On Mon, Jun 17, 2024 at 03:33:05PM +0200, Martin Uecker wrote:
> > I've done that and that was because build_qualified_type uses that
> > predicate, where qualified types created by build_qualified_type have
> > as TYPE_CANONICAL the qualified type of the main variant of the canonical
> > type, while in all other cases TYPE_CANONICAL is just the main variant of
> > the type.
> > Guess we could also just do
> > if (TYPE_QUALS (x) == TYPE_QUALS (t))
> > TYPE_CANONICAL (x) = TYPE_CANONICAL (t);
> > else if (TYPE_CANONICAL (t) != t
> > || TYPE_QUALS (x) != TYPE_QUALS (TYPE_CANONICAL (t)))
> > TYPE_CANONICAL (x)
> > = build_qualified_type (TYPE_CANONICAL (t), TYPE_QUALS (x));
> > else
> > TYPE_CANONICAL (x) = x;
> >
>
> Ok, that works. I think the final "else" is then then impossible to reach
> and can be eliminated as well, because if TYPE_CANONICAL (t) == t then
> TYPE_QUALS (x) == TYPE_QUALS (TYPE_CANONICAL (t)) is identical to
> TYPE_QUALS (x) == TYPE_QUALS (t) which is the first case.
If c_update_type_canonical is only ever called for the main variants of the
type and they always have !TYPE_QUALS (t), then yes.
But if we rely on that, perhaps we should gcc_checking_assert that.
So
gcc_checking_assert (t == TYPE_MAIN_VARIANT (t) && !TYPE_QUALS (t));
or something similar at the start of the function.
Then we could also change the
for (tree x = TYPE_MAIN_VARIANT (t); x; x = TYPE_NEXT_VARIANT (x))
to
for (tree x = t; x; x = TYPE_NEXT_VARIANT (x))
and
if (TYPE_QUALS (x) == TYPE_QUALS (t))
...
to
if (!TYPE_QUALS (x))
TYPE_CANONICAL (x) = TYPE_CANONICAL (t);
else
build_qualified_type (TYPE_CANONICAL (t), TYPE_QUALS (x));
Jakub
