[PPCMIPeephole] Fix splat elimination
Summary:
In PPCMIPeephole, when we see two splat instructions, we can't simply do the following transformation:
B = Splat A
C = Splat B
=>
C = Splat A
because B may still be used between these two instructions. Instead, we should make the second Splat a PPC::COPY and let later passes decide whether to remove it or not:
B = Splat A
C = Splat B
=>
B = Splat A
C = COPY B
Fixes PR30663.
Reviewers: echristo, iteratee, kbarton, nemanjai
Subscribers: mehdi_amini, llvm-commits
Differential Revision: https://reviews.llvm.org/D25493
git-svn-id: https://llvm.org/svn/llvm-project/llvm/trunk@283961 91177308-0d34-0410-b5e6-96231b3b80d8
diff --git a/lib/Target/PowerPC/PPCMIPeephole.cpp b/lib/Target/PowerPC/PPCMIPeephole.cpp
index 3360e74..ee62bb3 100644
--- a/lib/Target/PowerPC/PPCMIPeephole.cpp
+++ b/lib/Target/PowerPC/PPCMIPeephole.cpp
@@ -201,11 +201,13 @@
// Splat fed by another splat - switch the output of the first
// and remove the second.
if (SameOpcode) {
- DefMI->getOperand(0).setReg(MI.getOperand(0).getReg());
+ DEBUG(dbgs() << "Changing redundant splat to a copy: ");
+ DEBUG(MI.dump());
+ BuildMI(MBB, &MI, MI.getDebugLoc(), TII->get(PPC::COPY),
+ MI.getOperand(0).getReg())
+ .addOperand(MI.getOperand(OpNo));
ToErase = &MI;
Simplified = true;
- DEBUG(dbgs() << "Removing redundant splat: ");
- DEBUG(MI.dump());
}
// Splat fed by a shift. Usually when we align value to splat into
// vector element zero.
