Why Should You Avoid Using `std::enable_if` in Function Return Types?

Why Strive to Avoid std::enable_if in Function Signatures?

std::enable_if, a versatile tool in C , can be utilized in multiple ways to conditionally define functions or classes based on specified criteria. However, Scott Meyers has cautioned against using it in function signatures. Understanding this recommendation requires an exploration of the different strategies available with std::enable_if and their implications.

Implementation Options:

1. Function Parameter:

<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>

2. Template Parameter:

<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>

3. 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() {
        return 42;
    }

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

Preferred Approach:

The most favorable approach is to place the enable_if in the template parameters. This technique offers advantages in both clarity and versatility.

Clarity:

The enable_if clause can be kept distinct from the return and argument types, resulting in more readable code. Even with the use of alias templates to reduce clutter, the merged clauses in the other approaches still combine separate and distinct concepts.

Versatility:

This technique can be universally applied to various contexts, including constructors without return types and operators that prohibit additional arguments. SFINAE, the key behind std::enable_if's conditional behavior, is applicable only to templates, further supporting this approach's universality.

Avoidance of std::enable_if in Return Types:

When using std::enable_if in return types, the concern is not with the function signature but rather with template specializations. Meyers advises against this practice for clarity and consistency reasons. Specifying return types with enable_if introduces inconsistency between template definitions and the base template:

<code class="cpp">template<typename T>
struct Check4
{
    typename std::enable_if<std::is_same<T, int>::value, int>::type read() {
        return 42;
    }

    int read() {    // error: redeclared without 'typename'
        return 3.14;  // never used
    }   
};</code>

Member vs. Non-Member Function Templates:

The concerns and recommendations discussed apply to both member and non-member function templates. There are no notable differences in approach.

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