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From: "muecker at gwdg dot de" <gcc-bugzilla@gcc.gnu.org>
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Subject: [Bug middle-end/104800] reodering of potentially trapping operations
 and volatile stores
Date: Wed, 09 Mar 2022 07:40:08 +0000
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https://gcc.gnu.org/bugzilla/show_bug.cgi?id=3D104800

--- Comment #11 from Martin Uecker <muecker at gwdg dot de> ---
(In reply to Richard Biener from comment #9)
> (In reply to Martin Uecker from comment #8)
> > The standard specifies in 5.1.2.3p6 that
> >=20
> > "=E2=80=94 Volatile accesses to objects are evaluated strictly
> > according to the rules of the abstract machine."
> >=20
> > and
> >=20
> > "This is the observable behavior of the program."
> >=20
> >=20
> > If a trap is moved before a volatile access so that the access never
> > happens, than this changes the observable behavior because the volatile
> > access was then not evaluated strictly according to the abstract machin=
e.
>=20
> Well, the volatile access _was_ evaluated strictly according to the abstr=
act
> machine.=20

Not if there is a trap.

> Can't your argument be stretched in a way that for
>=20
>  global =3D 2;
>  *volatile =3D 1;
>=20
> your reasoning says that since the volatile has to be evaluated strictly
> according to the abstract machine that the full abstract machine status
> has to be reflected at the point of the volatile and thus the write of
> the global (non-volatile) memory has to be observable at that point
> and so we may not move accesses to global memory across (earlier or later)
> volatile accesses?

The state of the global variables is not directly observable.

> IMHO the case with the division is similar, you just introduce the extra
> twist of a trap.

The point is that the trap prevents the volatile store to happen.

> The two volatile accesses in your example are still evaluated according
> to the abstract machine, just all non-volatile (and non-I/O) statements
> are not necessarily.

The problem is that the volatile store might not be evaluated if there is a
trap.=