Bug in C program written to check prime number

Question

I have written following program to check if given number is prime or not. But somehow program gives multiples of 5 like 15,25 as prime numbers. Where is bug?

#include <stdio.h>

int main() {
    int number, i;

    printf("Enter the number to be checked\n");
    scanf("%d", &number);
    for (i = 2; i < number / 2; i++) {
        if (number % i == 0) {
            printf("The given number %d is not a prime number \n", number);
            break;
        } else {
            printf("The given number %d is a prime number \n", number);
            break;
        }
    }
    return 0; 
}

Answer 1

You should use a flag to know when it's a prime number (Or it will break as soon as it test a single number that's not divisor of number)

bool isPrime = true;
for(i=2;i<number/2;i++){
    if(number%i==0)
    {
      isPrime = false;
      break;
    }
}
if (isPrime){
    printf("The given number %d is a prime number \n",number);
}
else {
    printf("The given number %d is not a prime number \n",number);
}

Answer 2

Remove the else part from for and put these lines after the loop

if( i == number/2)
    printf("The given number %d is a prime number \n",number);

Answer 3

This code has lot of problems

#include<stdio.h>
int main()
{
  int number,i;
  printf("Enter the number to be checked\n");
  scanf("%d",&number);
  for(i=2;i<number/2;i++){
    if(number%i==0)
    {
      printf("The given number %d is not a prime number \n",number);
      break;
    }
    else
    {
      printf("The given number %d is a prime number \n",number);
      break;
    }
  }
  return 0; 
}

Say you entered 5, so it enterd for loop first iteration i=2

if(number%i==0) i.e 5%2==0 which is obviously false now your code goes to else condition and print and break

So what this break do is, it stops execution of innermost loop, as u have only one loop your execution stops and hence you get wrong answer not just for multiples of 5 but for many other inputs.

else condition should not be inside loop

A simple code would be

 #include <stdio.h>
    int main()
    {
      int n, i, flag=0;
      printf("Enter a positive integer: ");
      scanf("%d",&n);
      for(i=2;i<=n/2;++i)
      {
          if(n%i==0)
          {
              flag=1;
              break;
          }
      }
      if (flag==0)
          printf("%d is a prime number.",n);
      else
          printf("%d is not a prime number.",n);
      return 0;
    }

You should learn how to use if-else conditioning and use of break statements

Answer 4

There are several bugs in the program.

The first one is that you do not check whether the entered number is positive. So if the user will enter a negative number the program will report nothing.

The second bug is that if the user will enter 1, 2, 3, or 4 then the program again will report nothing.

The third bug is that you exit the loop after its first iteration.

Take into account that according to the C Standard function main without parameters shall be declared like

int main( void )

The program can look the following way

#include <stdio.h>

int main( void )
{
    unsigned int number = 0;

    printf("Enter the number to be checked: ");
    scanf( "%u", &number );


    int prime = number == 2 || ( number % 2 && number != 1 );

    for ( unsigned int i = 3; prime && i <= number / i; i += 2 )
    {
        prime = number % i;
    }

    if ( prime )
    {
        printf( "The given number %u is a prime number \n", number );
    }
    else
    {
        printf( "The given number %u is not a prime number \n", number );
    }

    return 0; 
}

Answer 5

So, a couple of things...

The main problem is that you're breaking out of your loop early whether the number is divisible by i or not:

if ( num % i == 0 )
  break;
else
  break;

You only want to break out of the loop early if the number is evenly divisible by i; otherwise, you check the against the next value of i:

for ( i = 2 ; i < number/2; i++ )
{
  if ( num % i == 0 )
    break;
}

After the loop, check to see if i is less than number / 2; if it is, then number is not prime:

if ( i < number / 2 )
  // number is not prime
else
  // number is prime

There are a couple of ways you can speed up your primality test. First of all, you only need to test up to the square root of number, rather than number / 2. Secondly, you can eliminate checks against even numbers; if a number is divisible by 2, then it's divisible by any even number, so you only need to check against 2 once. In code:

int prime = 0; // initially false

if ( number == 2 ) // 2 is a special case, check separately
{
  prime = 1;
}
else if ( number < 2 || number % 2 == 0 )
{
  prime = 0; // negative numbers, 0, 1, and even numbers are not prime
}
else
{
  prime = 1; // assume prime to begin with
  for ( int i = 3; prime && i * i < number; i += 2 ) // continue while prime is true
  {                                                  // and i is less than sqrt(number)
    prime = number % i;
  }
}

printf( "%d is%s prime\n", number, prime ? "" : " not" );