On Wed, 2024-05-01 at 18:29 +0100, Alan Maguire wrote:
[...]
> > > +/* Check if a member of a split BTF struct/union refers to a base BTF
> > > + * struct/union. Members can be const/restrict/volatile/typedef
> > > + * reference types, but if a pointer is encountered, type is no longer
> > > + * considered embedded.
> > > + */
> > > +static int btf_find_embedded_composite_type_ids(__u32 *id, void *ctx)
> > > +{
> > > + struct btf_distill *dist = ctx;
> > > + const struct btf_type *t;
> > > + __u32 next_id = *id;
> > > +
> > > + do {
> > > + if (next_id == 0)
> > > + return 0;
> > > + t = btf_type_by_id(dist->pipe.src, next_id);
> > > + switch (btf_kind(t)) {
> > > + case BTF_KIND_CONST:
> > > + case BTF_KIND_RESTRICT:
> > > + case BTF_KIND_VOLATILE:
> > > + case BTF_KIND_TYPEDEF:
> >
> > I think BTF_KIND_TYPE_TAG is missing.
> >
>
> It's implicit in the default clause; I can't see a case for having a
> split BTF type tag base BTF types, but I might be missing something
> there. I can make all the unexpected types explicit if that would be
> clearer?
I mean, this skips a series of modifiers, e.g.:
struct buz {
// next_id will get to 'struct bar' eventually
const volatile struct bar foo;
}
Now, it is legal to have this chain like below:
struct buz {
const volatile __type_tag("quux") struct bar foo;
}
In which case the traversal does not have to stop.
Am I confused?
(Note: at the moment type tags are only applied to pointers but that
would change in the future, I have a stalled LLVM change for this).
[...]
