template specialization using enable_if in function definition

February 13, 2023

if I have the following function

struct Struct { template<T> void Foo(); }

How can I use enable_if in the function definition without repeating the declaration above?

template<T> typename enable_if<is_class<T>,void>::type Struct::Foo() { ... } // error: Struct has no member `Foo<T>`
template<T> typename enable_if<!is_class<T>,void>::type Struct::Foo() { ... } // error: Struct has no member `Foo<T>`

enable_if<is_class<T>,void> is just an example but is there a way to not repeat the declaration with multiple enable_if definitions?

it seems I’m forced to do this

struct Struct
{ 
   template<T> typename enable_if<is_class<T>,void>::type Foo();
   template<T> typename enable_if<!is_class<T>,void>::type Foo();
}

>Solution :

The easiest way to not have to repeat yourself too much is to define them inline:

#include <type_traits>

struct Struct {
    template<class T>
    std::enable_if_t<std::is_class_v<T>> Foo() { /* ... */ }
    
    template<class T>
    std::enable_if_t<!std::is_class_v<T>> Foo() { /* ... */ }    
};

or with constexpr-if:

struct Struct {
    template<class T>
    void Foo() {
        if constexpr (std::is_class_v<T>) {
            /* ... */
        } else {
            /* ... */
        }
    }
};

which makes separating the declaration and definition slightly less verbose:

struct Struct {
    template<class T>
    void Foo();
};

template<class T>
void Struct::Foo() {
    if constexpr (std::is_class_v<T>) {
    } else {
    }
}