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

Another option is to simulate void_t(to appear officially in C++17), which makes use of expression SFINAE to make sure your function is call-able on a const instance, regardless of its return type.

#include 
#include 

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

struct Bar {
    void ham() {}
};

template
using void_t = void;

template
struct has_const_ham: std::false_type{};

template // specialization, instantiated when there is ham() const
struct has_const_ham().ham())>> : 
    std::true_type{};

int main()
{
    std::cout << std::boolalpha;
    std::cout << has_const_ham::value << std::endl;
    std::cout << has_const_ham::value << std::endl;
}

EDIT

If you want to enforce the return type, then derive the specialization from std::is_same, like

template // specialization, instantiated when there is ham() const
struct has_const_ham().ham())>> : 
    std::is_same().ham()), void> // return must be void
{}; 

Live on Coliru