[InstCombine] remove an icmp fold that is already handled by InstSimplify

Specifically, this is done near the end of "SimplifyICmpInst" using 
computeKnownBits() as the broader solution. There are even vector
tests (yay!) for this in test/Transforms/InstSimplify/compare.ll.

I considered putting an assert here instead of just deleting, but
then we could assert every possible fold in InstSimplify in 
InstCombine, so...less is more?

llvm-svn: 279300

1 files changed, 0 insertions, 10 deletions

diff --git a/llvm/lib/Transforms/InstCombine/InstCombineCompares.cpp b/llvm/lib/Transforms/InstCombine/InstCombineCompares.cpp
index 25587739806..f9a7eb2a512 100644
--- a/llvm/lib/Transforms/InstCombine/InstCombineCompares.cpp
+++ b/llvm/lib/Transforms/InstCombine/InstCombineCompares.cpp
@@ -2009,16 +2009,6 @@ Instruction *InstCombiner::foldICmpShlConstant(ICmpInst &Cmp, Instruction *Shl,
   ICmpInst::Predicate Pred = Cmp.getPredicate();
   Value *X = Shl->getOperand(0);
   if (Cmp.isEquality()) {
-    // If we are comparing against bits always shifted out, the comparison
-    // cannot succeed.
-    Constant *Comp =
-        ConstantExpr::getShl(ConstantExpr::getLShr(RHS, ShAmt), ShAmt);
-    if (Comp != RHS) { // Comparing against a bit that we know is zero.
-      bool IsICMP_NE = Pred == ICmpInst::ICMP_NE;
-      Constant *Cst = Builder->getInt1(IsICMP_NE);
-      return replaceInstUsesWith(Cmp, Cst);
-    }
-
     // If the shift is NUW, then it is just shifting out zeros, no need for an
     // AND.
     if (cast<BinaryOperator>(Shl)->hasNoUnsignedWrap())