std::condition_variable 是否真的在阻塞之前解锁了给定的 unique_lock 对象？

Question

正如参考所说：1) Atomically unlocks lock, blocks the current executing thread, and...

我有以下代码：

#include <iostream>
#include <thread>
#include <condition_variable>

std::mutex mutex_;
std::condition_variable condVar;
std::unique_lock<std::mutex> lck(mutex_); //purposely global to check return of owns_lock() in main

void waitingForWork()
{
    std::cout << "Before wait, lck.owns_lock() = " << lck.owns_lock() << '\n';

    condVar.wait(lck);

    std::cout << "After wait, lck.owns_lock() = " << lck.owns_lock() << '\n';
}

int main()
{
  std::thread t1(waitingForWork);

  std::this_thread::sleep_for(std::chrono::seconds(10));

  std::cout << "In main, lck.owns_lock() = " << lck.owns_lock() << '\n';
  condVar.notify_one();

  t1.join();
  return 0;
}

编译使用：g++ with c++17 on ubuntu

输出：

Before wait, lck.owns_lock() = 1
In main, lck.owns_lock() = 1
After wait, lck.owns_lock() = 1

但根据参考，我希望互斥锁在等待时解锁，即：

In main, lck.owns_lock() = 0

谁能告诉我为什么会这样？

Answer 1

您必须进一步阅读：

解锁后，无论出于何种原因，都会重新获取锁定并等待退出。如果此函数因异常退出，也会重新获取锁。

因此始终保证在退出等待时重新获取锁。