c++ - 使 C++ 类成为监视器（在并发意义上）

Question

我想确保一次只有一个线程可以运行我的 C++ 类的方法。换句话说，让这个类表现得像一个Monitor。

有没有一种模式、模板化的方式来做到这一点，或者我可以使用一些 Boost 类？因为到目前为止我唯一的想法是添加一个关键部分成员，并在每个方法的开头获取它并在最后释放它（当然使用 RAII）。但这似乎非常多余，我不能将它重用于其他课程。

Answer 1

您可以通过明智地使用operator->现代 c++ 来实现这一点，它提供了比以前接受的答案更简洁的语法：

template<class T>
class monitor
{
public:
    template<typename ...Args>
    monitor(Args&&... args) : m_cl(std::forward<Args>(args)...){}

    struct monitor_helper
    {
        monitor_helper(monitor* mon) : m_mon(mon), m_ul(mon->m_lock) {}
        T* operator->() { return &m_mon->m_cl;}
        monitor* m_mon;
        std::unique_lock<std::mutex> m_ul;
    };

    monitor_helper operator->() { return monitor_helper(this); }
    monitor_helper ManuallyLock() { return monitor_helper(this); }
    T& GetThreadUnsafeAccess() { return m_cl; }

private:
    T           m_cl;
    std::mutex  m_lock;
};

这个想法是您使用箭头运算符来访问底层对象，但这会返回一个辅助对象，该对象锁定然后解锁函数调用周围的互斥锁。然后通过反复应用语言的魔力，operator->您可以获得对底层对象的引用。

用法：

monitor<std::vector<int>> threadSafeVector {5};

threadSafeVector->push_back(0);
threadSafeVector->push_back(1);
threadSafeVector->push_back(2);

// Create a bunch of threads that hammer the vector
std::vector<std::thread> threads;
for(int i=0; i<16; ++i)
{
    threads.push_back(std::thread([&]()
    {
        for(int i=0; i<1024; ++i)
        {
            threadSafeVector->push_back(i);
        }
    }));
}

// You can explicitely take a lock then call multiple functions
// without the overhead of a relock each time. The 'lock handle'
// destructor will unlock the lock correctly. This is necessary
// if you want a chain of logically connected operations 
{
    auto lockedHandle = threadSafeVector.ManuallyLock();
    if(!lockedHandle->empty())
    {
        lockedHandle->pop_back();
        lockedHandle->push_back(-3);
    }
}

for(auto& t : threads)
{
    t.join();
}

// And finally access the underlying object in a raw fashion without a lock
// Use with Caution!

std::vector<int>& rawVector = threadSafeVector.GetThreadUnsafeAccess();
rawVector.push_back(555);

// Should be 16393 (5+3+16*1024+1)
std::cout << threadSafeVector->size() << std::endl;

Answer 2

首先制作通用监视器类。借助 C++11 的强大功能，您可以像这样简单地做到这一点：

template <class F>
struct FunctionType;
template <class R, class Object, class... Args>
struct FunctionType<R (Object::*)(Args...)> {
  typedef R return_type;
};
template <class R, class Object, class... Args>
struct FunctionType<R (Object::*)(Args...) const> {
  typedef R return_type;
};

template <class Object_>
class Monitor {
public:
   typedef Object_ object_type;
   template <class F, class... Args >
   typename FunctionType<F>::return_type operation(const F& f, Args... args)
   {
       critical_section cs;
       return (object.*f)(args...);
   }
   template <class F, class... Args >
   typename FunctionType<F>::return_type operation(const F& f, Args... args) const
   {
       critical_section cs;
       return (object.*f)(args...);
   }
private:
  object_type object;
  class critical_section {};
};

当然，critical_section实施取决于您。我推荐 POSIX 或一些 BOOST。

它现在可以使用了：

Monitor<std::vector<int> > v;
v.operation((void (std::vector<int>::*)(const int&)) &std::vector<int>::push_back, 1);
v.operation((void (std::vector<int>::*)(const int&)) &std::vector<int>::push_back, 2);
size = v.operation(&std::vector<int>::size);
std::cout << size << std::endl;

正如您所看到的，有时您需要明确说明要调用的成员函数 - std::vector<> 有多个 push_back...

---

对于仍然不支持可变参数模板的编译器 - 下面没有它的解决方案 - 我有最多两个参数的时间 - 这非常不方便 - 如果需要 - 添加具有更多参数的函数：

template <class F>
struct FunctionType;
template <class R, class Object>
struct FunctionType<R (Object::*)()> {
  typedef R return_type;
};
template <class R, class Object>
struct FunctionType<R (Object::*)() const> {
  typedef R return_type;
};
template <class R, class Object, class Arg1>
struct FunctionType<R (Object::*)(Arg1)> {
  typedef R return_type;
};
template <class R, class Object, class Arg1>
struct FunctionType<R (Object::*)(Arg1) const> {
  typedef R return_type;
};
template <class R, class Object, class Arg1, class Arg2>
struct FunctionType<R (Object::*)(Arg1,Arg2)> {
  typedef R return_type;
};
template <class R, class Object, class Arg1, class Arg2>
struct FunctionType<R (Object::*)(Arg1,Arg2) const> {
  typedef R return_type;
};

template <class Object_>
class Monitor {
public:
   typedef Object_ object_type;
   template <class F>
   typename FunctionType<F>::return_type operation(const F& f)
   {
       critical_section cs;
       return (object.*f)();
   }
   template <class F>
   typename FunctionType<F>::return_type operation(const F& f) const
   {
       critical_section cs;
       return (object.*f)();
   }
   template <class F, class Arg1>
   typename FunctionType<F>::return_type operation(const F& f, Arg1 arg1)
   {
       critical_section cs;
       return (object.*f)(arg1);
   }
   template <class F, class Arg1>
   typename FunctionType<F>::return_type operation(const F& f, Arg1 arg1) const
   {
       critical_section cs;
       return (object.*f)(arg1);
   }
   template <class F, class Arg1, class Arg2>
   typename FunctionType<F>::return_type operation(const F& f, Arg1 arg1, Arg2 arg2)
   {
       critical_section cs;
       return (object.*f)(arg1, arg2);
   }
   template <class F, class Arg1, class Arg2>
   typename FunctionType<F>::return_type operation(const F& f, Arg1 arg1, Arg2 arg2) const
   {
       critical_section cs;
       return (object.*f)(arg1, arg2);
   }
private:
  object_type object;
  class critical_section {};
};