On Tue, 9 May 2023 at 13:10, Sean Mayard via Std-Proposals <
std-proposals_at_[hidden]> wrote:

> I read [expr.ref#2] which includes the following sentence:
>
> > *The expression E1->E2 is converted to the equivalent form (*(E1)).E2*
>
> My question is, does this mean that "every" expression that has the form
> *E1->E2* will be converted to *(*(E1)).E2. *I mean it is not clear to me
> in what situations/contexts does the above sentence hold?
>
> For example, suppose we have a class that overloads *operator* *but does
> not overload *operator->* then if we use *operator-> *with an object of
> that class type, then that expression will be first converted to
> *(*(E1)).E2* form?
>
> #include <iostream> struct A { int foo = 1, bar =2;};struct B { A a{}; A& operator* () { return a; }}; int main(){ B b{}; int x = b->foo; //i know this will give error because there is no overloaded operator-> //but according to [expr.ref#2] shouldn't b->foo be first converted to (*(b)).foo //and then the overloaded operator* should be used }
>
>
> As you can see in the above program, `b->foo;` gives an error because the
> class `B` has not overloaded *operator-> *. But according to [expr.ref#2]
> the expression *b->foo *should be converted to *(*(b)).foo *and then the *operator*
> *could be used.
>
> So I want to know what exactly does [expr.ref#2] mean when it said " *The
> expression E1->E2 is converted to the equivalent form (*(E1)).E2* ". Does
> this conversion supposed to happen only after `E1->E2` is valid or before.
> Is a CWG issue required for this or I just did not read/interpret it
> correctly.
>

See [expr.pre] which explains that the rules in Clause [expr] apply to the
built-in operators, not to overloaded operators.

"Subclause 7.6 defines the effects of operators when applied to types for
which they have not been overloaded." (and also the note before that).