Proposed resolution (approved by CWG 2024-03-20):

1. Change in 6.6 [basic.link] paragraph 4 as follows:

   ... The name of an entity that belongs to a namespace scope that has not been given internal linkage above and that is the name of

   • a variable; or
   • a function; or
   • a named class (11.1 [class.pre]), or an unnamed class defined in a typedef declaration in which the class has the typedef name for linkage purposes (9.2.4 [dcl.typedef]); or
   • a named enumeration (9.7.1 [dcl.enum]), or an unnamed enumeration defined in a typedef declaration in which the enumeration has the typedef name for linkage purposes (9.2.4 [dcl.typedef]); or
   • an unnamed enumeration that has an enumerator as a name for linkage purposes (9.7.1 [dcl.enum]); or
   • a template

   has its linkage determined as follows:

   • if the entity is a function or function template first declared in a friend declaration and that declaration is a definition and the enclosing class is defined within an export-declaration, the name has the same linkage, if any, as the name of the enclosing class (11.8.4 [class.friend]);
   • otherwise, if the entity is a function or function template declared in a friend declaration and a corresponding non-friend declaration is reachable, the name has the linkage determined from that prior declaration,
   • otherwise, if the enclosing namespace has internal linkage, the name has internal linkage;
   • otherwise, if the declaration of the name is attached to a named module (10.1 [module.unit]) and is not exported (10.2 [module.interface]), the name has module linkage;
   • otherwise, the name has external linkage.

2. Remove 11.8.4 [class.friend] paragraph 4:

   A function first declared in a friend declaration has the linkage of the namespace of which it is a member (6.6 [basic.link]). Otherwise, the function retains its previous linkage (9.2.2 [dcl.stc]).

Proposed resolution (June, 2022) (approved by CWG 2023-01-27) [SUPERSEDED]:

1. Change in 6.6 [basic.link] paragraph 4 as follows:

   ... The name of an entity that belongs to a namespace scope that has not been given internal linkage above and that is the name of

   • a variable; or
   • a function; or
   • a named class (11.1 [class.pre]), or an unnamed class defined in a typedef declaration in which the class has the typedef name for linkage purposes (9.2.4 [dcl.typedef]); or
   • a named enumeration (9.7.1 [dcl.enum]), or an unnamed enumeration defined in a typedef declaration in which the enumeration has the typedef name for linkage purposes (9.2.4 [dcl.typedef]); or
   • an unnamed enumeration that has an enumerator as a name for linkage purposes (9.7.1 [dcl.enum]); or
   • a template

   has its linkage determined as follows:

   • if the entity is a function or function template first declared in a friend declaration and that declaration is a definition, the name has the same linkage, if any, as the name of the enclosing class (11.8.4 [class.friend]);
   • otherwise, if the entity is a function or function template declared in a friend declaration and a corresponding non-friend declaration is reachable, the name has the linkage determined from that prior declaration,
   • otherwise, if the enclosing namespace has internal linkage, the name has internal linkage;
   • otherwise, if the declaration of the name is attached to a named module (10.1 [module.unit]) and is not exported (10.2 [module.interface]), the name has module linkage;
   • otherwise, the name has external linkage.

2. Remove 11.8.4 [class.friend] paragraph 4:

   A function first declared in a friend declaration has the linkage of the namespace of which it is a member (6.6 [basic.link]). Otherwise, the function retains its previous linkage (9.2.2 [dcl.stc]).

EWG electronic poll 2022-06

Consensus for "A friend's linkage should be affected by the presence/absence of export on the containing class definition itself, but ONLY if the friend is a definition (option #2, modified by Jason's suggestion). This resolves CWG2588." See vote.

Additional note (May, 2022):

Forwarded to EWG with paper issue 1253, by decision of the CWG chair.

EWG telecon 2022-06-09

Consensus: "A friend's linkage should be affected by the presence/absence of export on the containing class definition itself, but ONLY if the friend is a definition", pending confirmation by electronic polling.

[Accepted as a DR at the June, 2024 meeting.]

Consider:

  export module Foo;
  class X {
    friend void f(X); // #1 linkage?
  };

Subclause 11.8.4 [class.friend] paragraph 4 gives #1 external linkage:

A function first declared in a friend declaration has the linkage of the namespace of which it is a member (6.6 [basic.link]).

(There is no similar provision for friend classes first declared in a class.)

However, 6.6 [basic.link] bullet 4.8 gives it module linkage:

... otherwise, if the declaration of the name is attached to a named module (10.1 [module.unit]) and is not exported (10.2 [module.interface]), the name has module linkage;

Subclause 10.2 [module.interface] paragraph 2 does not apply:

A declaration is exported if it is declared within an export-declaration and inhabits a namespace scope or it is

• a namespace-definition that contains an exported declaration, or
• a declaration within a header unit (10.3 [module.import]) that introduces at least one name.

Also consider this related example:

  export module Foo;
  export class Y;
  // maybe many lines later, or even a different partition of Foo
  class Y {
    friend void f(Y); // #2 linkage?
  };

• Should the friend's linkage be affected by the linkage of the befriending class? In this example, #2 would therefore have external linkage, as Y is exported.
• Or should the friend's linkage be affected by the presence or absence of export on the class definition itself? In this example, #2 would thus have module linkage.
• Or should the friend's linkage be determined ignoring any enclosing export and ignoring whether the enclosing class is exported, per 11.8.4 [class.friend] paragraph 4 (alone)?
• Or should the friend's linkage be as-if the declaration inhabited its nearest enclosing namespace scope, without the friend?

See issue 2607 for a similar question about enumerators.