[DAG] Make it clear mul(x,x) knownbits bit[1] == 0 check should be fo…

…r x is undef only As raised on rGffd0e464b4b9, if x is poison, this fold is still ok.
llvm · Feb 1, 2022 · d83a96f · d83a96f
1 parent f131d4d
commit d83a96f
Show file tree

Hide file tree

Showing 2 changed files with 2 additions and 1 deletion.
diff --git a/llvm/lib/CodeGen/SelectionDAG/SelectionDAG.cpp b/llvm/lib/CodeGen/SelectionDAG/SelectionDAG.cpp
@@ -3082,6 +3082,7 @@ KnownBits SelectionDAG::computeKnownBits(SDValue Op, const APInt &DemandedElts,
     Known = computeKnownBits(Op.getOperand(1), DemandedElts, Depth + 1);
     Known2 = computeKnownBits(Op.getOperand(0), DemandedElts, Depth + 1);
     bool SelfMultiply = Op.getOperand(0) == Op.getOperand(1);
+    // TODO: SelfMultiply can be poison, but not undef.
     SelfMultiply &= isGuaranteedNotToBeUndefOrPoison(
         Op.getOperand(0), DemandedElts, false, Depth + 1);
     Known = KnownBits::mul(Known, Known2, SelfMultiply);

diff --git a/llvm/test/CodeGen/X86/combine-mul.ll b/llvm/test/CodeGen/X86/combine-mul.ll
@@ -393,7 +393,7 @@ define <4 x i32> @combine_mul_self_knownbits_vector(<4 x i32> %x) {
   ret <4 x i32> %2
 }
 
-; mul(x,x) - bit[1] is 0, but if demanding the other bits the source must not be undef/poison
+; mul(x,x) - bit[1] is 0, but if demanding the other bits the source must not be undef
 
 define i64 @combine_mul_self_demandedbits(i64 %x) {
 ; SSE-LABEL: combine_mul_self_demandedbits: