[llvm-commits] [llvm] r92458 - in /llvm/trunk: lib/Target/README.txt lib/Transforms/Scalar/InstructionCombining.cpp test/Transforms/InstCombine/or.ll
Bill Wendling
wendling at apple.com
Mon Jan 4 18:46:18 PST 2010
On Jan 3, 2010, at 10:04 PM, Chris Lattner wrote:
> Author: lattner
> Date: Mon Jan 4 00:03:59 2010
> New Revision: 92458
>
> URL: http://llvm.org/viewvc/llvm-project?rev=92458&view=rev
> Log:
> implement an instcombine xform needed by clang's codegen
> on the example in PR4216. This doesn't trigger in the testsuite,
> so I'd really appreciate someone scrutinizing the logic for
> correctness.
>
> --- llvm/trunk/lib/Transforms/Scalar/InstructionCombining.cpp (original)
> +++ llvm/trunk/lib/Transforms/Scalar/InstructionCombining.cpp Mon Jan 4 00:03:59 2010
> @@ -5213,12 +5213,30 @@
> return ReplaceInstUsesWith(I, B);
> }
> }
> - V1 = 0; V2 = 0; V3 = 0;
> +
> + // ((V | N) & C1) | (V & C2) --> (V|N) & (C1|C2)
> + // iff (C1&C2) == 0 and (N&~C1) == 0
> + if ((C1->getValue() & C2->getValue()) == 0) {
> + if (match(A, m_Or(m_Value(V1), m_Value(V2))) &&
> + ((V1 == B && MaskedValueIsZero(V2, ~C1->getValue())) || // (V|N)
> + (V2 == B && MaskedValueIsZero(V1, ~C1->getValue())))) // (N|V)
> + return BinaryOperator::CreateAnd(A,
> + ConstantInt::get(A->getContext(),
> + C1->getValue()|C2->getValue()));
> + // Or commutes, try both ways.
> + if (match(B, m_Or(m_Value(V1), m_Value(V2))) &&
> + ((V1 == A && MaskedValueIsZero(V2, ~C2->getValue())) || // (V|N)
> + (V2 == A && MaskedValueIsZero(V1, ~C2->getValue())))) // (N|V)
> + return BinaryOperator::CreateAnd(B,
> + ConstantInt::get(B->getContext(),
> + C1->getValue()|C2->getValue()));
> + }
> }
>
Hi Chris,
I'm having trouble verifying the logic here. I'm probably doing something wrong. First a comment, if C1 and C2 are both zero, then this is zero. It can also be simplified if either one is zero. I don't know if those situations are caught before it gets to this point.
Okay. Here's my derivation of your transformation:
[(V | N) & C1] | (V & C2)
= {[(V | N) & C1] | V} & {[(V | N) & C1] | C2}
= {[(V | N | V) & (C1 | V)]} & {[(V | N | C2) & (C1 | C2)]}
= (V | N) & (V | C1) & (V | N | C2) & (C1 | C2)
Note that
A & (A | B) = A
So, (V | N) & [(V | N) | C2] = (V | N)
Therefore, we have
(V|N) & (V|C1) & (C1|C2)
Here's where I get stuck. I can expand out the (V|C1) term, but it doesn't appear to get me closer to your result. I freely admit that I probably made an error. :-)
Could you provide more insight into the result you got?
-bw
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