How Does `std::enable_if` Work: Unraveling the Mysteries of Its Implementation and Usage?

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:

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

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

Unveiling the Usage

Consider the following declaration:

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

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:

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

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.

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