Array of Pointers in C

Just like we can declare an array of int, float or char etc, we can also declare an array of pointers, here is the syntax to do the same.

Syntax: datatype *array_name[size];

Let's take an example:

int *arrop[5];

Here arrop is an array of 5 integer pointers. It means that this array can hold the address of 5 integer variables, or in other words, you can assign 5 pointer variables of type pointer to int to the elements of this array.

The following program demonstrates how to use an array of pointers.

#define SIZE 10

int main()
    int *arrop[3];
    int a = 10, b = 20, c = 50, i;

    arrop[0] = &a;
    arrop[1] = &b;
    arrop[2] = &c;

    for(i = 0; i < 3; i++)
        printf("Address = %d\t Value = %d\n", arrop[i], *arrop[i]);

    return 0;

Expected Output:

Address = 2686764 Value = 10
Address = 2686760 Value = 20
Address = 2686756 Value = 50

How it works ?

Notice how we are assigning the addresses of a, b and c. In line 9, we are assigning the address of variable a to the 0th element of the of the array. Similarly, the address of b and c is assigned to 1st and 2nd element respectively. At this point arr_of_pointer looks something like this:


arrop[i] gives the address of ith element of the array. So arrop[0] returns address of variable a, arrop[1] returns address of b and so on. To get the value at address use indirection operator (*).


So *arrop[0] gives value at address[0], Similarly *arrop[1] gives the value at address arrop[1] and so on.