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From: "redi at gcc dot gnu.org" <gcc-bugzilla@gcc.gnu.org>
To: gcc-bugs@gcc.gnu.org
Subject: [Bug libstdc++/109305] Allocator copy in basic_string::operator=
Date: Mon, 27 Mar 2023 21:03:14 +0000
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https://gcc.gnu.org/bugzilla/show_bug.cgi?id=3D109305

Jonathan Wakely <redi at gcc dot gnu.org> changed:

           What    |Removed                     |Added
----------------------------------------------------------------------------
             Status|UNCONFIRMED                 |WAITING
     Ever confirmed|0                           |1
   Last reconfirmed|                            |2023-03-27

--- Comment #3 from Jonathan Wakely <redi at gcc dot gnu.org> ---
(In reply to Marc-Andr=C3=A9 Laverdi=C3=A8re from comment #0)
> Created attachment 54773 [details]
> Self-contained preprocessed reproducer
>=20
> In basic_string& operator=3D(const basic_string& __str), I noticed this l=
ine:
>=20
> auto __alloc =3D __str._M_get_allocator();
>=20
> This creates a local copy of the allocator object. I doubt this is
> intentional, but I could be wrong.

It's completely intentional. The function needs to allocate new storage, so=
 it
needs a non-const allocator that compares equal to __str.get_allocator().

As the comment hints at, the order of the operations is important. We don't
want to update the allocator in *this until after we've allocated the stora=
ge,
because that could throw. So we make a copy of the allocator, use it to
allocate memory, then only if that was successful we replace the allocator =
in
*this.

Why do you think this local copy is a problem?=