C program for prime number

Prime number program in c: c program for prime number, this code prints prime numbers using c programming language. To check whether a number is prime or not see another code below. Prime number logic: a number is prime if it is divisible only by one and itself. Remember two is the only even and also the smallest prime number. First few prime numbers are 2, 3, 5, 7, 11, 13, 17....etc. Prime numbers have many applications in computer science and mathematics. A number greater than one can be factorized into prime numbers, For example 540 = 2²*3³*5¹

Prime number program in c language

#include<stdio.h>
 
int main()
{
   int n, i = 3, count, c;
 
   printf("Enter the number of prime numbers required\n");
   scanf("%d",&n);
 
   if ( n >= 1 )
   {
      printf("First %d prime numbers are :\n",n);
      printf("2\n");
   }
 
   for ( count = 2 ; count <= n ;  )
   {
      for ( c = 2 ; c <= i - 1 ; c++ )
      {
         if ( i%c == 0 )
            break;
      }
      if ( c == i )
      {
         printf("%d\n",i);
         count++;
      }
      i++;
   }
 
   return 0;
}

Output of program:

C program for prime number or not

#include<stdio.h>
 
main()
{
   int n, c = 2;
 
   printf("Enter a number to check if it is prime\n");
   scanf("%d",&n);
 
   for ( c = 2 ; c <= n - 1 ; c++ )
   {
      if ( n%c == 0 )
      {
         printf("%d is not prime.\n", n);
  break;
      }
   }
   if ( c == n )
      printf("%d is prime.\n", n);
 
   return 0;
}

C program for prime number using function

#include<stdio.h>
 
int check_prime(int);
 
main()
{
   int n, result;
 
   printf("Enter an integer to check whether it is prime or not.\n");
   scanf("%d",&n);
 
   result = check_prime(n);
 
   if ( result == 1 )
      printf("%d is prime.\n", n);
   else
      printf("%d is not prime.\n", n);
 
   return 0;
}
 
int check_prime(int a)
{
   int c;
 
   for ( c = 2 ; c <= a - 1 ; c++ )
   { 
      if ( a%c == 0 )
  return 0;
   }
   if ( c == a )
      return 1;
}

There are many logic to check prime numbers, one given below is more efficient then above method.
for ( c = 2 ; c <= (int)sqrt(n) ; c++ )
Only checking from 2 to square root of number is sufficient.
There are many more efficient logic available.