Can anyone tell me what is the problem here, considering I'm a beginner here? It is not showing Runtime error or something.

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int checkPrime(int n){
       for (int i = 2; i * i < n; i  ) {
        if(n % i == 0 && i!=n)
            return 0;
    }
    return n;
}

int main(){
    int t; 
    scanf("%d",&t);
    while(t--){
        int n,sum=0; 
        scanf("%d",&n);
        for(int i=n;i>1;i--){
              if ( checkPrime(i) != 0 )
              sum  = i;
        }
        printf("%d\n",sum);
    }
    return 0;
}

CodePudding user response：

In checkPrime, you have a loop

for(int i=0;i*i<n;i  ){
    if(n%i==0){

But in the first iteration, you have i=0 and then n%i, but mod 0 is undefined behavior in C. I think what you want is

for (i = 2; i * i <= n; i  )

UPDATE: In checkPrime, you can eliminate count entirely, as follows:

int checkPrime(int n) {
    for (int i = 2; i * i <= n; i  ) {
        if(n % i == 0)
            return 0;
    }
    return n;
}

This way, as soon as you find a divisor, you know the number isn't prime so you just return a 0. If you get all the way through the loop, then no divisor was found so the number is prime.

CodePudding user response：

OP's checkPrime() has various troubles.

int checkPrime(int n){
       for (int i = 2; i * i < n; i  ) {
        if(n % i == 0 && i!=n)
            return 0;
    }
    return n;
}

Risks overflow

i * i overflows when n is a prime near INT_MAX.

Returns non-zero (is a prime) for non-primes

checkPrime(1), checkPrime(negative values) should return 0.

Useless code

i!=n in if(n % i == 0 && i!=n) is not needed.

Repaired

int checkPrime(int n){
  for (int i = 2; i < n/i; i  ) {
    if (n % i == 0) {
      return 0;
    }
  }
  return n > 1;
}

Sum may overflow

//int n,sum=0;
...
//sum  = i;
...
//printf("%d\n",sum);

int n;
long long sum=0;
...
sum  = i;
...
printf("%lld\n", sum);

OP's overall code may have additional issues.