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[CodeGen] Make MachineInstr::isIdenticalTo() symmetric.

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Summary:
MachineInstr::isIdenticalTo() is for some reason not
symmetric when comparing bundles, which gives us the
property:

  I1->isIdenticalTo(*I2) != I2->isIdenticalTo(*I1)

when comparing bundles where one bundle is longer than
the other.

This patch makes sure that bundles of different length
always are considered as not being identical. Thus, the
result of the comparison will be the same regardless of
which side that happens to be to the left.

Reviewers: dexonsmith, jonpa, andrew.w.kaylor

Subscribers: llvm-commits, mehdi_amini

Differential Revision: https://reviews.llvm.org/D27508

git-svn-id: https://llvm.org/svn/llvm-project/llvm/trunk@290095 91177308-0d34-0410-b5e6-96231b3b80d8
This commit is contained in:
Bjorn Pettersson 2016-12-19 11:20:57 +00:00
parent 62e99e136f
commit 0fd1a2cc65
1 changed files with 13 additions and 5 deletions
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18
lib/CodeGen/MachineInstr.cpp
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@ -1000,16 +1000,24 @@ bool MachineInstr::isIdenticalTo(const MachineInstr &Other,
return false;
if (isBundle()) {
// Both instructions are bundles, compare MIs inside the bundle.
// We have passed the test above that both instructions have the same
// opcode, so we know that both instructions are bundles here. Let's compare
// MIs inside the bundle.
assert(Other.isBundle() && "Expected that both instructions are bundles.");
MachineBasicBlock::const_instr_iterator I1 = getIterator();
MachineBasicBlock::const_instr_iterator E1 = getParent()->instr_end();
MachineBasicBlock::const_instr_iterator I2 = Other.getIterator();
MachineBasicBlock::const_instr_iterator E2 = Other.getParent()->instr_end();
while (++I1 != E1 && I1->isInsideBundle()) {
// Loop until we analysed the last intruction inside at least one of the
// bundles.
while (I1->isBundledWithSucc() && I2->isBundledWithSucc()) {
++I1;
++I2;
if (I2 == E2 || !I2->isInsideBundle() || !I1->isIdenticalTo(*I2, Check))
if (!I1->isIdenticalTo(*I2, Check))
return false;
}
// If we've reached the end of just one of the two bundles, but not both,
// the instructions are not identical.
if (I1->isBundledWithSucc() || I2->isBundledWithSucc())
return false;
}
// Check operands to make sure they match.