How does `std::enable_if` help conditionally define function return types and enable function resolution?

How std::enable_if Facilitates Conditional Function Resolution

Understanding Substitution Failure Is Not An Error is crucial for comprehending 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>

The key is that the type is only defined when the condition is true.

Consider the following function:

<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>

std::enable_if is employed to conditionally define the function's return type, causing a compilation error if the condition is false.

In the code snippet:

<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 default value (0) for the second template parameter is provided solely to enable the call foo(1); without it, the function would require two template parameters instead of one.

Note: In C 14, enable_if_t is a defined type that should be used for clarity. Thus, the return type can be condensed to std::enable_if_t::is_integer>. In older versions of Visual Studio, default template parameters are not supported, so std::enable_if can only be used on the function return, as seen in the "std::numeric_limits as a Condition" example.

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