Mutex 作为 class 的成员

Question

我正在尝试制作一个简单的订阅者-生产者模式，其中多个订阅者和一个生产者运行在不同的线程中（原因是要了解有关线程的更多信息）。然而，我正在努力使互斥锁成为生产者 class 的成员。代码如下：

class Producer {
private:
    vector<Subscriber* > subs;
    thread th;
    int counter;
    mutex mux;

public:
    Producer() : counter(counter), th(&Producer::run, this) {};
    void addSubscriber(Subscriber* s);
    void notify();
    void incrementCounter();
    void run();
    void callJoin(){th.join(); } 
};

void Producer::run() {
    for (int i = 0; i < 10; i++) {
        incrementCounter();
        this_thread::sleep_for(std::chrono::milliseconds(3000));
    }
}

void Producer::addSubscriber(Subscriber* s) {
    lock_guard<mutex> lock(mux);
    subs.push_back(s); 
}

void Producer::notify() {
    lock_guard<mutex> lock(mux);
    for (auto it = subs.begin(); it != subs.end(); ++it) {
        (*it)->setCounterCopy(counter);
    }
}

void Producer::incrementCounter() {
    counter++;
    notify(); 
}

订户class：

class Subscriber {
private:
    string name;
    thread th;
    atomic<int> counterCopy = 0;

public:

    Subscriber(string name) : name(name),  th(&Subscriber::run, this) {};
    void run() {
        while (true) {
            cout << name << ": " << counterCopy << endl; 
            this_thread::sleep_for(std::chrono::milliseconds(1000));            
        }
    }
    void callJoin() { th.join(); }
    void setCounterCopy(int counterCopy) { this->counterCopy = counterCopy; };
};

主线：

int main() {
    Producer p;
    Subscriber s1("Sub1");
    p.addSubscriber(&s1);
    s1.callJoin();
    p.callJoin();
    return 0;
}

lock_guard 的目的是防止生产者在将订户添加到矢量时同时通知矢量中的订户。然而这个异常是在通知 Exception thrown at 0x59963734 (msvcp140d.dll) in Project1.exe: 0xC0000005: Access violation reading location 0x00000000. 的 lock_guard 中抛出的有谁知道这个异常的原因是什么？如果将互斥量设置为全局参数，则效果很好。也可以随意评论代码的其他问题。线程对我来说是全新的。

Answer 1

所以这里发生的是一个古怪的初始化顺序。

Class 成员按照它们在 class 中声明的顺序构造。在您的情况下，这首先意味着订阅者向量，然后是线程，然后是计数器（！），最后是互斥体。在构造函数中指定初始值设定项的顺序无关紧要。

但是！构造线程对象需要启动线程，运行它的功能。最终，可能会在 Producer 构造函数到达它实际初始化它的位置之前使用互斥锁。所以你最终使用了一个尚未构造的互斥锁，并且（不是它是你问题的原因）也是一个尚未初始化的计数器。

一般来说，当你有一个提到 this 或不同的 class 成员的成员初始值设定项时，你应该小心，包括调用成员函数。它设置了访问未初始化对象的场景。

在您的情况下，只需将互斥锁和计数器成员移动到线程成员之前就足够了。

Answer 2

我只想将此添加到@Sneftel 给出的答案中以供参考：

来自 CPP 标准 (N4713)，突出显示相关部分：

15.6.2 Initializing bases and members [class.base.init]

...

13 In a non-delegating constructor, initialization proceeds in the following order:
(13.1) — First, and only for the constructor of the most derived class (6.6.2), virtual base classes are initialized in the order they appear on a depth-first left-to-right traversal of the directed acyclic graph of base classes, where “left-to-right” is the order of appearance of the base classes in the derived class base-specifier-list.
(13.2) — Then, direct base classes are initialized in declaration order as they appear in the base-specifier-list (regardless of the order of the mem-initializers).
(13.3) — Then, non-static data members are initialized in the order they were declared in the class definition (again regardless of the order of the mem-initializers).
(13.4) — Finally, the compound-statement of the constructor body is executed.
[ Note: The declaration order is mandated to ensure that base and member subobjects are destroyed in the reverse order of initialization. —end note ]

Mutex 作为 class 的成员

Mutex as member of class

c++

multithreading

mutex