On Thu, Nov 12, 2020 at 11:39:07AM +0100, Jan Hubicka wrote:
> > On Thu, Nov 12, 2020 at 11:29:21AM +0100, Jan Hubicka wrote:
> > > > If OEP_ADDRESS is used also on non-addressable stuff, just to compare
> > > > that two COMPONENT_REFs access the same memory, then just comparing
> > > > DECL_BIT_FIELD_REPRESENTATIVE is not sufficient, you could have:
> > > > struct S { int c; int a : 7, b : 1; };
> > > > struct T { int c; int a : 7, b : 1; };
> > > > and compare s->a vs. t->b with OEP_ADDRESS and the offsets of their
> > > > DECL_BIT_FIELD_REPRESENATIVE is the same, yet we don't want to say
> > > > the two bit-fields are the same.
> > >
> > > You are right, I was just thinking of that. I suppose it indeed makes
> > > more sense to assert that there are no bitfields here and in the AO
> > > comparsion take care of stripping the last bitfield reference and
> > > handling it specially?
> >
> > Or just compare DECL_FIELD_OFFSET and DECL_FIELD_BIT_OFFSET of the fields
> > rather than their DECL_BIT_FIELD_REPRESENTATIVE?
>
> I think I will need to compare bitfields specially at the ao_ref_compare
> side anyway to distinguish
>
> struct S { int c; int a : 5, b : 1; };
> struct T { int c; int a : 5, b : 3; };
>
> s->b and t->b. Those have same base address (bitwise) but still we do
> not want to consider them equal.
How is that different from:
struct S { long long d; int e; };
struct T { long long d; long long e; };
s->e vs. t->e ?
One thing is comparison of the address (as it is comparing
DECL_FIELD_BIT_OFFSET too, it is essentially bit-address), and another thing
(unlrelated to OEP_ADDRESS comparisons) is if you need to ensure the access
has the same size, in that case you just compare the bit size of the access...
Jakub
