这段代码为什么发生死锁：pthread_mutex_t mutexA = PTHREAD_MUTEX_INITIALIZER; pthread_mutex_t mutexB = PTHREAD_MUTEX_INITIALIZER; pthread_mutex_t mutexC = PTHREAD_MUTEX_INITIALIZER; static int counterA = 0; static int counterB = 0; int func1() { pthread_mutex_lock(&mutexA); ++counterA; sleep(1); pthread_mutex_lock(&mutexB); ++counterB; pthread_mutex_unlock(&mutexB); pthread_mutex_unlock(&mutexA); return counterA; } int func2() { pthread_mutex_lock(&mutexB); ++counterB; sleep(1); pthread_mutex_lock(&mutexA); ++counterA; pthread_mutex_unlock(&mutexA); pthread_mutex_unlock(&mutexB); return counterB; } void* start_routine1(void* arg) { while (1) { int iRetValue = func1(); if (iRetValue == 100000) { pthread_exit(NULL); } } } void* start_routine2(void* arg) { while (1) { int iRetValue = func2(); if (iRetValue == 100000) { pthread_exit(NULL); } } } void* start_routine(void* arg) { while (1) { sleep(1); char szBuf[128]; memset(szBuf, 0, sizeof(szBuf)); strcpy(szBuf, (char*)arg); } } int main() { pthread_t tid[4]; if (pthread_create(&tid[0], NULL, &start_routine1, NULL) != 0) { _exit(1); } if (pthread_create(&tid[1], NULL, &start_routine2, NULL) != 0) { _exit(1); } if (pthread_create(&tid[2], NULL, &start_routine, "thread3") != 0) { _exit(1); } if (pthread_create(&tid[3], NULL, &start_routine, "thread3") != 0) { _exit(1); } sleep(5); //pthread_cancel(tid[0]); pthread_join(tid[0], NULL); pthread_join(tid[1], NULL); pthread_join(tid[2], NULL); pthread_join(tid[3], NULL); pthread_mutex_destroy(&mutexA); pthread_mutex_destroy(&mutexB); pthread_mutex_destroy(&mutexC); return 0; }

这段代码可能会发生死锁的原因是，当两个线程同时尝试获得多个互斥锁时，可能会发生死锁。例如，如果一个线程在获得 mutexA 的锁之后，尝试获得 mutexB 的锁，而另一个线程已经获得了 mutexB 的锁并正在尝试获得 mutexA 的锁，那么两个线程将会相互等待对方释放已经获取的锁，导致死锁。

解决死锁的方法是，保证所有线程获取多个互斥锁的顺序是相同的。例如，可以规定所有线程都先获取 mutexA 的锁，再获取 mutexB 的锁，最后获取 mutexC 的锁。这样做可以避免死锁的发生。