On Fri, Nov 03, 2023 at 07:07:36AM +0100, Martin Uecker wrote:
> Am Donnerstag, dem 02.11.2023 um 17:28 -0700 schrieb Bill Wendling:
> > On Thu, Nov 2, 2023 at 1:36 PM Qing Zhao <qing.z...@oracle.com> wrote:
> > >
> > > Thanks a lot for raising these issues.
> > >
> > > If I understand correctly, the major question we need to answer is:
> > >
> > > For the following example: (Jakub mentioned this in an early message)
> > >
> > > 1 struct S { int a; char b __attribute__((counted_by (a))) []; };
> > > 2 struct S s;
> > > 3 s.a = 5;
> > > 4 char *p = &s.b[2];
> > > 5 int i1 = __builtin_dynamic_object_size (p, 0);
> > > 6 s.a = 3;
> > > 7 int i2 = __builtin_dynamic_object_size (p, 0);
> > >
> > > Should the 2nd __bdos call (line 7) get
> > > A. the latest value of s.a (line 6) for it’s size?
> > > Or B. the value when the s.b was referenced (line 3, line 4)?
> > >
> > I personally think it should be (A). The user is specifically
> > indicating that the size has somehow changed, and the compiler should
> > behave accordingly.
>
>
> One potential problem for A apart from the potential impact on
> optimization is that the information may get lost more
> easily. Consider:
>
> char *p = &s.b[2];
> f(&s);
> int i = __bdos(p, 0);
>
> If the compiler can not see into 'f', the information is lost
> because f may have changed the size.
Why? It doesn't really matter. The options are
A. p is at &s.b[2] associated with &s.a and int type (or size of int
or whatever); .ACCESS_WITH_SIZE can't be pure, but sure, for aliasing
POV we can describe it with more detail that it doesn't modify anything
in the pointed structure, just escapes the pointer; __bdos can stay
leaf I believe; and when expanding __bdos later on, it would just
dereference the associated pointer at that point (note, __bdos is
pure, so it has vuse but not vdef and can load from memory); if
f changes s.a, no problem, __bdos will load the changed value in there
B. if .ACCESS_WITH_SIZE associates the pointer with the s.a value from that
point, .ACCESS_WITH_SIZE can be const, but obviously if f changes s.a,
__bdos later will use s.a value from the &s.b[2] spot
Jakub
