代碼同時從兩個不同線程獲取相同的互斥量。我知道應該發生死鎖。爲什麼沒有發生?爲什麼它不會發生死鎖?
#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;
}
我的輸出是:
啪......
鎖pop()方法
主線程將推動
鎖推()
推
鎖彈出清醒。項目:1
從彈出
彈出返回:2
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