c - 为什么在信号处理程序中使用互斥锁会出现问题？

Question

下面是有问题的伪代码:

int c;
pthread_mutex_t mtx;


void inc(int count)
{
    pthread_mutex_lock(&mtx);
    c += count;
    pthread_mutex_unlock(&mtx);
}


int main(void)
{
    pthread_mutex_init(&mtx);
    signal(SIGUSR1, inc);
    signal(SIGUSR2, inc);
    sleep(100000); // Sleep for long enough
    return 0;
}

这段代码如何以及为什么会导致死锁？

为什么这与以下场景不同:

1. Thread 1 acquires the mutex.
2. Context switch is made and Thread 2 tries to get the lock and put on waiting list.
3. Thread 1 finishes and releases the lock.
4. Thread 2 wakes up and continues its execution.
5. No deadlock.

Answer 1

您的信号处理程序都将在同一个线程中运行。如果第二个信号到达，而第一个信号的处理程序已锁定互斥量，则您的独立线程将再次尝试锁定互斥量和死锁:

time    thread 0
----    --------
  0      main:...
  1      main:sleep()
 ...     ...
 100     <<SIGUSR1>>
 101     inc:pthread_mutex_lock()
 102     inc:count += ...
 103     <<SIGUSR2>>
 104     inc:pthread_mutex_lock()  // deadlock