将参数传递给std :: thread包装器

我想实现一个小线程包装器,如果一个线程仍处于活动状态,或者该线程已完成其工作,它将提供信息.为此,我需要将函数及其参数传递给线程类到另一个函数.我有一个简单的实现应该可以工作,但不能让它编译,我无法弄清楚该怎么做才能使它工作.

这是我的代码:

#include 
#include 
#include 
#include 
#include 

class ManagedThread
{
public:
   template< class Function, class... Args> explicit ManagedThread( Function&& f, Args&&... args);
   bool isActive() const { return mActive; }
private:
   volatile bool  mActive;
   std::thread    mThread;
};

template< class Function, class... Args>
   void threadFunction( volatile bool& active_flag, Function&& f, Args&&... args)
{
   active_flag = true;
   f( args...);
   active_flag = false;
}

template< class Function, class... Args>
   ManagedThread::ManagedThread( Function&& f, Args&&... args):
      mActive( false),
      mThread( threadFunction< Function, Args...>, std::ref( mActive), f, args...)
{
}

static void func() { std::cout << "thread 1" << std::endl; }

int main() {
   ManagedThread  mt1( func);
   std::cout << "thread 1 active = " << std::boolalpha << mt1.isActive() << std::endl;
   ::sleep( 1);
   std::cout << "thread 1 active = " << std::boolalpha << mt1.isActive() << std::endl;

   return 0;
}

我得到的编译器错误:

In file included from /usr/include/c++/5/thread:39:0,
                 from prog.cpp:4:
/usr/include/c++/5/functional: In instantiation of 'struct std::_Bind_simple, void (*)()))(volatile bool&, void (&)())>':
/usr/include/c++/5/thread:137:59:   required from 'std::thread::thread(_Callable&&, _Args&& ...) [with _Callable = void (&)(volatile bool&, void (&)()); _Args = {std::reference_wrapper, void (&)()}]'
prog.cpp:28:82:   required from 'ManagedThread::ManagedThread(Function&&, Args&& ...) [with Function = void (&)(); Args = {}]'
prog.cpp:35:28:   required from here
/usr/include/c++/5/functional:1505:61: error: no type named 'type' in 'class std::result_of, void (*)()))(volatile bool&, void (&)())>'
       typedef typename result_of<_Callable(_Args...)>::type result_type;
                                                             ^
/usr/include/c++/5/functional:1526:9: error: no type named 'type' in 'class std::result_of, void (*)()))(volatile bool&, void (&)())>'
         _M_invoke(_Index_tuple<_Indices...>)
         ^

现场示例如下:https://ideone.com/jhBF1q

在错误消息,你可以看到其中的差别void (*)()VS void (&)().这是因为性病::线程的构造函数的参数std::decay编.

另外还添加std::ref到f:

template< class Function, class... Args>
   ManagedThread::ManagedThread( Function&& f, Args&&... args):
      mActive( false),
      mThread( threadFunction< Function, Args...>, std::ref(mActive), std::ref(f), std::forward(args)...)
{
}

@ O'Neil的回答是正确的,但我想提供一个简单的lambda方法,因为你已将其标记为C++14.

template
ManagedThread::ManagedThread(Function&& f, Args&&... args):
      mActive(false),
      mThread([&] /*()*/ { // uncomment if C++11 compatibility needed
        mActive = true;
        std::forward(f)(std::forward(args)...);
        mActive = false;
      })
{}

这将不再需要外部功能.