C++ 朋友 function 与模板参数 enable_if

Question

I'm struggling with a friend function for a struct that has a template argument with enable_if :

// foo.h
#ifndef FOO_H
#define FOO_H
#include <type_traits>

template<
    typename T,
    typename = typename std::enable_if<std::is_arithmetic<T>::value>::type
>
struct foo {
    foo(T bar) : bar(bar) {}

    T get() { return bar; }

    friend foo operator+(const foo& lhs, const foo& rhs);
    // Defining inside a body works:
    // {
    //     return foo(lhs.bar + rhs.bar);
    // }

private:
    T bar;
};

// None of these work:
// tempate<typename T, typename>
// tempate<typename T>
// tempate<typename T, typename = void>
template<
    typename T,
    typename = typename std::enable_if<std::is_arithmetic<T>::value>::type
>
foo<T> operator+(const foo<T>& lhs, const foo<T>& rhs)
{
    return foo<T>(lhs.bar + rhs.bar);
}
#endif /* ifndef FOO_H */

and

// main.cpp
#include <iostream>
#include "foo.h"

int main()
{
    foo<int> f{1};
    foo<int> g{2};
    std::cout << (f + g).get() << '\n';
    return 0;
}

If I try to compile, get the following linker error:

Undefined symbols for architecture x86_64:
  "operator+(foo<int, void> const&, foo<int, void> const&)", referenced from:
      _main in main-5fd87c.o
ld: symbol(s) not found for architecture x86_64
clang: error: linker command failed with exit code 1 (use -v to see invocation)

(with Apple clang version 11.0.3 (clang-1103.0.32.59) .)

I want the operator + to only work with types with the same template arguments, eg, foo only with foo, but not with foo or foo.

I think this is closely related to this question , but I'm having a hard time trying to figure out how to solve my problem. I tried many template definitions like tempate<typename T, typename> , tempate<typename T> , tempate<typename T, typename = typename std::enable_if...> but none of these work.

As commented in the code, defining inside a body works, but I want to learn how to work with template friend functions with type traits. Any help would be greatly appreciated!

Answer 1

The friend declaration refers to a non-template operator, while the definition out of the class definition refers to a template one, they don't match.

You might want

// forward declaration of the class template
template<
    typename T,
    typename X = typename std::enable_if<std::is_arithmetic<T>::value>::type
>
struct foo;

// declaration of the operator template
template<
    typename T,
    typename = typename std::enable_if<std::is_arithmetic<T>::value>::type
>
foo<T> operator+(const foo<T>& lhs, const foo<T>& rhs);

// definition of the class template
template<
    typename T,
    typename
>
struct foo {
    foo(T bar) : bar(bar) {}

    T get() { return bar; }

    friend foo operator+<T>(const foo& lhs, const foo& rhs);
    // or left the template parameters to be deduced as
    friend foo operator+<>(const foo& lhs, const foo& rhs);

private:
    T bar;
};

//definition of the operator template
template<
    typename T,
    typename
>
foo<T> operator+(const foo<T>& lhs, const foo<T>& rhs)
{
    return foo<T>(lhs.bar + rhs.bar);
}

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