Why should you avoid `std::enable_if` in function signatures?

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

std::enable_if is a powerful tool for conditional template metaprogramming, but its misuse in function signatures can lead to various pitfalls. This article explores why you should generally avoid using std::enable_if in function signatures and provides alternative approaches.

Function Parameter Enables

When used as a function parameter, std::enable_if can clutter function signatures with complex typename expressions. This degrades the readability and makes code harder to maintain. For example:

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

   template<typename U = T>
   U read(typename std::enable_if<
          std::is_same<U, double>::value >::type* = 0) { return 3.14; }   
};</code>

Template Parameter Enables

The recommended approach is to place std::enable_if in template parameters:

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

   template<typename U = T, typename std::enable_if<
            std::is_same<U, double>::value, int>::type = 0>
   U read() { return 3.14; }   
};</code>

This approach enhances readability by separating the template parameters from the return/argument types. It also provides universal applicability, as constructors and certain operators cannot have extra arguments or return types.

Return Type Enables

Using std::enable_if as a return type is not part of a typical function signature. However, it can be misleading and should generally be avoided.

Member vs. Non-Member Templates

The concerns mentioned in this article apply equally to both member and non-member function templates.

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