cannot access namespace scope friend explicitly

问题

I had an issue today where ADL wasn't finding a static member function for a type defined inside a class.

That is, in the below example, str(foo::Foo::Enum) isn't located via ADL without explicitly scoping it, foo::Foo::str(foo::Foo::Enum)

namespace foo {

struct Foo
{
    enum Enum
    {
        FOO1,
        FOO2
    };

    static const char* str(Enum e);
};

}

foo::Foo::Enum e = foo::Foo::FOO1;
const char* s = str(e);              // ADL doesn't work

I found this SO question, and as stated in the accepted answer, changing it to a friend function results in ADL now working.

namespace foo {

struct Foo
{
    enum Enum
    {
        FOO1,
        FOO2
    };

    friend const char* str(Enum e);  // note str is now a friend
};

}

foo::Foo::Enum e = foo::Foo::FOO1;
const char* s = str(e);              // ADL works now

Whilst this now helps ADL, I was surprised to find that I couldn't access str by scoping it with a namespace foo

foo::Foo::Enum e = foo::Foo::FOO1;
const char* s = foo::str(e);         // error: ‘str’ is not a member of ‘foo’

I ran a test, where I printed out the result of __PRETTY_FUNCTION__, and was even more surprised to see that the scope of str is apparently foo:::

__PRETTY_FUNCTION__: const char* foo::str(foo::Foo::Enum)

Working example below:

#include <iostream>

namespace foo {

struct Foo
{
    enum Enum
    {
        FOO1,
        FOO2
    };

    friend const char* str(Enum e)
    {
        return __PRETTY_FUNCTION__;
    }
};

}

int main()
{
    foo::Foo::Enum e = foo::Foo::FOO1;

    std::cout << str(e) << '\n';
    // std::cout << foo::str(e) << '\n'; // error: ‘str’ is not a member of ‘foo’

    return 0;
}

Output:

$ ./a.out
const char* foo::str(foo::Foo::Enum)

Question:

Why am I unable to locate str(..) explicitly scoping it with the enclosing namespace?
Why does __PRETTY_FUNCTION__ say it's in foo::, and yet I am unable to locate it as such?

回答1:

Why am I unable to locate str(..) explicitly scoping it with the enclosing namespace?

From the standard, [namespace.memdef]/3

If a friend declaration in a non-local class first declares a class, function, class template or function template the friend is a member of the innermost enclosing namespace. The friend declaration does not by itself make the name visible to unqualified lookup or qualified lookup. [ Note: The name of the friend will be visible in its namespace if a matching declaration is provided at namespace scope (either before or after the class definition granting friendship). — end note ]

That means str is not visible to name lookup; it can only be called via ADL.

Why does __PRETTY_FUNCTION__ say it's in foo::, and yet I am unable to locate it as such?

From [class.friend]/6,

A function can be defined in a friend declaration of a class if and only if the class is a non-local class ([class.local]), the function name is unqualified, and the function has namespace scope.

str does become member of namespace foo; it's just invisible.

Explanations from cppreference.com:

Names introduced by friend declarations within a non-local class X become members of the innermost enclosing namespace of X, but they do not become visible to lookup (neither unqualified nor qualified) unless a matching declaration is provided at namespace scope, either before or after the class definition. Such name may be found through ADL which considers both namespaces and classes.

来源：https://stackoverflow.com/questions/47889416/cannot-access-namespace-scope-friend-explicitly

标签

c++

language-lawyer

friend

argument-dependent-lookup

name-lookup