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enable_if Class

Conditionally makes an instance of a type for SFINAE overload resolution. The nested typedef enable_if<Condition,Type>::type exists—and is a synonym for Type—if and only if Condition is true.

template<bool B, class T = void>
    struct enable_if;

B

The value that determines the existence of the resulting type.

T

The type to instantiate if B is true.

If B is true, enable_if<B, T> has a nested typedef named "type" that's a synonym for T.

If B is false, enable_if<B, T> doesn't have a nested typedef named "type".

This alias template is provided:

template< bool B, class T = void >
using enable_if_t = typename enable_if<B,T>::type;

In C++, substitution failure of template parameters is not an error in itself—this is referred to as SFINAE (substitution failure is not an error). Typically, enable_if is used to remove candidates from overload resolution—that is, it culls the overload set—so that one definition can be rejected in favor of another. This conforms to SFINAE behavior. For more information about SFINAE, see Substitution failure is not an error on Wikipedia.

Here are four example scenarios:

  • Scenario 1: Wrapping the return type of a function:

    template <your_stuff> typename enable_if<your_condition, your_return_type>::type
    yourfunction(args) {
        // ...
    }
    
    // The alias template makes it more concise:
    template <your_stuff>
    enable_if_t<your_condition, your_return_type> yourfunction(args) {
        // ...
    }
    
  • Scenario 2: Adding a function parameter that has a default argument:

    template <your_stuff> your_return_type_if_present
    yourfunction(args, enable_if_t<your condition, FOO> = BAR) {
        // ...
    }
    
  • Scenario 3: Adding a template parameter that has a default argument:

    template <your_stuff, typename Dummy = enable_if_t<your_condition>>
    rest_of_function_declaration_goes_here
    
  • Scenario 4: If your function has a non-templated argument, you can wrap its type:

    template <typename T>
    void your_function(const T& t,
        enable_if_t<is_something<T>::value, const string&> s) {
        // ...
    }
    

Scenario 1 doesn't work with constructors and conversion operators because they don't have return types.

Scenario 2 leaves the parameter unnamed. You could say ::type Dummy = BAR, but the name Dummy is irrelevant, and giving it a name is likely to trigger an "unreferenced parameter" warning. You have to choose a FOO function parameter type and BAR default argument. You could say int and 0, but then users of your code could accidentally pass to the function an extra integer that would be ignored. Instead, we recommend that you use void ** and either 0 or nullptr because almost nothing is convertible to void **:

template <your_stuff> your_return_type_if_present 
yourfunction(args, typename enable_if<your_condition, void **>::type = nullptr) {
    // ...
}

Scenario 2 also works for ordinary constructors. However, it doesn't work for conversion operators because they can't take extra parameters. It also doesn't work for variadic constructors because adding extra parameters makes the function parameter pack a non-deduced context and thereby defeats the purpose of enable_if.

Scenario 3 uses the name Dummy, but it's optional. Just "typename = typename" would work, but if you think that looks weird, you can use a "dummy" name—just don't use one that might also be used in the function definition. If you don't give a type to enable_if, it defaults to void, and that's perfectly reasonable because you don't care what Dummy is. This works for everything, including conversion operators and variadic constructors.

Scenario 4 works in constructors that don't have return types, and thereby solves the wrapping limitation of Scenario 1. However, Scenario 4 is limited to non-templated function arguments, which aren't always available. (Using Scenario 4 on a templated function argument prevents template argument deduction from working on it.)

enable_if is powerful, but also dangerous if it's misused. Because its purpose is to make candidates vanish before overload resolution, when it's misused, its effects can be very confusing. Here are some recommendations:

  • Do not use enable_if to select between implementations at compile-time. Don't ever write one enable_if for CONDITION and another for !CONDITION. Instead, use a tag dispatch pattern—for example, an algorithm that selects implementations depending on the strengths of the iterators they're given.

  • Do not use enable_if to enforce requirements. If you want to validate template parameters, and if the validation fails, cause an error instead of selecting another implementation, use static_assert.

  • Use enable_if when you have an overload set that makes otherwise good code ambiguous. Most often, this occurs in implicitly converting constructors.

This example explains how the STL template function std::make_pair() takes advantage of enable_if.

void func(const pair<int, int>&);
void func(const pair<string, string>&);

func(make_pair("foo", "bar"));

In this example, make_pair("foo", "bar") returns pair<const char *, const char *>. Overload resolution has to determine which func() you want. pair<A, B> has an implicitly converting constructor from pair<X, Y>. This isn't new—it was in C++98. However, in C++98/03, the implicitly converting constructor's signature always exists, even if it's pair<int, int>(const pair<const char *, const char *>&). Overload resolution doesn't care that an attempt to instantiate that constructor explodes horribly because const char * isn't implicitly convertible to int; it's only looking at signatures, before function definitions are instantiated. Therefore, the example code is ambiguous, because signatures exist to convert pair<const char *, const char *> to both pair<int, int> and pair<string, string>.

C++11 solved this ambiguity by using enable_if to make sure pair<A, B>(const pair<X, Y>&) exists only when const X& is implicitly convertible to A and const Y& is implicitly convertible to B. This allows overload resolution to determine that pair<const char *, const char *> is not convertible to pair<int, int> and that the overload that takes pair<string, string> is viable.

Header: <type_traits>

Namespace: std

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