On Tue, May 13, 2025 at 10:49:59PM +0800, laokz wrote:
> On 5/10/2025 4:17 AM, Josh Poimboeuf wrote:
> > +
> > +#define sym_for_each_reloc(elf, sym, reloc)
> > \
> > + for (reloc = find_reloc_by_dest_range(elf, sym->sec, \
> > + sym->offset, sym->len); \
> > + reloc && reloc_offset(reloc) < sym->offset + sym->len; \
> > + reloc = rsec_next_reloc(sym->sec->rsec, reloc))
>
> This macro intents to walk through ALL relocations for the 'sym'. It seems
> we have the assumption that, there is at most one single relocation for the
> same offset and find_reloc_by_dest_range only needs to do 'less than' offset
> comparison:
>
> elf_hash_for_each_possible(reloc, reloc, hash,
> sec_offset_hash(rsec, o)) {
> if (reloc->sec != rsec)
> continue;
> if (reloc_offset(reloc) >= offset &&
> reloc_offset(reloc) < offset + len) {
> less than ==> if (!r || reloc_offset(reloc) < reloc_offset(r))
> r = reloc;
>
> Because if there were multiple relocations for the same offset, the returned
> one would be the last one in section entry order(hash list has reverse order
> against section order), then broken the intention.
Right. Is that a problem? I don't believe I've ever seen two
relocations for the same offset.
--
Josh
