[ValueTracking] Correct early out in computeKnownBitsFromOperator to work with non...

There's an early out that's trying to detect when we don't know any bits that make up the legal range of a shift. The code subtracts one from BitWidth which creates a mask in the lower bits for power of 2 bit widths. This is then ANDed with the known bits to see if any of those bits are known. If the bit width isn't a power of 2 this creates a non-sensical mask.

This patch corrects this by rounding up to a power of 2 before doing the subtract and mask.

Differential Revision: https://reviews.llvm.org/D34165

git-svn-id: https://llvm.org/svn/llvm-project/llvm/trunk@305400 91177308-0d34-0410-b5e6-96231b3b80d8

diff --git a/lib/Analysis/ValueTracking.cpp b/lib/Analysis/ValueTracking.cpp
index 7cb8c8d8aaebff18ac7039244bce240a434d5e96..b065f427b06cba146282bbcf75bc6f4ff51a12da 100644 (file)
--- a/lib/Analysis/ValueTracking.cpp
+++ b/lib/Analysis/ValueTracking.cpp
@@ -852,7 +852,8 @@ static void computeKnownBitsFromShiftOperator(
   Optional<bool> ShifterOperandIsNonZero;
 
   // Early exit if we can't constrain any well-defined shift amount.
-  if (!(ShiftAmtKZ & (BitWidth - 1)) && !(ShiftAmtKO & (BitWidth - 1))) {
+  if (!(ShiftAmtKZ & (PowerOf2Ceil(BitWidth) - 1)) &&
+      !(ShiftAmtKO & (PowerOf2Ceil(BitWidth) - 1))) {
     ShifterOperandIsNonZero =
         isKnownNonZero(I->getOperand(1), Depth + 1, Q);
     if (!*ShifterOperandIsNonZero)

diff --git a/test/Transforms/InstCombine/shift.ll b/test/Transforms/InstCombine/shift.ll
index ce8e2fcd38b9bb674aa8914a1adb0fc7bea9c545..68bbf35d1e65ac46f36738927cfca00a1261701c 100644 (file)
--- a/test/Transforms/InstCombine/shift.ll
+++ b/test/Transforms/InstCombine/shift.ll
@@ -1306,3 +1306,13 @@ define <2 x i8> @lshr_demanded_bits_splat(<2 x i8> %x) {
   ret <2 x i8> %shr
 }
 
+; Make sure known bits works correctly with non power of 2 bit widths.
+define i7 @test65(i7 %a, i7 %b) {
+; CHECK-LABEL: @test65(
+; CHECK-NEXT:    ret i7 0
+;
+  %shiftamt = and i7 %b, 6 ; this ensures the shift amount is even and less than the bit width.
+  %x = lshr i7 42, %shiftamt ; 42 has a zero in every even numbered bit and a one in every odd bit.
+  %y = and i7 %x, 1 ; this extracts the lsb which should be 0 because we shifted an even number of bits and all even bits of the shift input are 0.
+  ret i7 %y
+}