------- Additional Comments From squell at alumina dot nl 2005-07-23 03:22 ------- ptr->f is a constant expression, I believe, since it doesn't access any object (see 5.19/4, C++ Std), because the expression will resolve to a static member. But whether I am right or not:
I got to this testcase because I was reading C++'s 9.3.1/2 to decipher g++3.3's error message. This paragraph reads (relevant text shown); "When an id-expression [..] not used to form a pointer to member is used in the body of a non-static member function [..], if name lookup resolves the name in the id-expression to a non-static non-type member [..], the id-expression is transformed into a class member access expression using (*this) as the postfix-expression to the left of the . operator. The member name then refers to the member of the object for which the function is called." Since 'f' and 'foo::f' resolve to a non-static member (name lookup is performed before overload resolution, after all), this applies. So, the original code snippet: f_obj<&foo::f> (g++ accepted this one) -- &foo::f is pointer to member syntax, do nothing! f_obj<foo::f> -- transforms to f_obj<this->foo::f> f_obj<f> -- transforms to f_obj<this->foo::f> f_obj<&f> -- transforms to f_obj<&this->foo::f> Which seams to rather nicely match the results! :) This also explains why f_obj<foo::f> is OK outside the member function (no transformation done), but an error inside of it. If you are correct about the constant expression, then the original behaviour I reported is absolutely correct. Surprisingly. But I think ptr->f is a constant expression, because in this expression, the expression 'ptr' (just an lvalue) is evaluated, not (*ptr). This still leaves the problem that 9.3.1/2 defines operations in terms of "(*this)", not "this->", but then again, "this" isn't an object, right? -- http://gcc.gnu.org/bugzilla/show_bug.cgi?id=22621
