Breaking out of a loop from within a function called in that loop

Question

I\'m currently trying to figure out a way to break out of a for loop from within a function called in that loop. I\'m aware of the possibility to just have the

Answer 1

(Note: the question has been edited since I originally wrote this)

Because of the way C is compiled it must know where to break to when the function is called. Since you can call it from anywhere, or even somewhere a break makes no sense, you cannot have a break; statement in your function and have it work like this.

Other answers have suggested terrible solutions such as setting a global variable, using a #define or longjumping(!) out of the function. These are extremely poor solutions. Instead, you should use the solution you wrongly dismiss in your opening paragraph and return a value from your function that indicates the state that you want to trigger a break in this case and do something like this:

#include 

bool checkAndDisplay(int n)
{
    printf("%d\n", n);
    return (n == 14);
}

int main(void) {
    for (int i = 0; i <= 100; i++) {
        if (checkAndDisplay(i))
            break;
    }
    return 0;
}

Trying to find obscure ways to achieve things like this instead of using the correct way to achieve the same end result is a surefire way to generate dire quality code that is a nightmare to maintain and debug.

You mention, hidden in a comment, that you must use a void return, this is not a problem, just pass the break parameter in as a pointer:

#include 

void checkAndDisplay(int n, bool* wantBreak)
{
    printf("%d\n", n);
    if (n == 14)
        wantBreak = true;
}

int main(void) {
    bool wantBreak = false;
    for (int i = 0; i <= 100; i++) {
        checkAndDisplay(i, &wantBreak);
        if (wantBreak)
            break;
    }
    return 0;
}

Since your parameters are fixed type I suggest you use a cast to pass in the pointer to one of the parameters, e.g. foo(a, b, (long)&out);