template <class T, class = enable_if_t<!is_reference<T>::value>>
auto f(T&& x); auto foo(int = 0);
template <class T>
auto foo(T&&) {
static_assert(!is_reference<T>::value, "call me with an r-value");
...
}First, "class = enable_if_t<!is_reference<T>::value>" is just "class Dummy = enable_if_t<!is_reference<T>::value>" that doesn't bother to name Dummy. It's providing a default template argument for a function template (this is a C++11 feature; in C++03, only class templates could have this).
Next, "enable_if_t<!is_reference<T>::value>" is a C++14 library feature (enable_if_t is an alias template; the Core Language gained alias templates in C++11, the library gained these in C++14). It is a synonym for "typename enable_if<!is_reference<T>::value>::type", which is more verbose (and what one had to type in C++11).
"is_reference<T>::value" is a C++11 type trait, that is true when T is X& or X&&, and false otherwise. (It's implemented with partial specializations, and a "static const bool value = true;" data member - actually constexpr these days, but no difference there. This is type traits 101.)
enable_if is a library helper for SFINAE. (Not necessarily C++11 Expression SFINAE; here, just C++03 classic SFINAE is being used.) Basically, enable_if<true>::type is void. enable_if<false>::type doesn't exist. This works with the Core Language's SFINAE rules to make function templates vanish from the overload set (if you don't know what SFINAE is, look it up; it's a little complicated, but it exists for good reasons, and it goes back to C++98).
So, this is how one writes a function template that vanishes under certain criteria. This is often desirable compared to doing different things for different types (that can often be done with "tag dispatch") or emitting compiler errors for certain types (that's a static_assert).