c++ - pthread_mutex_trylock？ Windows 中的线程

Question

/////////////////////////////////*

pthread_mutex_t stop = PTHREAD_MUTEX_INITIALIZER;
int a = 1;

void* decrement(void* arg)
{ 
    pthread_mutex_trylock(&stop);
    if(a > 0) { a--; } 
    cout << "Esecuzione thread tid" << endl;
    pthread_mutex_unlock(&stop);
    pthread_exit(NULL);
}

int main()
{
    pthread_t tid;

    pthread_attr_t tattr;
    pthread_attr_init(&tattr);
    pthread_attr_setdetachstate(&tattr, PTHREAD_CREATE_DETACHED);

    pthread_create(&tid, &tattr, decrement, NULL);

    pthread_mutex_lock(&stop);
    if(a > 0) { a--; } 
    cout << "Esecuzione thread main" << endl;

    cout << a << endl;

    pthread_exit(NULL);
    return 0;
}

为什么分离到主线程的线程继续执行而不是返回到 EBUSY 的调用者？

Answer 1

您的问题中没有特定于 Windows 的内容。您实际上误解了 pthread_mutex_trylock() 的工作原理。

• http://pubs.opengroup.org/onlinepubs/9699919799/functions/pthread_mutex_lock.html

The pthread_mutex_trylock() function shall be equivalent to pthread_mutex_lock(), except that if the mutex object referenced by mutex is currently locked (by any thread, including the current thread), the call shall return immediately.
...
The pthread_mutex_trylock() function shall fail if:
[EBUSY]
The mutex could not be acquired because it was already locked.

返回(可能返回)EBUSY 的不是decrement 线程而是pthread_mutex_trylock()(您没有检查... )

顺便说一句，decrement 线程也有可能比 main() 中的 pthread_mutex_lock(&stop) 更早完成执行线。这完全是不确定的。