From a3080591addcc999523bc288614d85758fa89f79 Mon Sep 17 00:00:00 2001
From: Craig Topper <craig.topper@intel.com>
Date: Fri, 25 Aug 2017 18:39:40 +0000
Subject: [PATCH] [InstCombine] Don't fall back to only calling
 computeKnownBits if the upper bit of Add/Sub is demanded.

Just create an all 1s demanded mask and continue recursing like normal. The recursive calls should be able to handle an all 1s mask and do the right thing.

The only time we should care about knowing whether the upper bit was demanded is when we need to know if we should clear the NSW/NUW flags.

Now that we have a consistent path through the code for all cases, use KnownBits::computeForAddSub to compute the known bits at the end since we already have the LHS and RHS.

My larger goal here is to move the code that turns add into xor if only 1 bit is demanded and no bits below it are non-zero from InstCombiner::OptAndOp to here. This will allow it to be more general instead of just looking for 'add' and 'and' with constant RHS.

Differential Revision: https://reviews.llvm.org/D36486

llvm-svn: 311789
---
 .../InstCombineSimplifyDemanded.cpp           | 47 ++++++++++---------
 1 file changed, 24 insertions(+), 23 deletions(-)

diff --git a/lib/Transforms/InstCombine/InstCombineSimplifyDemanded.cpp b/lib/Transforms/InstCombine/InstCombineSimplifyDemanded.cpp
index 461ab35e83c..39d1dfe27a0 100644
--- a/lib/Transforms/InstCombine/InstCombineSimplifyDemanded.cpp
+++ b/lib/Transforms/InstCombine/InstCombineSimplifyDemanded.cpp
@@ -396,38 +396,39 @@ Value *InstCombiner::SimplifyDemandedUseBits(Value *V, APInt DemandedMask,
     /// If the high-bits of an ADD/SUB are not demanded, then we do not care
     /// about the high bits of the operands.
     unsigned NLZ = DemandedMask.countLeadingZeros();
-    if (NLZ > 0) {
-      // Right fill the mask of bits for this ADD/SUB to demand the most
-      // significant bit and all those below it.
-      APInt DemandedFromOps(APInt::getLowBitsSet(BitWidth, BitWidth-NLZ));
-      if (ShrinkDemandedConstant(I, 0, DemandedFromOps) ||
-          SimplifyDemandedBits(I, 0, DemandedFromOps, LHSKnown, Depth + 1) ||
-          ShrinkDemandedConstant(I, 1, DemandedFromOps) ||
-          SimplifyDemandedBits(I, 1, DemandedFromOps, RHSKnown, Depth + 1)) {
+    // Right fill the mask of bits for this ADD/SUB to demand the most
+    // significant bit and all those below it.
+    APInt DemandedFromOps(APInt::getLowBitsSet(BitWidth, BitWidth-NLZ));
+    if (ShrinkDemandedConstant(I, 0, DemandedFromOps) ||
+        SimplifyDemandedBits(I, 0, DemandedFromOps, LHSKnown, Depth + 1) ||
+        ShrinkDemandedConstant(I, 1, DemandedFromOps) ||
+        SimplifyDemandedBits(I, 1, DemandedFromOps, RHSKnown, Depth + 1)) {
+      if (NLZ > 0) {
         // Disable the nsw and nuw flags here: We can no longer guarantee that
         // we won't wrap after simplification. Removing the nsw/nuw flags is
         // legal here because the top bit is not demanded.
         BinaryOperator &BinOP = *cast<BinaryOperator>(I);
         BinOP.setHasNoSignedWrap(false);
         BinOP.setHasNoUnsignedWrap(false);
-        return I;
       }
-
-      // If we are known to be adding/subtracting zeros to every bit below
-      // the highest demanded bit, we just return the other side.
-      if (DemandedFromOps.isSubsetOf(RHSKnown.Zero))
-        return I->getOperand(0);
-      // We can't do this with the LHS for subtraction, unless we are only
-      // demanding the LSB.
-      if ((I->getOpcode() == Instruction::Add ||
-           DemandedFromOps.isOneValue()) &&
-          DemandedFromOps.isSubsetOf(LHSKnown.Zero))
-        return I->getOperand(1);
+      return I;
     }
 
-    // Otherwise just hand the add/sub off to computeKnownBits to fill in
-    // the known zeros and ones.
-    computeKnownBits(V, Known, Depth, CxtI);
+    // If we are known to be adding/subtracting zeros to every bit below
+    // the highest demanded bit, we just return the other side.
+    if (DemandedFromOps.isSubsetOf(RHSKnown.Zero))
+      return I->getOperand(0);
+    // We can't do this with the LHS for subtraction, unless we are only
+    // demanding the LSB.
+    if ((I->getOpcode() == Instruction::Add ||
+         DemandedFromOps.isOneValue()) &&
+        DemandedFromOps.isSubsetOf(LHSKnown.Zero))
+      return I->getOperand(1);
+
+    // Otherwise just compute the known bits of the result.
+    bool NSW = cast<OverflowingBinaryOperator>(I)->hasNoUnsignedWrap();
+    Known = KnownBits::computeForAddSub(I->getOpcode() == Instruction::Add,
+                                        NSW, LHSKnown, RHSKnown);
     break;
   }
   case Instruction::Shl: {