How to call a function object differently, depending on its arity (or other information known at compile time)?

问题

In a function template, I'd like to call a function, or function object differently, depending on its arity (how many arguments it takes). In pseudocode:

if arity(f) == 1:
    f(x)
if arity(f) == 2:
    f(x, y)
if arity(f) == 3:
    f(x, y, z)

How can this be done in C++?

Edit To clarify the difficulty: f(x, y, z) won't compile if f only takes 2 arguments, and vice versa, f(x, y) won't compile when f needs 3 arguments.

回答1:

With C++11:

#include <iostream>

template <typename F> struct Traits;

template <typename R, typename... A>
struct Traits<R (A...)>
{
    static constexpr unsigned Arity = sizeof...(A);
};

void f(int, int, int);

int main() {
    std::cout
        << Traits<void()>::Arity
        << Traits<void(int)>::Arity
        << Traits<void(int, int)>::Arity
        << Traits<decltype(f)>::Arity
        << '\n';
    return 0;
}

Otherwise you might lookup boost::function: http://www.boost.org/doc/libs/1_55_0b1/doc/html/function.html

来源：https://stackoverflow.com/questions/21349158/how-to-call-a-function-object-differently-depending-on-its-arity-or-other-info