C++ enable_if via return type
I found SFINAE or “Substitution Failure Is Not An Error” quite fascinating. At first, it looked kind of cryptic to me (what do all these “typenames” mean?), so I tended to avoid it.
But when used right, std::enable_if (which leverages SFINAE) really helps simplifying the code. So I started to depend on it.
Recently I wrote a function based on the example provided by Cppreference std::void_t article.
Basically, I wanted to reset a variable, which can be either a scalar type (int, float, etc) or a container. If it is a container, I wanted to call Container::clear().
Otherwise, I can simply set it to zero.
template <typename T, typename = void>
struct is_clearable : std::false_type {};
template <typename T>
struct is_clearable<T, std::void_t<decltype(std::declval<T>().clear())> > : std::true_type {};
template <typename T>
inline constexpr bool is_clearable_v = is_clearable<T>::value;
template <typename T>
typename std::enable_if_t<is_clearable_v<T> > reset(T* t) { // #1
t->clear();
}
template <typename T>
typename std::enable_if_t<!is_clearable_v<T> > reset(T* t) { // #2
*t = 0;
}
In my example, std::void_t is used to detect whether T has the member function clear(). is_clearable<T>::value yields true or false based on the result. is_clearable_v<T> is defined as a compile-time boolean constant that takes the is_clearable<T>::value.
Then, the reset(T* t) function is defined separately for the two cases. The first version is enabled (via the return type) when T has the member function clear(); the second version is enabled when T does not.
It turns out to work as advertised for me. But to apply this enable_if idiom, one would have to figure out what any of these (decltype, declval, constexpr, void_t, enable_if) are, and I think that’s not trivial without the help of some good examples.