c++ - 具有互斥成员的默认 move 构造函数

Question

我有一个带有删除的复制构造函数的类，我正试图将互斥成员放在这样的地方:

struct A {
    A(const A &other) = delete;
    A& operator=(const A &other) = delete;
    A(A&& other) = default;
    A& operator=(A &&other) = default;

    std::mutex lock;
};

编译器提示我正在尝试调用已删除的复制构造函数，我总结这是由于 std::mutex 类型不可 move 。我怎样才能使互斥成员以最少的麻烦与 move 构造函数一起玩？我实际上并不想将互斥锁成员本身 move 到新构造的对象中，实际上我希望每个 move 的对象只构造它自己的互斥锁

Answer 1

I don't actually want to move the mutex member itself into the newly constructed object, and would actually like each moved object to just construct it's own mutex

然后简单地定义你的 move 构造函数来构造一个新的互斥量:

struct A {
    A(const A &other) = delete;
    A& operator=(const A &other) = delete;
    A(A&& other)
        : lock()
    {
    }

    A& operator=(A &&other) = delete;

    std::mutex lock;
};

move 分配仍然是一个问题，可能应该被删除。除非你能回答这个问题:当你被分配一个新值时，现有的互斥锁成员会发生什么？特别是:如果在现有互斥​​体被锁定的情况下为您分配了一个新值怎么办？