Passing pointer to any member function as class template argument

Question

template<typename T, typename M, M Method>
class ProxyObject
{
public:

    template<typename... Args>
    void Invoke (T& Object, _In_ Args&&... A)
    {
        (void)(Object.*Method)(std::forward<Args>(A)...);
    }
};

class Object
{
public:

    int MyMethod (int Val)
    {
        wcout << L"Hello!" << endl;
        return Val;
    }
};


int wmain ()
{
    Object myObj;
    ProxyObject<Object, decltype(&Object::MyMethod), &Object::MyMethod> obj;

    obj.Invoke(myObj, 10);

    return 0;
}

The decltype(&Object::MyMethod) seems redundant in the definition of obj. Is there any way to make the ProxyObject automatically infer the type of the pointer-to-member-function being passed, so that I can define obj as follows:

ProxyObject<Object, &Object::MyMethod> obj;

@cpplearner That's nice. The timing is almost uncanny. Still hoping for a solution that I can use today though 😊 — TripShock
– TripShock, Commented Oct 10, 2015 at 3:01

songyuanyao · Accepted Answer · 2015-10-10 03:09:36Z

2

I think it's impossible for class template, because you have to specify the member function type explicitly.

Template function could help you lot with the argument deduction:

template<typename T, typename M, typename... Args>
void invoke (T& Object, M Method, Args&&... A)
{
    (void)(Object.*Method)(std::forward<Args>(A)...);
}

then

invoke(myObj, &Object::MyMethod, 10);

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answered Oct 10, 2015 at 3:09

songyuanyao

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2 Comments

TripShock Over a year ago

That's unfortunate. I do want the member function to be part of the class type though. Thanks for the suggestion.

songyuanyao Over a year ago

Class template need specify the template parameter explicity, and there're both type template parameter and nontype template parameter, so...

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