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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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