search for: remove_instr

...NewMI = BUILD_INS(TEST8ri, AH, 1 << (imm - 8)); else NewMI = BUILD_INS(BT32ri8, EAX, imm); break; // ... } } break; } // NewMI has been inserted before OldMI if (NewMI != NULL) { // MBB->remove_instr(OldMI); // I've tried these (and others) // OldMI->removeFromParent(); // MBB.erase(OldMI); NumX86Peepholes++; } } -------------- next part -------------- An HTML attachment was scrubbed... URL: <http://lists.llvm.org/pipermail/llvm-dev/attachments/201...

...; This works and it correctly places the new instruction just before the old instruction in the assembly output. So far so good. Now I have to remove the old instruction. But everything I try crashes LLVM, either immediately or eventually. Various incantations which haven't worked. MBB->remove_instr(OldMI); OldMI->removeFromParent(); MBB.erase(OldMI); I should add that there are flags // %EFLAGS<imp-def> is getting copied automatically // %RDX<imp-use,kill> is not getting copied (when it appears) So, any advice to peephole adding/deleting or just replacing M...

...well) and the new instruction is present in the assembly output. The problem is removing the old instruction correctly. > The loop header needs to be modified, because MBBI will be invalidated when you remove the instruction: So if I remove the old instruction with something like: MBB->remove_instr(OldMI); I could just start the loop over or is there a more better way to invalidate the MBBI iterator? Basic blocks aren't that long (on average 4-6 instructions) and this peephole isn't that common either. -------------- next part -------------- An HTML attachment was scrubbed... URL: &l...