Determining number of arguments of C++ template functor - c++

Question

Suppose we have:
template<typename F, typename T1, typename T2>
void my_magic_method(F func, T1 t1, T2 t2)
{
if (???)
func(t1);
else
func(t1,t2);
}
What can help me to determine:
Number of arguments
Maybe types of each argument
Type of return value
I cant use variadic templates because of MSVS 2010...
UPDATE
My first solution:
template<typename F>
auto my_magic_func(F f) -> decltype(f(1))
{
return f(1);
}
template<typename F>
auto my_magic_func(F f, void * fake = NULL) -> decltype(f(2,3))
{
return f(2,3);
}
int main()
{
auto x1 = my_magic_func([](int a){ return a+100; });
auto x2 = my_magic_func([](int a, int b){ return a*b; });
// x1 == 1+100
// x2 == 2*3
}
Thats my way of function-type overloading.
It works, but maybe a better solutuion?

Answer 1

Not exactly what you asked for, but if I understand your intent correctly, in VC++ 2010 it's possible (but ugly) via simple overloading based on arity:
#include <utility>
#include <string>
#include <iostream>
template<typename F, typename T1>
auto my_magic_method(F&& func, T1&& t1) ->
decltype(std::forward<F>(func)(std::forward<T1>(t1)))
{
return std::forward<F>(func)(std::forward<T1>(t1));
}
template<typename F, typename T1, typename T2>
auto my_magic_method(F&& func, T1&& t1, T2&& t2) ->
decltype(std::forward<F>(func)(std::forward<T1>(t1), std::forward<T2>(t2)))
{
return std::forward<F>(func)(std::forward<T1>(t1), std::forward<T2>(t2));
}
struct string_to_float_functor
{
float operator ()(std::string const& s) const
{
return std::stof(s);
}
};
int main()
{
auto a = my_magic_method([](std::string const& x) { return x + x; }, "foo");
auto b = my_magic_method([](double x, int y) { return x * y; }, 21.5, 3);
auto c = my_magic_method(string_to_float_functor(), "3.14159265");
std::cout << a << '\n' << b << '\n' << c << '\n';
}
This supports unary and binary functors – continue the pattern and add overloads for other arities as needed.

Answer 2

Here's a few approaches; all assume C++11. Tested on clang++ 3.2 with -std=c++11.
//Taking a function pointer argument (template types inferred)
template <typename ret, typename ... Args>
constexpr int arg_count(ret (*f)(Args...)) {
return sizeof...(Args);
}
//Taking a function type (or related) directly
template <typename T>
struct ArgCount {
static const int value = 0;
};
template <typename Ret, typename ... Args>
struct ArgCount<Ret(Args...)> {
static const int value = sizeof...(Args);
};
template <typename Ret, typename ... Args>
struct ArgCount<Ret(*)(Args...)> {
static const int value = sizeof...(Args);
};
template <typename Ret, typename ... Args>
struct ArgCount<Ret(&)(Args...)> {
static const int value = sizeof...(Args);
};
//Using the latter for dispatch
template <int N>
struct helper {
template<typename F, typename T1, typename T2>
static void call(F func, T1 t1, T2 t2);
};
template <>
struct helper<1> {
template<typename F, typename T1, typename T2>
static void call(F func, T1 t1, T2 t2) {
func(t1);
}
};
template <>
struct helper<2> {
template<typename F, typename T1, typename T2>
static void call(F func, T1 t1, T2 t2) {
func(t1, t2);
}
};
template<typename F, typename T1, typename T2>
void my_magic_method(F func, T1 t1, T2 t2)
{
helper<ArgCount<F>::value>::call(func, t1, t2);
}
//Testing
#include <cstdio>
void a(int a, int b) { printf("%i\n", a + b); }
void b(int x) { printf("%i\n", x); }
int main() {
printf("%i %i\n", arg_count(a), arg_count(b));
printf("%i %i\n", ArgCount<decltype(a)>::value, ArgCount<decltype(b)>::value);
my_magic_method(a, 1, 2);
my_magic_method(b, 1, 2);
}