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C Program to Search an Element in an array using Pointers

Get an element and find the location of element in array, print -1 if element is not found.

Sample Input 1:

5 5 7 9 3 1 9

Sample Output 1:

2

Sample Input 2:

5 5 7 9 3 1 4

Sample Output 2:

-1

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(void)

{
	
	int n,i,pos=-1,*a,element;

	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);

	}

	printf("Enter a element to find:");
	scanf("%d",&element);

	
	for(i=0;i<n;i++)
	
	{

		if(*(a+i)==element)
		{
			pos=i;
			break;
		}
	}

	printf("%d is in %d",element,pos);	
	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., i is initialized to 0 and incremented by 1 at each iteration of both the for loops.

 First for loop reads n input numbers from user and stores them in array a[] from location 0 to n-1  using Second for loop the element is searched in each location from 0 to the location of element or location n-1.

if element is found in a location that location i is stored as pos and exit from the for loop using break statement.

 if element is not found from 0 to n-1 location, pos remains -1.

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