# Array of Pointers in C

Last updated on July 27, 2020

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. 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.

  1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 #include #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:

 1 2 3 Address = 387130656 Value = 10 Address = 387130660 Value = 20 Address = 387130664 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, the arrop 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 (*).

*arrop[i]


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