# C Program for Linear Search

In this article, you will learn and get code about searching of a number or an element from given array using linear search technique. But before going through the program, if you want to check out the algorithm used for linear search, then refer to Linear Search.

## Linear Search in C

Linear search is an easiest way to search an element from an array. It is also easy to learn. Let's first create a program on it. The question is Write a Program in C that asks from user to enter any 10 array elements, and then ask to enter a number to search from the given array. The answer to this question is:

```#include<stdio.h>
#include<conio.h>
int main()
{
int arr, i, num, pos;
printf("Enter any 10 Numbers: ");
for(i=0; i<10; i++)
scanf("%d", &arr[i]);
printf("\nEnter a Number to Search: ");
scanf("%d", &num);
for(i=0; i<10; i++)
{
if(arr[i]==num)
{
pos=i;
break;
}
}
printf("\nFound at Index No.%d", pos);
getch();
return 0;
}```

This program was build and run under Code::Blocks IDE. Here is its sample run:

Now supply any 10 numbers say 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 and then enter a number say 7 to search. On pressing the `ENTER`. Here is the output produced:

As you can see from the above code, the number (stored in num) gets compared with each and every number one by one. That is, first number (number at 0th index) gets compared to num. If found, then initial the index to a variable say pos and with the help of break keyword, exit the loop. Print the value of pos as number's index. Otherwise check for the number at next index. Continue the process until the number gets found.

Note - Previous program has two limitations. The first one is, if user enters a number that does not available in the list. And the second one is, if user enters a number that present more than one time in the list or array. To overcome this problem, we have created another program.

## Linear Search with Duplicates in C

This program find and prints index number at which the number is available. If the given number present in the given array in repeated order. Then this program also prints its all the indexes where it is available.

```#include<stdio.h>
#include<conio.h>
int main()
{
int arr, size, i, num, arrTemp, j=0, count=0;
printf("Enter Array Size: ");
scanf("%d", &size);
printf("\nEnter any %d Array Elements: ", size);
for(i=0; i<size; i++)
scanf("%d", &arr[i]);
printf("\nEnter a Number to Search: ");
scanf("%d", &num);
for(i=0; i<size; i++)
{
if(arr[i]==num)
{
arrTemp[j] = i;
j++;
count++;
}
}
if(count>0)
{
printf("\nNumber Found at Index No.");
size = count;
for(i=0; i<size; i++)
printf("%d  ", arrTemp[i]);
}
else
getch();
return 0;
}```

Let's suppose user has entered array size as 10 and its 10 elements as 1, 2, 3, 4, 5, 4, 6, 4, 7, 4. And 4 is the number to be search. Therefore here is its sample output you will see:

Here the main logic used is:

• If number gets matched. Then just initialize the current index number to another array say arrTemp[]
• Increment a variable say count every time, initializing the index number to the array arrTemp[]
• The array arrTemp[] holds index number(s) where the number is available and the variable count holds the size of arrTemp[]

#### Same Program in Other Languages

C Online Test

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