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?
1. p+q
2. p+n
3. p-q
4. q-n
5. 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.
1. ```
a_ptr = a;
```
2. ```
a_ptr ++;
*a_ptr = 3.5;
```
3. ```
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]?

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.
1. p+q - This is invalid. We cannot add two pointers.
2. p+n - This is valid. It increments the memory address p points to by n.
3. p-q - This is valid. The subtraction of the two pointers gives the number of integer variables stored between the two locations
4. q-n - This is valid. It decrements the memory address q points to by n
5. 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].
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