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http://www.open-std.org/jtc1/sc22/wg21/
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The reference of a conversion function(template) with a deduced return type #4671

Closed xmh0511 closed 3 years ago

xmh0511 commented 3 years ago

In N4861, the relevant rule temp.mem#5 is:

A specialization of a conversion function template is referenced in the same way as a non-template conversion function that converts to the same type.

After P1787, the above normative rule is radically changed to be a note

[Note 1: A specialization of a conversion function template is referenced in the same way as a non-template conversion function that converts to the same type ([class.conv.fct]).

There's no normative rule based on to interpret why we should use a.operator int*(); to refer to the specialization. I also feel that this issue is similar with How to explicitly call a conversion function whose conversion-type-id contains a placeholder specifier that I have posted it to wmm@edg.com, which has became the issue 2493 that will be published on http://open-std.org/jtc1/sc22/wg21/

jensmaurer commented 3 years ago

Duplicate of #4811 CWG2493.

xmh0511 commented 3 years ago

@jensmaurer It is a bit similar to #4811 but not completely. One is template function, the other is a function with place-holder return type.

jensmaurer commented 3 years ago

Agreed.

The rationale for the conversion function template case is that the deduction that needs to happen here is part of name lookup. [class.member.lookup] p7

xmh0511 commented 3 years ago

@jensmaurer So, I think it should be an individual issue that should be processed. After all, the deduction of a conversion function template occurs as a part of name lookup and that is deduced with the actual argument. Instead, the conversion function with the place-holder is directly deduced from its return statement.

jensmaurer commented 3 years ago

I'm saying the normative rule for the template case is in [class.member.lookup] p7. If you feel that's insufficient, please give more details. And fix the heading if this particular issue is not about deduced return types, but just about conversion function templates.

xmh0511 commented 3 years ago

@jensmaurer After further consideration, you're right. [class.member.lookup] p7 has clarified this.

struct A {
  template <class T> operator T*();
};
A a;
int* ip = a.operator int*();

a.operator int*() may be the only way to refer to that specialization.