Member function wrapper with a single argument template?

Question

I made a template function which takes a member function as a parameter.

However, since the class has to be declared before it can be used as part of the member function parameter, I have to make it a separate parameter:

template<typename C, void (C::*Method)(void)>
function<void(C*)> methodWrap()
{
}

Which means when explicitly instantiate the template (I want these wrappers to be generated at compile time, not passing in a member pointer as an argument) I have to type it twice when I use it:

function<void(C*)> someFunc = wrapMethod<SomeClass, &SomeClass::someMethod>();

Why can't I just write something like tis:

template<void (C::*Method)(void)>
function<void(C*)> methodWrap()
{
}

and let it capture the type of C and its member function pointer without having to type SomeClass twice?

Or why can't I wrapper it in an outer template that declares C as a "free variable" and then has an inner template argument that performs the deduction

template<typename C>
template<void (C::*Method)(void)>
function<void(C*)> methodWrap()
{
}

Answer 1

If you can live with having a normal function parameter for the member function pointer instead of making it a template parameter, you can do

#include <functional>

template<typename R, typename C, typename... Args>
struct MemberFunctionPointer
{
    typedef R Return;
    typedef C Class;
};

template<typename R, typename C>
constexpr auto inferMemberFunctionPointer(R (C::*method)())
{
    return MemberFunctionPointer<R,C>{};
}

template<typename T> 
constexpr auto methodWrap(T m)
{
    typedef typename decltype(inferMemberFunctionPointer(m))::Class Class;
    typedef typename decltype(inferMemberFunctionPointer(m))::Return Return;

        return std::function<Return (Class*)>();
}

struct B {};

struct A
{
    B f();   
};

void foo()
{
    auto const m = methodWrap( &A::f );
}

std::function is not a constexpr type, so we cannot use methodWrap to initialize a constexpr variable. You can bypass this by creating your own simple constexpr member function wrapper. I've also added a static_assert to get better error messages.

#include <functional>

template<typename R, typename C, typename... Args>
struct MemberFunctionPointer
{
    typedef R Return;
    typedef C Class;
};

template<typename R, typename C>
constexpr auto inferMemberFunctionPointer(R (C::*method)() const)
{
    return MemberFunctionPointer<R,C>{};
}

template<typename R, typename C>
struct MemberFunction
{
    constexpr explicit MemberFunction(R (C::*g)() const): f(g) {}

    constexpr R operator()(C const* obj) const
    {
        return (obj->*f)();
    }

    R (C::*f)() const;
};

template<typename T> 
constexpr auto methodWrap(T m)
{
    static_assert( std::is_member_function_pointer<T>::value, 
                   "Member function pointer expected!");

    typedef typename decltype(inferMemberFunctionPointer(m))::Class Class;
    typedef typename decltype(inferMemberFunctionPointer(m))::Return Return;

        return MemberFunction<Return, Class>{ MemberFunctionPointer<Return,Class>{} };
}

struct B {};

struct A
{
    constexpr B f() const;   
};

void foo()
{
    auto constexpr m = methodWrap( &A::f );
    auto constexpr a = A{};
    m( &a );

    auto b = A{};
    m( &b );
}