How can I check whether a member function has const overload?

Question

Lets say I have

struct foo {
    void ham() {}
    void ham() const {}
};

struct bar {
    void ham() {}
};

Assuming I have a templated fu

Answer 1

SFINAE over and over again. Here is another option which is unspecified about the return types but lets you specify the arguments.

(For comparison: the approach by @Jarod42 checks the exact signature, return type + arguments, the other void_t expression sfinae stuff up to now checks only whether ham() can be called.)

Plus, it works with the current MSVC 2015 Update 1 version (unlike the usual void_t stuff).

template
struct is_callable_impl
{
    template static constexpr auto test(int) -> decltype(std::declval().ham(std::declval() ...), bool{}) { return true; }
    template static constexpr auto test(...) { return false; }
    static constexpr bool value = test(int{});
    using type = std::integral_constant;
};

template
using is_callable = typename is_callable_impl::type;

Use it as

struct foo
{
    void ham() {}
    void ham() const {}
    int ham(int) const {}
};

int main()
{
     std::cout
      <::value                //true
      <::value          //true
      <::value     //true
      <::value  //also true, double is converted to int
      <::value  //false, can't call foo::ham(std::string) const
      <

Demo on Coliru

For the "newest" sfinae stuff, however, I suggest you have a look at boost.hana.