C Program to Reverse the Elements in array using pointers

Get array size n and n elements of array, then reverse the n elements.

Sample Input 1:

5 5 7 9 3 1

Sample Output 1:

1 3 9 7 5

Try your Solution

Strongly recommended to Solve it on your own, Don't directly go to the solution given below.

#include<stdio.h> int main() { //write your code here }

Program or Solution


							
			
					
#include <stdio.h>


#include <stdlib.h>
int main()

{
	
	int *a,n,i,j,temp;

	printf("Enter size of array:");
	scanf("%d",&n);

	a=calloc(sizeof(int),n);
	printf("Enter %d Elements:",n);
	for(i=0;i<n;i++)

	{

		scanf("%d",a+i);

	}

	
		
	for(i=0,j=n-1;i<j;i++,j--)
	
	{

		temp=*(a+i);
		*(a+i)=*(a+j);
		*(a+j)=temp;
	}

	printf("After reversing the array:\n");
	for(i=0;i<n;i++)

	{

		printf("%d",*(a+i));

	}

	
	return 0;

}

Program Explanation

calloc() is predefined function allocates memory of specified bytes

Number of bytes is specified as (4,n), it means n 4 bytes.

Since we are using integers, specified as 4 bytes.

*a denotes first four bytes, *(a+1) denotes second four bytes, *(a+2) denotes third four bytes and so on., initialize i to first location of array and j to last location of array using i=0 j=n-1

swap the elements in location i and j, then increment i by 1 and decrement j by 1.

t=*(a+i)*(a+i)=*(a+j) *(a+j)=temp repeat the above step till i is less than j