c++ concurrency problem using semaphore: print in order

As an c++ concurrency exercise, I want to print letters A B C in that order, using semaphore. Here's my code:

binary_semaphore sem[] = { binary_semaphore(1), binary_semaphore(0), binary_semaphore(0) };
int cnt = 0;

void func(int c) {
    while (true) {
        sem[c].acquire();
        printf("%c ", c + 'A');
        // if (++cnt == 10) {
        //     break;
        // }
        sem[(c + 1) % 3].release();
    }
}

int main() {
    vector<thread> pool;
    for (int i = 0; i < 3; i++) {
        pool.push_back(thread(func, i));
    }
    for (auto& t : pool) {
        t.join();
    }
    return 0;
}

It works. But it's strange that, if I enable the three commented lines

        if (++cnt == 10) {
            break;
        }

Something seems blocked and nothing is printed out.

Can anyone please give me some hints, how does these lines make semaphore broken? Thanks.

With that break;, one thread finishes whereas the 2 others are still blocked by the acquisition. As printf might be buffered, you don't see actual output.

if you flush output, you see the output Demo.

If you move the break;:

std::atomic<int> cnt = 0;

void func(int c) {
    while (true) {
        sem[c].acquire();
        printf("%c ", c + 'A');
        fflush(stdout);
        sem[(c + 1) % 3].release();
        if (++cnt >= 10) {
             break;
        }
    }
}

other threads are no longer blocked Demo (incrementation is no longer protected though).