c++ - 为什么它不会发生死锁？

Question

代码同时从两个不同的线程获取相同的互斥体。 我知道应该发生死锁。为什么没有发生？

#include <iostream>
#include <thread>
#include <queue>
#include <mutex>
#include <condition_variable>

template <typename T>
class SafeQueue
{
 public:
  T pop()
  {
    std::unique_lock<std::mutex> mlock(mutex_);
    std::cout << "lock pop()" << std::endl;
    while (queue_.empty())
    {
      cond_.wait(mlock);
      std::cout << "lock pop awake. Items: " << queue_.size() << std::endl;
    }
    auto item = queue_.front();
    queue_.pop();
    std::cout << "returning from pop" << std::endl;
    return item;
  }

  void push(const T& item)
  {
    std::unique_lock<std::mutex> mlock(mutex_);
    std::cout << "lock push()" << std::endl;
    queue_.push(item);
    mlock.unlock();
    cond_.notify_one();
  }
 private:
  std::queue<T> queue_;
  mutable std::mutex mutex_;
  std::condition_variable cond_;
};

SafeQueue<int> queue;

void pop()
{
    std::cout << "popping..." << std::endl;
    std::cout << "popped: " << queue.pop() << std::endl;
}

int main()
{
    std::thread consumerThread(pop);
    std::this_thread::sleep_for(std::chrono::seconds(3));
    std::cout << "main thread will push" << std::endl;
    queue.push(2);
    std::cout << "pushed" << std::endl;
    consumerThread.join();
    std::cout << "end" << std::endl << std::endl;
}

我的输出是:

popping...

lock pop()

main thread will push

lock push()

pushed

lock pop awake. Items: 1

returning from pop

popped: 2

end

Answer 1

这个声明:

 cond_.wait(mlock);

实际上在等待期间解锁互斥锁，并在发出信号后重新获取锁。这就是您没有任何死锁的原因。