如何将包含对象的互斥体从一个函数传输到另一个函数？

Question

#include <mutex>

class ConcurrentQueue {
    // This class contains a queue and contains functions push to the queue and pop from the queue done in a thread safe manner.
    std::mutex m;
};

class Producer {
    // This class contains several methods which take some ConcurrentQueue objects and then schedule tasks onto it.
public:
    void func(ConcurrentQueue a) {}
};

class Consumer {
    // This class contains several methods which take the same ConcurrentQueue objects and then remove the tasks and complete them one by one.
public:
    void func(ConcurrentQueue a) {}
};

int main() {
    // Here I want to generate the necessary ConcurrentQueue objects and then start threads for producer and consumer methods where I supply it with the required queue objects.
    ConcurrentQueue a;
    Producer b;

    // :( Unfortunately I cannot pass in any of my ConcurrentQueue objects to the methods as apparantly I cannot copy transfer a mutex.
    b.func(a); // This line gives compiler error saying the copy constructor is deleted.
    
    return 0;
}

上面的代码通过注释说明了整个情况。我如何更好地设计它才能实现这一目标？

Answer 1

如果您不希望 ConcurrentQueue 类可复制，则不要按值传递它；请改用引用传递（即类型const ConcurrentQueue &或的参数ConcurrentQueue &）。

OTOH 如果您确实希望您的 ConcurrentQueue 类是可复制的（并且您应该仔细考虑允许复制 ConcurrentQueue 对象是否对您的目标有帮助或有害），您可以简单地向您的 ConcurrentQueue 类添加一个复制构造函数复制其他成员变量，但不尝试复制 std::mutex：

class ConcurrentQueue {
    // This class contains a queue and contains functions push to the queue and pop from the queue done in a thread safe manner.
    std::mutex m;
    std::queue<int> q;

public:
    ConcurrentQueue()
    {
       // default constructor
    }

    ConcurrentQueue(const ConcurrentQueue & rhs)
    {
       // serialize access to rhs.q so we can read its 
       // contents safely while copying them into this->q
       const std::lock_guard<std::mutex> lock(const_cast<std::mutex &>(rhs.m));
       q = rhs.q;
    }
};

请注意，我还需要添加一个默认构造函数，因为一旦您向类添加任何类型的构造函数，编译器将不再自动为您创建默认构造函数。如果您使用的是 C++11 或更高版本，则可以改为通过声明默认构造函数ConcurrentQueue() = default;。