Fix some bad fall-throughs in a switch statement. Both the 'Q' and 'R' cases

should fall through to the 'H' case, but instead 'Q' was falling through to 'R'
so that it would do the wrong thing for a big-endian ARM target.


git-svn-id: https://llvm.org/svn/llvm-project/llvm/trunk@104883 91177308-0d34-0410-b5e6-96231b3b80d8
This commit is contained in:
Bob Wilson 2010-05-27 20:23:42 +00:00
parent 4b7416b75d
commit d984eb6073

View File

@ -1064,21 +1064,27 @@ bool ARMAsmPrinter::PrintAsmOperand(const MachineInstr *MI, unsigned OpNum,
printOperand(MI, OpNum, O);
return false;
case 'Q':
if (TM.getTargetData()->isLittleEndian())
break;
// Fallthrough
case 'R':
// Print the least significant half of a register pair.
if (TM.getTargetData()->isBigEndian())
break;
// Fallthrough
case 'H': // Write second word of DI / DF reference.
// Verify that this operand has two consecutive registers.
if (!MI->getOperand(OpNum).isReg() ||
OpNum+1 == MI->getNumOperands() ||
!MI->getOperand(OpNum+1).isReg())
return true;
++OpNum; // Return the high-part.
printOperand(MI, OpNum, O);
return false;
case 'R':
// Print the most significant half of a register pair.
if (TM.getTargetData()->isLittleEndian())
break;
printOperand(MI, OpNum, O);
return false;
case 'H':
break;
}
// Print the second half of a register pair (for 'Q', 'R' or 'H').
// Verify that this operand has two consecutive registers.
if (!MI->getOperand(OpNum).isReg() ||
OpNum+1 == MI->getNumOperands() ||
!MI->getOperand(OpNum+1).isReg())
return true;
++OpNum;
}
printOperand(MI, OpNum, O);