Understanding enable_if implementation in C++98

Question

I have seen this given as a self-explanatory implementation of enable_if for C++98 :

template<bool b, typename T = void>
struct enable_if {
    typedef T type;
};

template<typename T>
struct enable_if<false, T> {};

But alas I personally don't understand it. I don't see where the boolean kicks into play. Would really appreciate if someone would unwrap it for me.

Answer 1

First consider this:

template<bool b>
struct foo {
    static const bool B = b;
};

template <>
struct foo<false> {
    static const bool B = false;
};

Its a primary template and a specialization. In the general case foo<b>::B is just b. In the special case when b == false the specialization kicks in and foo<false>::B is false.

Your example of std::enable_if is different for two reasons: A) It is using partial specialization. The specialization is for any type T and b == false;. B) in the specialization there is no type member alias. And thats the whole purpose of std::enable_if. When the condition is false then std::enable_if< condition, T>::type is a substitution failure, because the specialization has no type. When the condition is true then std::enable_if<condition,T>::type is just T.