std::enable_if to conditionally compile a member function

Question

I am trying to get a simple example to work to understand how to use std::enable_if. After I read this answer, I thought it shouldn\'t be too hard to come up wi

Answer 1

The boolean needs to depend on the template parameter being deduced. So an easy way to fix is to use a default boolean parameter:

template< class T >
class Y {

    public:
        template < bool EnableBool = true, typename = typename std::enable_if<( std::is_same<T, double>::value && EnableBool )>::type >
        T foo() {
            return 10;
        }

};

However, this won't work if you want to overload the member function. Instead, its best to use TICK_MEMBER_REQUIRES from the Tick library:

template< class T >
class Y {

    public:
        TICK_MEMBER_REQUIRES(std::is_same<T, double>::value)
        T foo() {
            return 10;
        }

        TICK_MEMBER_REQUIRES(!std::is_same<T, double>::value)
        T foo() {
            return 10;
        }

};

You can also implement your own member requires macro like this(just in case you don't want to use another library):

template<long N>
struct requires_enum
{
    enum class type
    {
        none,
        all       
    };
};


#define MEMBER_REQUIRES(...) \
typename requires_enum<__LINE__>::type PrivateRequiresEnum ## __LINE__ = requires_enum<__LINE__>::type::none, \
class=typename std::enable_if<((PrivateRequiresEnum ## __LINE__ == requires_enum<__LINE__>::type::none) && (__VA_ARGS__))>::type