http://gcc.gnu.org/bugzilla/show_bug.cgi?id=51821
--- Comment #14 from Uros Bizjak <ubizjak at gmail dot com> 2012-01-12 11:42:00 UTC --- (In reply to comment #12) > I'm not sure I understand. If the peephole matches, then the insn pattern is > present in the insn stream with instantiated registers, so it's sufficient to > scan the insn stream. AFAICS the code doesn't see that edx is live after the > instruction because DF reports that only eax is; of course it's eax in DImode > so edx is "implicitly" live, but it's nevertheless live. A comment to "implicitly" live: my second example (Comment #9) marks ax register as available, but IT IS SET in the pattern! This supports the theory outlined in the Comment #13, but df_simulate_one_insn_forwards now marks unused register ax as available. Again, live analysis is not the correct tool to determine which registers are clobbered by the insn.
