PointersArray of pointersArray of pointers in CArray of PointersAn array in which each element is a pointer to a variable of a particular data type.In other words, each element of the array points to a memory location where a variable of a specific type is stored.Array of Pointers - Example#include <stdio.h>int main(){ int num1 = 10, num2 = 20, num3 = 30; int *arr[3]; arr[0] = &num1; arr[1] = &num2; arr[2] = &num3; printf("Elements of array:\n"); for (int i = 0; i < 3; i++) { printf("%d\n", *arr[i]); } return 0;}Run Example >>Output:ExplanationWe have declared an array of three integer pointers arr, and we have initialized each pointer to the address of an integer variable. We then print the elements of the array by dereferencing each pointer using the * operator.Example 2#include <stdio.h>int main(){ int arr[] = {10, 20, 30, 40, 50}; int *ptr[5]; int sum = 0; // Initialize the pointers in the array for (int i = 0; i < 5; i++) { ptr[i] = &arr[i]; } // Add up the elements using the pointers for (int i = 0; i < 5; i++) { sum += *ptr[i]; } printf("The sum is: %d\n", sum); return 0;}Run Example >>Output:ExplanationWe start by initializing an integer array arr with five elements.We then declare an array of five integer pointers ptr.We use a loop to initialize each pointer in the ptr array to the address of the corresponding element in the arr array.We then use another loop to iterate over the ptr array and add up the values of each element using the pointers.Finally, we print out the sum.