The scope of the built-in friend function?

Question

A few days ago I asked a question about the scope of functions defined in a class ( What scope does a function defined in a class belong to? ), And I find out that the function is in the scope of the encompassing namespace, but will not be searchable until it explicitly declared out of class (ADL exception).

Today I found some relevant statements in the C ++ standard (section 11.3):

A function can be defined in the declaration of a friend of the class if and only if the class is a non-local class (9.8), the name of the function is unqualified, and the function has a namespace scope. [Example:

class M { friend void f() { } // definition of global f, a friend of M, // not the definition of a member function }; —end example ] 

Such a function is implicitly inline . The friend function defined in the class is in the (lexical) scope of the class in which it is defined. A friend function defined outside a class is not (3.4.1).

We see that there are two operators associated with the scope: “has a namespace scope and “ in the (lexical) scope of the class in which it is defined . ” I am confused here. If the first one relates to my previous question ( Which scope is a function defined in a class belonging to? ), then what does the latter mean?

Answer 1

A “namespace function” is a function that is a member of a namespace (ie, “scope” here means “scope” of a function).

A later statement refers to 3.4.1, which should say

The search for the name of the name used in the definition of a friend's function (11.3), defined in a string in the class that provides friendships, should be performed as described for searching in the definitions of member functions.