This is an unofficial snapshot of the ISO/IEC JTC1 SC22 WG21 Core Issues List revision 110d. See for the official list.


156. Name lookup for conversion functions

Section: _N4868_.6.5.6  [basic.lookup.classref]     Status: NAD     Submitter: Derek Inglis     Date: 18 Aug 1999

Paragraph 7 of _N4868_.6.5.6 [basic.lookup.classref] says,

If the id-expression is a conversion-function-id, its conversion-type-id shall denote the same type in both the context in which the entire postfix-expression occurs and in the context of the class of the object expression (or the class pointed to by the pointer expression).
Does this mean that the following example is ill-formed?
    struct A { operator int(); } a;
    void foo() {
      typedef int T;
      a.operator T(); // 1) error T is not found in the context
		      // of the class of the object expression?
The second bullet in paragraph 1 of [class.qual] says,
a conversion-type-id of an operator-function-id is looked up both in the scope of the class and in the context in which the entire postfix-expression occurs and shall refer to the same type in both contexts
How about:
    struct A { typedef int T; operator T(); };
    struct B : A { operator T(); } b;
    void foo() {
      b.A::operator T(); // 2) error T is not found in the context
			 // of the postfix-expression?
Is this interpretation correct? Or was the intent for this to be an error only if T was found in both scopes and referred to different entities?

If the intent was for these to be errors, how do these rules apply to template arguments?

    template <class T1> struct A { operator T1(); }
    template <class T2> struct B : A<T2> {
      operator T2();
      void foo() {
	T2 a = A<T2>::operator T2(); // 3) error? when instantiated T2 is not
				     // found in the scope of the class
	T2 b = ((A<T2>*)this)->operator T2(); // 4) error when instantiated?

(Note bullets 2 and 3 in paragraph 1 of [class.qual] refer to postfix-expression. It would be better to use qualified-id in both cases.)

Erwin Unruh: The intent was that you look in both contexts. If you find it only once, that's the symbol. If you find it in both, both symbols must be "the same" in some respect. (If you don't find it, its an error).

Mike Miller: What's not clear to me in these examples is whether what is being looked up is T or int. Clearly the T has to be looked up somehow, but the "name" of a conversion function clearly involves the base (non-typedefed) type, not typedefs that might be used in a definition or reference (cf 6.1 [basic.pre] paragraph 7 and 11.4.8 [class.conv] paragraph 5) . (This is true even for types that must be written using typedefs because of the limited syntax in conversion-type-ids — e.g., the "name" of the conversion function in the following example

    typedef void (*pf)();
    struct S {
	operator pf();
is S::operator void(*)(), even though you can't write its name directly.)

My guess is that this means that in each scope you look up the type named in the reference and form the canonical operator name; if the name used in the reference isn't found in one or the other scope, the canonical name constructed from the other scope is used. These names must be identical, and the conversion-type-id in the canonical operator name must not denote different types in the two scopes (i.e., the type might not be found in one or the other scope, but if it's found in both, they must be the same type).

I think this is all very vague in the current wording.

Rationale (February, 2021):

This issue was resolved by the resolution of issue 1111.