Explain array of pointers in C programming language

CServer Side ProgrammingProgramming

Pointer is a variable that stores the address of another variable.


  • Pointer saves the memory space.
  • Execution time of pointer is faster because of direct access to memory location.
  • With the help of pointers, the memory is accessed efficiently, i.e., memory is allocated and deallocated dynamically.
  • Pointers are used with data structures.

Pointer declaration and initialization

Consider the following statement −

int qty = 179;

In memory, the variable can be represented as follows −

Declaring a pointer

It means ‘p’ is a pointer variable, which holds the address of another integer variable, as shown below −

Int *p;

Initialization of a pointer

Address operator (&) is used to initialise a pointer variable.

For Example − int qty = 175;

        int *p;

        p= &qty;

Array of pointers

It is collection of addresses (or) collection of pointers


Following is the declaration for array of pointers −

datatype *pointername [size];

For example,

int *p[5];

It represents an array of pointers that can hold 5 integer element addresses.


‘&’ is used for initialisation.

For Example,

int a[3] = {10,20,30};
int *p[3], i;
for (i=0; i<3; i++) (or) for (i=0; i<3,i++)
p[i] = &a[i];
p[i] = a+i;


Indirection operator (*) is used for accessing.

For Example,

for (i=0, i<3; i++)
printf ("%d", *p[i]);

Example program

Given below is the program for array of pointers −

 Live Demo

main ( ){
   int a[3] = {10,20,30};
   int *p[3],i;
   for (i=0; i<3; i++)
      p[i] = &a[i];
   printf ("elements of the array are \n");
   for (i=0; i<3; i++)
      printf ("%d \t", *p[i]);


When the above program is executed, it produces the following result −

elements at the array are : 10 20 30

Example 2

Given below is the program for the array of pointers to strings −

 Live Demo

#include <stdio.h>
#include <stdlib.h>
int main(void){
   char *a[5] = {"one", "two", "three", "four", "five"};
   int i;
   printf ( "the strings are at locations:\n");
   for (i=0; i<5; i++)
      printf ("%d\n", a[i]);
   return 0;


When the above program is executed, it produces the following result −

The strings are at locations:
Published on 11-Mar-2021 17:06:43