13 Derived classes [class.derived]

13.2 Member name lookup [class.member.lookup]

Member name lookup determines the meaning of a name (id-expression) in a class scope ([basic.scope.class]).

Name lookup can result in an ambiguity, in which case the program is ill-formed.

For an id-expression, name lookup begins in the class scope of this; for a qualified-id, name lookup begins in the scope of the nested-name-specifier.

Name lookup takes place before access control ([basic.lookup], Clause [class.access]).

The following steps define the result of name lookup for a member name f in a class scope C.

The lookup set for f in C, called

S (f, C)

, consists of two component sets: the declaration set, a set of members named f; and the subobject set, a set of subobjects where declarations of these members (possibly including using-declarations) were found.

In the declaration set, using-declarations are replaced by the set of designated members that are not hidden or overridden by members of the derived class ([namespace.udecl]), and type declarations (including injected-class-names) are replaced by the types they designate.

S (f, C)

is calculated as follows:

If C contains a declaration of the name f, the declaration set contains every declaration of f declared in C that satisfies the requirements of the language construct in which the lookup occurs.

[ Note

Looking up a name in an elaborated-type-specifier ([basic.lookup.elab]) or base-specifier (Clause [class.derived]), for instance, ignores all non-type declarations, while looking up a name in a nested-name-specifier ([basic.lookup.qual]) ignores function, variable, and enumerator declarations.

As another example, looking up a name in a using-declaration ([namespace.udecl]) includes the declaration of a class or enumeration that would ordinarily be hidden by another declaration of that name in the same scope.

— end note

]

If the resulting declaration set is not empty, the subobject set contains C itself, and calculation is complete.

Otherwise (i.e., C does not contain a declaration of f or the resulting declaration set is empty),

S (f, C)

is initially empty.

If C has base classes, calculate the lookup set for f in each direct base class subobject

B_{i}

, and merge each such lookup set

S (f, B_{i})

in turn into

S (f, C)

The following steps define the result of merging lookup set

S (f, B_{i})

into the intermediate

S (f, C)

(6.1)
If each of the subobject members of $S (f, B_{i})$ is a base class subobject of at least one of the subobject members of $S (f, C)$ , or if $S (f, B_{i})$ is empty, $S (f, C)$ is unchanged and the merge is complete.

Conversely, if each of the subobject members of $S (f, C)$ is a base class subobject of at least one of the subobject members of $S (f, B_{i})$ , or if $S (f, C)$ is empty, the new $S (f, C)$ is a copy of $S (f, B_{i})$ .
(6.2)
Otherwise, if the declaration sets of $S (f, B_{i})$ and $S (f, C)$ differ, the merge is ambiguous: the new $S (f, C)$ is a lookup set with an invalid declaration set and the union of the subobject sets.

In subsequent merges, an invalid declaration set is considered different from any other.
(6.3)
Otherwise, the new $S (f, C)$ is a lookup set with the shared set of declarations and the union of the subobject sets.

The result of name lookup for f in C is the declaration set of

S (f, C)

If it is an invalid set, the program is ill-formed.

[ Example

struct A { int x; };                    // S(x,A) = { { A::x }, { A } }
struct B { float x; };                  // S(x,B) = { { B::x }, { B } }
struct C: public A, public B { };       // S(x,C) = { invalid, { A in C, B in C } }
struct D: public virtual C { };         // S(x,D) = S(x,C)
struct E: public virtual C { char x; }; // S(x,E) = { { E::x }, { E } }
struct F: public D, public E { };       // S(x,F) = S(x,E)
int main() {
  F f;
  f.x = 0;                              // OK, lookup finds E::x
}

S (x, F)

is unambiguous because the A and B base class subobjects of D are also base class subobjects of E, so

S (x, D)

is discarded in the first merge step.

— end example

]

If the name of an overloaded function is unambiguously found, overload resolution ([over.match]) also takes place before access control.

Ambiguities can often be resolved by qualifying a name with its class name.

[ Example

struct A {
  int f();
};

struct B {
  int f();
};

struct C : A, B {
  int f() { return A::f() + B::f(); }
};

— end example

]

[ Note

A static member, a nested type or an enumerator defined in a base class T can unambiguously be found even if an object has more than one base class subobject of type T.

Two base class subobjects share the non-static member subobjects of their common virtual base classes.

— end note

]

[ Example

struct V {
  int v;
};
struct A {
  int a;
  static int   s;
  enum { e };
};
struct B : A, virtual V { };
struct C : A, virtual V { };
struct D : B, C { };

void f(D* pd) {
  pd->v++;          // OK: only one v (virtual)
  pd->s++;          // OK: only one s (static)
  int i = pd->e;    // OK: only one e (enumerator)
  pd->a++;          // error, ambiguous: two as in D
}

— end example

]

[ Note

When virtual base classes are used, a hidden declaration can be reached along a path through the subobject lattice that does not pass through the hiding declaration.

This is not an ambiguity.

The identical use with non-virtual base classes is an ambiguity; in that case there is no unique instance of the name that hides all the others.

— end note

]

[ Example

struct V { int f();  int x; };
struct W { int g();  int y; };
struct B : virtual V, W {
  int f();  int x;
  int g();  int y;
};
struct C : virtual V, W { };

struct D : B, C { void glorp(); };

Figure 6 — Name lookup

The names declared in V and the left-hand instance of W are hidden by those in B, but the names declared in the right-hand instance of W are not hidden at all.

void D::glorp() {
  x++;              // OK: B::x hides V::x
  f();              // OK: B::f() hides V::f()
  y++;              // error: B::y and C's W::y
  g();              // error: B::g() and C's W::g()
}

— end example

]

An explicit or implicit conversion from a pointer to or an expression designating an object of a derived class to a pointer or reference to one of its base classes shall unambiguously refer to a unique object representing the base class.

[ Example

struct V { };
struct A { };
struct B : A, virtual V { };
struct C : A, virtual V { };
struct D : B, C { };

void g() {
  D d;
  B* pb = &d;
  A* pa = &d;       // error, ambiguous: C's A or B's A?
  V* pv = &d;       // OK: only one V subobject
}

— end example

]

[ Note

Even if the result of name lookup is unambiguous, use of a name found in multiple subobjects might still be ambiguous ([conv.mem], [expr.ref], [class.access.base]).

— end note

]

[ Example

struct B1 {
  void f();
  static void f(int);
  int i;
};
struct B2 {
  void f(double);
};
struct I1: B1 { };
struct I2: B1 { };

struct D: I1, I2, B2 {
  using B1::f;
  using B2::f;
  void g() {
    f();                        // Ambiguous conversion of this
    f(0);                       // Unambiguous (static)
    f(0.0);                     // Unambiguous (only one B2)
    int B1::* mpB1 = &D::i;     // Unambiguous
    int D::* mpD = &D::i;       // Ambiguous conversion
  }
};

— end example

]