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Re: [std-proposals] Why [expr.ref] says that E1->E2 is converted to (*(E1)).E2

From: Jonathan Wakely <cxx_at_[hidden]>
Date: Tue, 9 May 2023 17:06:25 +0100
On Tue, 9 May 2023 at 15:42, Sean Mayard <seanmayard_at_[hidden]> wrote:

> Look at the previous sentence.
> Okay, so the previous sentence says "For the first option (dot) the first
> expression shall be a glvalue.
> <https://timsong-cpp.github.io/cppwp/n4868/expr.ref#2.sentence-1> For the
> second option (arrow) the first expression shall be a prvalue having *pointer
> type*. <https://timsong-cpp.github.io/cppwp/n4868/expr.ref#2.sentence-2>
> " By pointer type they mean built-in pointer type and not smart pointer
> type.

Yes, smart pointers have class type. There are no "smart pointer types" in
the type taxonomy of C++, that's just an idiomatic name for class types
with a pointer-like API.

But it's irrelevant, [expr.pre] means the whole paragraph only applies to
the built-in operator, not an overloaded operator. There are no built-in ->
operators for smart pointer types, only operator->() overloads.

> On Tue, 9 May 2023 at 18:35, Jonathan Wakely <cxx_at_[hidden]> wrote:
>> 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).

Received on 2023-05-09 16:06:40