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< // Option 1: As function parameter
std::is_same<U, int>::value >::type* = 0) { return 42; }
};</code>
Template Parameter:
<code class="cpp">template<typename T>
struct Check2
{
template<typename U = T, typename std::enable_if< // Option 2: As template parameter
std::is_same<U, int>::value, int >::type = 0>
U read() { return 42; }
};</code>
Return Type:
<code class="cpp">template<typename T>
struct Check3
{
template<typename U = T>
typename std::enable_if<std::is_same<U, int>::value, U>::type read() { // Option 3: As return type
return 42;
}
};</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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