Understanding std::enable_if: Deciphering Its Purpose and Implementation

While the nature of std::enable_if is grasped in certain contexts, its intricacies, particularly the second argument and the assignment to std::enable_if within the template statement, remain enigmatic. Delving deeper into its workings will unravel these mysteries.

The Essentials of std::enable_if

std::enable_if is a specialized template defined as follows:

template struct enable_if {};
template struct enable_if { typedef T type; };

Crucially, the type alias typedef T type is only defined when Cond is true.

Unveiling the Usage

Consider the following declaration:

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

Here, the return type of foo is defined by std::enable_if<:numeric_limits>::is_integer, void>::type. Since std::numeric_limits::is_integer is a boolean condition, this return type will only be defined if the condition is true.

Clarifying the Second Argument

In the notation:

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

The = 0 is utilized to default the second template parameter. This allows both options to be invoked using foo(1), as opposed to requiring two template parameters if the std::enable_if parameter were not defaulted.

Noteworthy Details

• Explicitly typing out typename std::enable_if<:condition t>::type enhances clarity.
• In C 14, enable_if_t is an established type that should be employed for the return type, simplifying it to std::enable_if_t.
• For Visual Studio versions prior to 2013, only the return type can employ enable_if.