https://gcc.gnu.org/bugzilla/show_bug.cgi?id=104800
--- 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 > > > > "— Volatile accesses to objects are evaluated strictly > > according to the rules of the abstract machine." > > > > and > > > > "This is the observable behavior of the program." > > > > > > 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 machine. > > Well, the volatile access _was_ evaluated strictly according to the abstract > machine. Not if there is a trap. > Can't your argument be stretched in a way that for > > global = 2; > *volatile = 1; > > 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.
