destroy pthread_mutex_lock 会怎么样？？？
请问会有什么后果？锁会被打开么？ 线程会怎么走：
一个线程中阻塞于 pthread_mutex_lock(&mutexSync);
另一个线程中 pthread_mutex_destroy(&mutexSync);
一个线程中阻塞于 sem_wait(&Sync);
另一个线程中 sem_destroy(&Sync); 线程 互斥量 信号量
[解决办法]
好吧, 第二个可以容忍, 但第一个恐怕行不通噢.
[解决办法]
EBUSY The implementation has detected an attempt to destroy the object referenced by mutex while it is locked or referenced (for example, while being used
in a pthread_cond_timedwait() or pthread_cond_wait()) by another thread.
[解决办法]
It is safe to destroy an initialized semaphore upon which no threads are currently blocked. The effect of destroying a semaphore upon which other threads
are currently blocked is undefined.
[解决办法]
不好意思，我回答错误了，关于释放使用中的互斥锁和信号量确实可以正确返回，我测试的行为也是正常的，但文档确实明文说了是会导致不确定行为。我查询了内核代码发现确实会导致不确定行为，I am sorry 可能误导了你。
