RPCSX/llvm
1
0
Fork 0
mirror of https://github.com/RPCSX/llvm.git synced 2024-12-03 09:21:13 +00:00
Code Issues Actions Packages Projects Releases Wiki Activity

[ValueTracking] Correct early out in computeKnownBitsFromOperator to work with non power of 2 bit widths

Browse Source 
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
This commit is contained in:
Craig Topper 2017-06-14 17:04:59 +00:00
parent b76b866ea1
commit 996d9b5480
2 changed files with 12 additions and 1 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

3
lib/Analysis/ValueTracking.cpp
View File

@ -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)

10
test/Transforms/InstCombine/shift.ll
View File

@ -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
}