How Does std::enable_if Work with Conditional Return Types and Template Parameters?

Understanding std::enable_if

Understanding std::enable_if requires a grasp of Substitution Failure Is Not An Error.

Definition of std::enable_if

std::enable_if is a specialized template defined as:

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

Crucially, the typedef T type definition is only triggered when bool Cond is true.

Usage in Conditional Return Types

Consider the example:

<code class="cpp">template<typename T>
typename std::enable_if<std::numeric_limits<T>::is_integer, void>::type foo(const T &bar) { isInt(bar); }</code>

Here, the return type is defined by:

<code class="cpp">std::enable_if<std::numeric_limits<T>::is_integer, void>::type</code>

The usage of enable_if ensures that foo has a valid return type only if is_integer is true for T.

Defaulting the Second Template Parameter

In the example:

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

The second template parameter is defaulted to 0. This allows foo to be called with a single template parameter, e.g. foo(1);.

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