CODEWITHSUNDEEP
c
Rojet
Rojet

Reputation: 9

I cannot find what is wrong with this C program to check whether a number is prime or not using function

I know this may be a dumb question, but I cannot find what is wrong with this code. Here I have initialized a function which checks whether the number entered is prime or not.

#include <stdio.h>
int check_prime(int n)
{
    int i,count=0;
    for(i=2;i<n/2;i++)
     {
         if(n%i==0)
          {
              count++;
          }
     }
     if(count==0)
      {
          printf("the number is prime");
      }
    
}
int main()
{
    int n;
    printf("enter the number,\n");
    scanf("%d",&n);
    int check_prime(int n);
    return 0;
}

the output only shows:

enter the number

after I enter the number it does not display anything.

Upvotes: 0

Views: 164

Answers (4)

chux
chux

Reputation: 153447

after I enter the number it does not display anything

Aside from the non-call to check_prime() should be replaced with a call

// int check_prime(int n);  // This is only a function declaration.
check_prime(n);

Prime detection is very slow for large values and can be significantly sped up.

//for(i=2;i<n/2;i++) {
//  if(n%i==0) {
//    count++;
//  }
//}

//         vvvv 
for(i=2; i<=n/i; i++) {
  if(n%i==0) {
    count++;
    break; // add
  }
}

Upvotes: 0

L.Grozinger
L.Grozinger

Reputation: 2398

There appears to be 3 problems.

The return type of check_prime

You declare int check_prime(int n), which means the function should return an int. In this case, your function never returns anything, it just prints stuff, so a return type of void is more suitable.

You declare check_prime again in main instead of calling it

In main, when you say int check_prime(int n); you are actually declaring a check_prime function again, not calling it. You would call it as check_prime(n);.

The algorithm has a bug in it

As it is, your algorithm thinks 4 is a prime number (which it is not).

In your for loop, you check if n is divisible by the integers from 2 to (n / 2) - 1, because the termination condition is i < n/2, which is true when i == n / 2. For the example of 4, this mean it never runs the for loop.

The termination condition in the loop should be changed to i <= n / 2, as in:

for(i = 2; i <= n / 2; i++)

Upvotes: 1

Mihir Joshi
Mihir Joshi

Reputation: 450

The answer above is perfect. Since you said you are new to the programming world, I'd love to explain a bit more.

Philosophical Answer:

  • In C/C++, when you declare the function, you tell the compiler, "Hey, I have created a function foo(), watch out".
  • And when you call the function, you tell the compiler, "Yo, do you remember that thing I told you to watch out for? I need to access it now".

Technical Answer:

  • Syntactically, int check_prime(int n); is how you declare a function.
  • And check_prime(n); //n is an integer is how you call the function.

Upvotes: 2

MikeCAT
MikeCAT

Reputation: 75062

int check_prime(int n);

Is not a call of function but a declaration of function.

It should be

check_prime(n);

to call the function.

Upvotes: 3

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