tmp/tmp1ldf632c/{from.md → to.md}
RENAMED
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#### Namespace member definitions <a id="namespace.memdef">[[namespace.memdef]]</a>
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A declaration in a namespace `N` (excluding declarations in nested
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scopes) whose *declarator-id* is an *unqualified-id* [[dcl.meaning]],
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whose *class-head-name* [[class.pre]] or *enum-head-name* [[dcl.enum]]
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is an *identifier*, or whose *elaborated-type-specifier* is of the form
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*class-key* *attribute-specifier-seq*ₒₚₜ *identifier*
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[[dcl.type.elab]], or that is an *opaque-enum-declaration*, declares (or
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redeclares) its *unqualified-id* or *identifier* as a member of `N`.
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[*Note 1*: An explicit instantiation [[temp.explicit]] or explicit
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specialization [[temp.expl.spec]] of a template does not introduce a
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name and thus may be declared using an *unqualified-id* in a member of
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the enclosing namespace set, if the primary template is declared in an
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inline namespace. — *end note*]
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[*Example 1*:
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``` cpp
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namespace X {
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void f() { ... } // OK: introduces X::f()
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namespace M {
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void g(); // OK: introduces X::M::g()
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}
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using M::g;
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void g(); // error: conflicts with X::M::g()
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}
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```
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— *end example*]
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Members of a named namespace can also be defined outside that namespace
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by explicit qualification [[namespace.qual]] of the name being defined,
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provided that the entity being defined was already declared in the
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namespace and the definition appears after the point of declaration in a
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namespace that encloses the declaration’s namespace.
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[*Example 2*:
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``` cpp
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namespace Q {
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namespace V {
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void f();
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}
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void V::f() { ... } // OK
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void V::g() { ... } // error: g() is not yet a member of V
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namespace V {
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void g();
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}
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}
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namespace R {
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void Q::V::g() { ... } // error: R doesn't enclose Q
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}
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```
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— *end example*]
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If a friend declaration in a non-local class first declares a class,
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function, class template or function template[^10] the friend is a
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member of the innermost enclosing namespace. The friend declaration does
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not by itself make the name visible to unqualified lookup
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[[basic.lookup.unqual]] or qualified lookup [[basic.lookup.qual]].
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[*Note 2*: The name of the friend will be visible in its namespace if a
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matching declaration is provided at namespace scope (either before or
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after the class definition granting friendship). — *end note*]
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If a friend function or function template is called, its name may be
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found by the name lookup that considers functions from namespaces and
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classes associated with the types of the function arguments
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[[basic.lookup.argdep]]. If the name in a friend declaration is neither
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qualified nor a *template-id* and the declaration is a function or an
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*elaborated-type-specifier*, the lookup to determine whether the entity
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has been previously declared shall not consider any scopes outside the
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innermost enclosing namespace.
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[*Note 3*: The other forms of friend declarations cannot declare a new
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member of the innermost enclosing namespace and thus follow the usual
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lookup rules. — *end note*]
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[*Example 3*:
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``` cpp
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// Assume f and g have not yet been declared.
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void h(int);
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template <class T> void f2(T);
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namespace A {
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class X {
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friend void f(X); // A::f(X) is a friend
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class Y {
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friend void g(); // A::g is a friend
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friend void h(int); // A::h is a friend
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// ::h not considered
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friend void f2<>(int); // ::f2<>(int) is a friend
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};
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};
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// A::f, A::g and A::h are not visible here
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X x;
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void g() { f(x); } // definition of A::g
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void f(X) { ... } // definition of A::f
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void h(int) { ... } // definition of A::h
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// A::f, A::g and A::h are visible here and known to be friends
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}
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using A::x;
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void h() {
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A::f(x);
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A::X::f(x); // error: f is not a member of A::X
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A::X::Y::g(); // error: g is not a member of A::X::Y
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}
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```
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— *end example*]
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