完美转发可变模板参数到成员函数

Perfect forwarding variadic template parameters to member function

我正在尝试使用示例用法实现观察者模式,如下所示:

class Widget
{
    class WidgetObserver
    {
        virtual void SomethingHappened(Widget&) { };
        virtual void TextChanged(Widget&, const char*) { };
        virtual void BoundsChanged(Widget&, const Rect&) { };
    }

    void AddObserver(WidgetObserver& observer)
    {
        observers.Add(observer);
    }

    virtual void OnSomthingHappened()
    {
        observers.Fire(&WidgetObserver::SomethingHappened, *this);
    }

    virtual void OnTextChanged(const char* text)
    {
        observers.Fire(&WidgetObserver::TextChanged, *this, text);
    }

    // Same for OnBoundsChanged(const Rect&)

    ObserverList<Widget> observers;
}

class AnotherWidget : WidgetObserver
{
    virtual void SomethingHappened(Widget& widget) override
    {
         std::cout << "Something happened with " widget.name() << std::endl;
    }

    // Implementations for TextChanged and BoundsChanged ...
}

void main()
{
    Widget w;
    AnotherWidget aw;
    w.AddObserver(aw);
    w.OnSomethingHappened();
}

现在我的 ObserverList 当前(部分)实现是:

template <typename TObserver>
class ObserverList
{
    std::vector<TObserver*> observer_list;

    void AddObserver(TObserver& observer) { observer_list.push_back(&observer); }

    template <typename ...TArgs>
    void Fire(void(TObserver::*func)(TArgs...), TArgs&& args)
    {
         for (TObserver* ob : observer_list)
         {
             (ob->*func)(std::forward<TArgs>(args)...);
         }
    }
}

这与函数 SomethingHappenedBoundsChanged 的预期一样工作,它们都通过引用而不是通过值传递其所有参数,但是 TextChanged 它的参数之一(const char* text) 按值而不是引用传递。当我将其更改为 const char*& text 时,它会编译,但它不会用 MSVC++ 14.1 (Visual Studio 2017)

编译
error C2672: 'ObserverList<Widget::WidgetObserver>::Fire': no matching overloaded function found
error C2782: 'void ObserverList<Widget::WidgetObserver>::Fire(void (__thiscall Widget::WidgetObserver::* )(TArgs...),TArgs &&...)': template parameter 'TArgs' is ambiguous

好像编译器不喜欢按值传递参数,有什么办法让它喜欢它们吗?

您可以声明 func 具有简单类型 F,即

template <typename F, typename ...TArgs> 
void Fire(F func, TArgs&&... args);