Instead of matching value and then blindly casting to BinaryOperator
just to get the opcode, just match instruction and do no cast.
Fixes https://bugs.llvm.org/show_bug.cgi?id=42962
llvm-svn: 368554
If one of the values being shifted is a constant, since the new shift
amount is known-constant, the new shift will end up being constant-folded
so, we don't need that one-use restriction then.
llvm-svn: 368519
That one-use restriction is not needed for correctness - we have already
ensured that one of the shifts will go away, so we know we won't increase
the instruction count. So there is no need for that restriction.
llvm-svn: 368518
Summary:
Given pattern:
`icmp eq/ne (and ((x shift Q), (y oppositeshift K))), 0`
we should move shifts to the same hand of 'and', i.e. rewrite as
`icmp eq/ne (and (x shift (Q+K)), y), 0` iff `(Q+K) u< bitwidth(x)`
It might be tempting to not restrict this to situations where we know
we'd fold two shifts together, but i'm not sure what rules should there be
to avoid endless combine loops.
We pick the same shift that was originally used to shift the variable we picked to shift:
https://rise4fun.com/Alive/6x1v
Should fix [[ https://bugs.llvm.org/show_bug.cgi?id=42399 | PR42399]].
Reviewers: spatel, nikic, RKSimon
Reviewed By: spatel
Subscribers: llvm-commits
Tags: #llvm
Differential Revision: https://reviews.llvm.org/D63829
llvm-svn: 364791
As discussed in https://reviews.llvm.org/D63829
*if* *both* shifts are one-use, we'd most likely want to produce `lshr`,
and not rely on ordering.
Also, there should likely be a *separate* fold to do this reordering.
llvm-svn: 364772
Roman Lebedev