Proper check for two-addressness.

git-svn-id: https://llvm.org/svn/llvm-project/llvm/trunk@31408 91177308-0d34-0410-b5e6-96231b3b80d8
This commit is contained in:
Evan Cheng 2006-11-03 03:04:46 +00:00
parent 24d9cf025d
commit bf105c8424

View File

@ -364,6 +364,25 @@ void LiveIntervals::printRegName(unsigned reg) const {
std::cerr << "%reg" << reg;
}
/// isReDefinedByTwoAddr - Returns true if the Reg re-definition is due to
/// two addr elimination.
static bool isReDefinedByTwoAddr(MachineInstr *MI, unsigned Reg,
const TargetInstrInfo *TII) {
for (unsigned i = 0, e = MI->getNumOperands(); i != e; ++i) {
MachineOperand &MO1 = MI->getOperand(i);
if (MO1.isRegister() && MO1.isDef() && MO1.getReg() == Reg) {
for (unsigned j = i+1; j < e; ++j) {
MachineOperand &MO2 = MI->getOperand(j);
if (MO2.isRegister() && MO2.isUse() && MO2.getReg() == Reg &&
TII->getOperandConstraint(MI->getOpcode(), j,
TargetInstrInfo::TIED_TO) == (int)i)
return true;
}
}
}
return false;
}
void LiveIntervals::handleVirtualRegisterDef(MachineBasicBlock *mbb,
MachineBasicBlock::iterator mi,
unsigned MIIdx,
@ -453,13 +472,9 @@ void LiveIntervals::handleVirtualRegisterDef(MachineBasicBlock *mbb,
} else {
// If this is the second time we see a virtual register definition, it
// must be due to phi elimination or two addr elimination. If this is
// the result of two address elimination, then the vreg is the first
// operand, and is a def-and-use.
if (mi->getOperand(0).isRegister() &&
mi->getOperand(0).getReg() == interval.reg &&
mi->getNumOperands() > 1 && mi->getOperand(1).isRegister() &&
mi->getOperand(1).getReg() == interval.reg &&
mi->getOperand(0).isDef() && mi->getOperand(1).isUse()) {
// the result of two address elimination, then the vreg is one of the
// def-and-use register operand.
if (isReDefinedByTwoAddr(mi, interval.reg, tii_)) {
// If this is a two-address definition, then we have already processed
// the live range. The only problem is that we didn't realize there
// are actually two values in the live interval. Because of this we