A code made to tell if a number is "Perfect Number"-CodePudding

This is a code I wrote in C to find if a given number is a Perfect number or Not.

A Perfect number is a number that is the sum of all its factors.

EXAMPLE - 6

6 has factors 2 , 3 and 1 (1 because it divisible by itself) and 2 3 1 = 6

#include <stdio.h>
int main(){
    int number;
    int sum =0,i;
    scanf("%d",&number);
    for(i=0;i<number;i  ){
       
        if(number % i==0){
            sum  = i;
        }else{ sum = sum;}
    }if(sum==number){
        printf("Perfect Number");
    }else{
        printf("Not a Perfect Number");
    }
       return 0;
}

The code is logically correct and its suppose to give the right output but the problem is that it is not giving any output.

Instead in CLion it terminates with a code " Process finished with exit code -1073741676 (0xC0000094)"

CodePudding user response：

Your loop must have i to start from 1, not 0, otherwise you'll have division by zero when you do number % i == 0:

#include <stdio.h>

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

  printf("Insert number: ");
  scanf("%d", &number);

  // NOTE: replaced condition `i < number` with `i <= number / 2`
  //   for improved performance
  for (int i = 1; i <= number / 2; i  ) {
    if (number % i == 0) {
      sum  = i;
    }
  }

  if (sum == number) {
    printf("Perfect Number\n");
  }
  else {
    printf("Not a Perfect Number\n");
  }

  return 0;
}

CodePudding user response：

Here is a solution with Cyclomatic Complexity of 2. I don't see a reason for this, but you can do it.

#include <stdio.h>
 
int main(void)
{
    int number = 33550336;
    int sum = 0;
 
    for(int i = 1; i <= number / 2; i  ) {
        sum  = i * (number % i == 0);
    }
 
    printf("%d is %s", number, "not a perfect number."   4 * (sum==number));
 
    return 0;
}