const MachineOperand &MO = getOperand(UseOpIdx);
if (!MO.isReg() || !MO.isUse() || MO.getReg() == 0)
return false;
- int FlagIdx = UseOpIdx - 1;
- if (FlagIdx < 1)
- return false;
- while (!getOperand(FlagIdx).isImm()) {
- if (--FlagIdx == 0)
- return false;
+
+ // Find the flag operand corresponding to UseOpIdx
+ unsigned FlagIdx, NumOps=0;
+ for (FlagIdx = 1; FlagIdx < UseOpIdx; FlagIdx += NumOps+1) {
+ const MachineOperand &UFMO = getOperand(FlagIdx);
+ assert(UFMO.isImm() && "Expecting flag operand on inline asm");
+ NumOps = InlineAsm::getNumOperandRegisters(UFMO.getImm());
+ assert(NumOps < getNumOperands() && "Invalid inline asm flag");
+ if (UseOpIdx < FlagIdx+NumOps+1)
+ break;
}
- const MachineOperand &UFMO = getOperand(FlagIdx);
- if (FlagIdx + InlineAsm::getNumOperandRegisters(UFMO.getImm()) < UseOpIdx)
+ if (FlagIdx >= UseOpIdx)
return false;
+ const MachineOperand &UFMO = getOperand(FlagIdx);
unsigned DefNo;
if (InlineAsm::isUseOperandTiedToDef(UFMO.getImm(), DefNo)) {
if (!DefOpIdx)
--- /dev/null
+; RUN: llvm-as < %s | llc -march=ppc32 -verify-machineinstrs
+
+; Machine code verifier will call isRegTiedToDefOperand() on /all/ register use
+; operands. We must make sure that the operand flag is found correctly.
+
+; This test case is actually not specific to PowerPC, but the (imm, reg) format
+; of PowerPC "m" operands trigger this bug.
+
+define void @memory_asm_operand(i32 %a) {
+ ; "m" operand will be represented as:
+ ; INLINEASM <es:fake $0>, 10, %R2, 20, -4, %R1
+ ; It is difficult to find the flag operand (20) when starting from %R1
+ call i32 asm "lbzx $0, $1", "=r,m" (i32 %a)
+ ret void
+}
+