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For me, the emphasized part both apply to id-expression and class member access , right?

Playing with my favorite compiler, I get the following.

✓ Accepted by the compiler

namespace N { void f() {} }
using type = decltype(N::f);
type* pf = N::f;
Ok I guess; N::f is an unparenthesized id-expression and does not name a set of overloaded functions.
✗ Rejected by the compiler
namespace N { void f() {} void f(int) {} }
using type = decltype(N::f); // error: decltype cannot resolve address of overloaded function
type* pf = N::f;
Ok; N::f does name a set of overloaded functions.
✗ Rejected by the compiler
struct S { void f(){} };
using type = decltype(S::f); // error: invalid use of non-static member function 'void S::f()'
type* pf = &S::f;
Hum? S::f would name a set of one overloaded function?
All in all, is my understanding of [dcl.type]/4.3 just bad? is gcc trunk wrong? both? none? kamoulox?
The simple reason is that the use of S::f is constrained for class members.
[expr.prim.id]
2 An id-expression that denotes a non-static data member or
non-static member function of a class can only be used:
as part of a class member access in which the object expression refers to the member's class or a class derived from that class, or
to form a pointer to member ([expr.unary.op]), or
if that id-expression denotes a non-static data member and it appears in an unevaluated operand.
The last bullet, the one related to your code, only applies to non-static data member. There is no provision for functions.
I can only speculate as to why it's not allowed, though I previously asked that question.
It's worth a note that decltype(&S::f) works here as pointer-to-member-function type,

again unless the f names a set of overloaded (member) functions.
The function type itself can be extracted from the pointer-to-member-function type.

If the member function is cv-or-ref qualified then it has an abominable function type.

The std traits are lacking here - a library like Boost.CallableTraits helps.
                Hi. Thank you for your help. Note though this question is tagged language-lawyer. It generally means we're not looking for a solution or a workaround, but for a reason/a quote from the Standard.
– YSC
                Sep 27, 2018 at 16:47
        
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