Question

假设我在下面的示例中有一个运行成员方法的线程，如runController：

class SomeClass {
public:
    SomeClass() { 
         // Start controller thread
         mControllerThread = std::thread(&SomeClass::runController, this) 
    }

    ~SomeClass() {
         // Stop controller thread
         mIsControllerThreadInterrupted = true;
         // wait for thread to die.
         std::unique_lock<std:::mutex> lk(mControllerThreadAlive); 
    }

    // Both controller and external client threads might call this
    void modifyObject() {
         std::unique_lock<std::mutex> lock(mObjectMutex);
         mObject.doSomeModification();
    }
    //...
private:
    std::mutex mObjectMutex;
    Object mObject;

    std::thread mControllerThread;
    std::atomic<bool> mIsControllerInterrupted;
    std::mutex mControllerThreadAlive;

    void runController() {        
        std::unique_lock<std::mutex> aliveLock(mControllerThreadAlive);
        while(!mIsControllerInterruped) {
            // Say I need to synchronize on mObject for all of these calls
            std::unique_lock<std::mutex> lock(mObjectMutex);
            someMethodA();
            modifyObject(); // but calling modifyObject will then lock mutex twice
            someMethodC();
        }
    }
    //...
};

runController中的一些（或所有）子例程需要修改线程之间共享并由互斥锁保护的数据。其中一些（或全部）也可能被需要修改此共享数据的其他线程调用。

凭借C ++ 11的所有荣耀，我怎样才能确保没有线程曾经两次锁定互斥锁？

现在，我将unique_lock引用作为参数传递给方法，如下所示。但这似乎很笨拙，难以维持，可能是灾难性的......等等

void modifyObject(std::unique_lock<std::mutex>& objectLock) {

    // We don't even know if this lock manages the right mutex... 
    // so let's waste some time checking that.
    if(objectLock.mutex() != &mObjectMutex)
         throw std::logic_error();

    // Lock mutex if not locked by this thread
    bool wasObjectLockOwned = objectLock.owns_lock();
    if(!wasObjectLockOwned)
        objectLock.lock();

    mObject.doSomeModification();

    // restore previous lock state
    if(!wasObjectLockOwned)
        objectLock.unlock();

}

谢谢！

Answer 1

有几种方法可以避免这种编程错误。我建议在类设计级别上进行：

将 public 和私有成员函数
只有公共成员函数锁定互斥锁，
和 public 成员函数永远不会被其他成员函数调用。

如果内部和外部都需要一个函数，请创建函数的两个变体，并从一个变为另一个：

public:
    // intended to be used from the outside
    int foobar(int x, int y)
    {
         std::unique_lock<std::mutex> lock(mControllerThreadAlive);
         return _foobar(x, y);
    }
private:
    // intended to be used from other (public or private) member functions
    int _foobar(int x, int y)
    {
        // ... code that requires locking
    }

确保线程没有两次锁定互斥锁？

1 个答案: