Wrap a function pointer in C++ with variadic template

Question

The Question

I have a number of C++ functions void f(), R g(T a), S h(U a, V b) and so on. I want to write a template functi

Answer 1

There's a duplicate somewhere here, I remember it, but I can't find it... The conclusion of which being that it was impossible to pass the pointer's type and its value at the same time.

Some hope lies in a suggestion for implicit type template parameters, which you can find here.

Answer 2

This is the best I've been able to do so far:

template<typename R, typename...A>
struct S<R(A...)>
{
    typedef R(*F)(A...);
    F f;
    constexpr S(F _f) : f(_f) { }
    inline R operator()(A... a)
    { return f(a...); }
};

#define wrapper(FUNC, ARGS...) (S<decltype(FUNC)>(FUNC))(ARGS)

int f(float g);

int main(void)
{
    return wrapper(f, 3.0f);
}

Sadly I can't make it compile under MSVC.

Answer 3

template<typename Fn, Fn fn, typename... Args>
typename std::result_of<Fn(Args...)>::type
wrapper(Args&&... args) {
    return fn(std::forward<Args>(args)...);
}
#define WRAPPER(FUNC) wrapper<decltype(&FUNC), &FUNC>

//Usage:

int min(int a, int b){
    return (a<b)?a:b;
}

#include<iostream>
#include<cstdlib>
int main(){
    std::cout<<WRAPPER(min)(10, 20)<<'\n';
    std::cout<<WRAPPER(rand)()<<'\n';
}

Alternatively, to get maybe quite less readable, but shorter syntax:

#define WRAPPER(FUNC, ...) wrapper<decltype(&FUNC), &FUNC>(__VA_ARGS__)

//Usage:

int main(){
    sdt::cout<<WRAPPER(min, 10, 20)<<'\n';
    std::cout<<WRAPPER(rand)<<'\n';
}