We can prove that a 'sub' can be a 'sub nsw' under certain conditions:
- The sign bits of the operands is the same.
- Both operands have more than 1 sign bit.
The subtraction cannot be a signed overflow in either case.
llvm-svn: 216037
This update was done with the following bash script:
find test/Transforms -name "*.ll" | \
while read NAME; do
echo "$NAME"
if ! grep -q "^; *RUN: *llc" $NAME; then
TEMP=`mktemp -t temp`
cp $NAME $TEMP
sed -n "s/^define [^@]*@\([A-Za-z0-9_]*\)(.*$/\1/p" < $NAME | \
while read FUNC; do
sed -i '' "s/;\(.*\)\([A-Za-z0-9_]*\):\( *\)@$FUNC\([( ]*\)\$/;\1\2-LABEL:\3@$FUNC(/g" $TEMP
done
mv $TEMP $NAME
fi
done
llvm-svn: 186268
This has the obvious advantage of being commutable and is always a win on x86 because
const - x wastes a register there. On less weird architectures this may lead to
a regression because other arithmetic doesn't fuse with it anymore. I'll address that
problem in a followup.
llvm-svn: 147254
David Majnemer