subtraction of the two pointers gives the number of integer variables stored between the two locations
1. q-n - This is valid. It decrements the memory address q points to by n
0. n-q - This is invalid. We cannot subtract a pointer from an integer.
    
1. This makes the pointer a_ptr point to the beginning of the array (element number 0 of array a)
2. This increments the pointer so it now points to a[1]. The contents of a[1] then becomes 3.5.
3. 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].
0. 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

---


 

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

---

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();
}

---

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

---

File Input and Output

Additional Exercises






---

1. 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.

---

 



File Input and Output

---

2. Declare a file pointer, fi
3. Using this file pointer, open the text file file.txt for reading only.
4. Add code to check that the file has been opened correctly.
5. How would the code for question 2 be changed if you wished to oen the file for writing rather than reading.
6. 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



---

7. ifstream fi;
8. fi.open("file.txt");

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

9. oftream fi;


 

 

fi.open("file.txt");

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

---


 

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
	Test Cases for Tutorial 9

---

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();
}

---

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.

---

Introduction to Classes

Additional Exercises






---

---



Introduction to Classes

---

0. Example the difference between a public and a private member of a class.
1. Example the difference between a constructor and a destructor
2. What name must a constructor have?
3. What name must a destructor have?


 

 
 
 

---

Answers



---

4. A public member is accessible from outside a class. A private member can only be accessed within the class.
5. 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)
6. A constructor always has the same name as the class