[llvm-commits] bug fix for infinite loop in InstCombine
Guo, Xiaoyi
Xiaoyi.Guo at amd.com
Mon Sep 19 14:27:49 PDT 2011
Thanks for reviewing.
The code you mentioned will eventually fall through to the code below and do the isa<Constant> check there:
+ if (A == tmpOp0 && !isa<Constant>(A)) // A&(A^B) -> A & ~B
+ return BinaryOperator::CreateAnd(A, Builder->CreateNot(B, "tmp"));
Xiaoyi
-----Original Message-----
From: Eli Friedman [mailto:eli.friedman at gmail.com]
Sent: Monday, September 19, 2011 2:21 PM
To: Guo, Xiaoyi
Cc: llvm-commits at cs.uiuc.edu
Subject: Re: [llvm-commits] bug fix for infinite loop in InstCombine
On Mon, Sep 19, 2011 at 2:12 PM, Guo, Xiaoyi <Xiaoyi.Guo at amd.com> wrote:
> Ping?
Oh, hmm... I somehow thought I already had reviewed this.
+ {
+ Value *tmpOp0 = Op0;
+ Value *tmpOp1 = Op1;
+ if (Op0->hasOneUse() &&
+ match(Op0, m_Xor(m_Value(A), m_Value(B)))) {
+ if (A == Op1 || B == Op1 ) {
+ tmpOp1 = Op0;
+ tmpOp0 = Op1;
+ // Simplify below
+ }
It looks like a check for isa<Constant>(Op1) check is necessary to handle some cases?
Otherwise, looks fine.
-Eli
> -----Original Message-----
> From: Guo, Xiaoyi
> Sent: Thursday, September 15, 2011 6:34 PM
> To: 'Eli Friedman'
> Cc: llvm-commits at cs.uiuc.edu
> Subject: RE: [llvm-commits] bug fix for infinite loop in InstCombine
>
> Thanks for reviewing it. Here's another try. See attached new diff.
>
> Xiaoyi
>
> -----Original Message-----
> From: Eli Friedman [mailto:eli.friedman at gmail.com]
> Sent: Thursday, September 15, 2011 5:55 PM
> To: Guo, Xiaoyi
> Cc: llvm-commits at cs.uiuc.edu
> Subject: Re: [llvm-commits] bug fix for infinite loop in InstCombine
>
> On Thu, Sep 15, 2011 at 4:45 PM, Guo, Xiaoyi <Xiaoyi.Guo at amd.com> wrote:
>> Please review the attached fix for a bug in InstCombiner. The patch also includes a new test case which will cause the current InstCombiner to go into an infinite loop.
>>
>> Please help to commit if acceptable.
>>
>> The problem is with the following code in InstCombiner::visitAnd() in InstCombineAndOrXor.cpp:
>>
>> Instruction *InstCombiner::visitAnd(BinaryOperator &I) {
>> bool Changed = SimplifyAssociativeOrCommutative(I);
>> ...
>> if (Op0->hasOneUse() &&
>> match(Op0, m_Xor(m_Value(A), m_Value(B)))) {
>> if (A == Op1) { // (A^B)&A ->
>> A&(A^B)
>> I.swapOperands(); // Simplify below
>> std::swap(Op0, Op1); <==========
>> } else if (B == Op1) { // (A^B)&B ->
>> B&(B^A)
>> cast<BinaryOperator>(Op0)->swapOperands();
>> I.swapOperands(); // Simplify below
>> std::swap(Op0, Op1); <==========
>> }
>> }
>>
>> if (Op1->hasOneUse() &&
>> match(Op1, m_Xor(m_Value(A), m_Value(B)))) {
>> if (B == Op0) { // B&(A^B) ->
>> B&(B^A)
>> cast<BinaryOperator>(Op1)->swapOperands();
>> std::swap(A, B);
>> }
>> // Notice that the patten (A&(~B)) is actually (A&(-1^B)), so if
>> // A is originally -1 (or a vector of -1 and undefs), then we
>> enter
>> // an endless loop. By checking that A is non-constant we ensure
>> that
>> // we will never get to the loop.
>> if (A == Op0 && !isa<Constant>(A)) // A&(A^B) -> A & ~B
>> return BinaryOperator::CreateAnd(A, Builder->CreateNot(B,
>> "tmp"));
>> }
>>
>>
>> With the given test case, op0 is a const and op1 is xor, so the two operands are be swapped in SimplifyAssociativeOrCommutative(). Then I's operands are swapped back at one of the statements marked "<===========" above. Then because A is a constant, a new instruction is not created. So we end up with Changed flag as true but the instruction is not changed and we go into an infinite loop in InstCombiner::DoOneIteration().
>>
>> My fix is to remove the swapOperands() calls, because I don't think they are necessary.
>
> The general idea seems fine. Your fix disables the transform in some cases where both Op0 and Op1 are Xor operators, though.
>
> -Eli
>
>
>
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