Should You Avoid `std::enable_if` in Function Signatures?

Procedural Impairment: Avoiding std::enable_if in Function Signatures

Scott Meyers' upcoming book, EC 11, advises avoiding std::enable_if in function signatures. Despite its versatility in conditionally removing functions or classes from overload resolution, certain limitations and readability issues warrant reconsideration of its usage.

Methods of Employing std::enable_if in Function Signatures

Function Parameter:

<code class="cpp">template<typename t>
struct Check1
{
   template<typename u="T">
   U read(typename std::enable_if::value >::type* = 0) { return 42; }
};</typename></typename></code>

Template Parameter:

<code class="cpp">template<typename t>
struct Check2
{
   template<typename u="T," typename std::enable_if option as template parameter std::is_same int>::value, int >::type = 0>
   U read() { return 42; }
};</typename></typename></code>

Return Type:

<code class="cpp">template<typename t>
struct Check3
{
   template<typename u="T">
   typename std::enable_if<:is_same int>::value, U>::type read() {  // Option 3: As return type
      return 42;
   }
};</:is_same></typename></typename></code>

Recommended Approach: Placement in Template Parameters

The optimal solution involves placing the enable_if in the template parameters. This approach offers several advantages:

Readability Enhancement:
The enable_if usage is separated from return/argument types, improving clarity and reducing clutter.

Universal Applicability:
Unlike the other options, the template parameter placement is applicable to constructors and operators without additional arguments.

Return Type Exclusion:
While "Avoid std::enable_if in function signatures" primarily addresses its use in normal function signatures, it does apply to return types in template specializations. The concern stems from readability issues when merging enable_if with return types.

Member vs. Non-Member Functions:
The principle holds true for both member and non-member function templates.

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