Why isn't break good enough?

Question

C++

When I first ran this code with a different input value of 881, 643, 743, etc... which are all primes numbers, I got a result of "True" but when I input a higher number like 804047277, it came back as "True" when it should have been "False"

#include <iostream>

int main(){

    int num;

    std::cin >> num;

    for(int i = 2; i < num; i++){
        if(num % i == 0){
            std::cout << "True" << std::endl;
            break;
        }
        else{
            std::cout << "False" << std::endl;
            break;
        }
    }
    return 0;
}

I corrected my code (The code below) and received the correct answer, which was "False"

#include <iostream>

int main(){

    int num;

    std::cin >> num;

    for(int i = 2; i < num; i++){
        if(num % i == 0){
            std::cout << "True" << std::endl;
            break;
            return 0;
        }
        else{
            std::cout << "False" << std::endl;
            break;
        }
    }
    return 0;
}

Shouldn't the break in the if statement stop the loop overall? I am just trying to understand why the break wasn't good enough, and I had to return 0;

Answer 1

I would correct your code like following (see description afterwards):

Try it online!

#include <iostream>

int main() {
    int num = 0;
    std::cin >> num;
    
    for (int i = 2; i < num; ++i)
        if (num % i == 0) {
            std::cout << "True (Composite)" << std::endl;
            return 0;
        }

    std::cout << "False (Prime)" << std::endl;
    return 0;
}

Input:

804047277

Output:

True (Composite)

As it is easy to understand, your program is intended to check primality and compositness of a number.

Mistake in your program is that you show False (Prime) when division by a very first i gives non-zero remainder. Instead, to actually check primality, you need to divide by all possible i and only if ALL of them give non-zero, then number is prime. It means that you shouldn't break or show False on very first non-zero remainder.

If ANY of i gives zero remainder then given number by definition is composite. So unlike the Prime case, this Composite case should break (or return ) on very first occurance of zero remainder.

In code above on very first zero remainder I finish program showing to console that number is composite (True).

And only if whole loop finishes (all possible divisors are tested) then I show False (that number is prime).

Regarding question if break; is enough to finish a loop, then Yes, after break loop finishes and you don't need to return 0; after break, this return statement never finishes.

Also it is well known fact that it is enough to check divisibility until divisor equal to Sqrt(num) , which will be much faster. So your loop for (int i = 2; i < num; ++i) should become for (int i = 2; i * i <= num; ++i) which is square times faster.