c++ - 线程安全堆栈 C++ 中的潜在死锁

Question

在“Concurrency in Action”一书中，有一个线程安全堆栈的实现，其中在进入 pop() 和 empty() 函数时获取/锁定互斥锁，如下所示:

class threadsafe_stack {
   private:
      std::stack<T> data;
      mutable std::mutex m;
   public:
      //...
      void pop(T& value) {
         std::lock_guard<std::mutex> lock(m);
         if(data.empty()) throw empty_stack();
         value = std::move(data.top());
         data.pop();
      }

      bool empty() const {
         std::lock_guard<std::mutex> lock(m);
         return data.empty();
      }
};

我的问题是，当进入 pop() 时获取锁的线程调用也受互斥锁保护的 empty() 时，这段代码如何不会陷入死锁？如果 lock() 被已经拥有互斥锁的线程调用，那不是未定义的行为吗？

Answer 1

how does this code not run into a deadlock, when a thread, that has acquired the lock upon entering pop() is calling empty() which is protected by mutex as well?

因为你不是在打电话 empty threadsafe_stack的成员函数但是您正在调用类 std::stack<T> 的 empty() .如果代码是:

void pop(T& value) 
{
    std::lock_guard<std::mutex> lock(m);
    if(empty()) // instead of data.empty()
        throw empty_stack();
    value = std::move(data.top());
    data.pop();
}

那么，它将是 undefined behavior :

If lock is called by a thread that already owns the mutex, the behavior is undefined: for example, the program may deadlock. An implementation that can detect the invalid usage is encouraged to throw a std::system_error with error condition resource_deadlock_would_occur instead of deadlocking.

了解 recursive和 shared互斥体。