Why is there a Seemingly Meaningless Assignment to std::enable_if in C Templates?

Understanding the Inner Workings of std::enable_if

In a recent question, the usage of std::enable_if as a conditional return type was discussed. While the first usage was clear, the second, which included a seemingly meaningless assignment to std::enable_if, remained puzzling.

Unlocking the Concept

To unravel the mystery, we must delve into the definition of std::enable_if:

template<bool Cond, class T = void> struct enable_if {};
template<class T> struct enable_if<true, T> { typedef T type; };

The key lies in the fact that typedef T type is only defined when bool Cond is true.

Applying to the Example

With this understanding in hand, let's revisit the code:

template<typename T>
typename std::enable_if<std::numeric_limits<T>::is_integer, void>::type foo(const T &bar) { isInt(bar); }

Here, std::enable_if is used to define the return type of the foo function. If T is an integer, the return type will be void; otherwise, the function will not compile.

The Role of Defaulting

In the second example:

template<typename T, typename std::enable_if<std::is_integral<T>::value, int>::type = 0>
void foo(const T& bar) { isInt(); }

The = 0 default parameter ensures that both options can be called with foo(1);. Without defaulting, calling foo would require two template parameters instead of just the int.

Evolving the Understanding

In C 14, the std::enable_if_t type is introduced, which should be used in place of the typedef form. This results in a more concise return type:

std::enable_if_t<std::numeric_limits<T>::is_integer>

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