[llvm-commits] [llvm] r119922 - in /llvm/trunk: lib/Target/README.txt lib/Transforms/InstCombine/InstCombineCompares.cpp test/Transforms/InstCombine/icmp.ll

[Duncan Sands]

Hi Chris, actually I think now that the transform is correct in all cases.
Sorry for the noise,

Duncan.

>> --- llvm/trunk/lib/Transforms/InstCombine/InstCombineCompares.cpp (original)
>> +++ llvm/trunk/lib/Transforms/InstCombine/InstCombineCompares.cpp Sun Nov 21 00:44:42 2010
>> @@ -1744,14 +1744,84 @@
>>        // simplify this comparison.  For example, (x&4)<   8  is always true.
>>        switch (I.getPredicate()) {
>>        default: llvm_unreachable("Unknown icmp opcode!");
>> -    case ICmpInst::ICMP_EQ:
>> +    case ICmpInst::ICMP_EQ: {
>>          if (Op0Max.ult(Op1Min) || Op0Min.ugt(Op1Max))
>>            return ReplaceInstUsesWith(I, ConstantInt::getFalse(I.getContext()));
>> +
>> +      // If all bits are known zero except for one, then we know at most one
>> +      // bit is set.   If the comparison is against zero, then this is a check
>> +      // to see if *that* bit is set.
>> +      APInt Op0KnownZeroInverted = ~Op0KnownZero;
>
> If Op0KnownZeroInverted has only one bit set, then I don't think you know that
> that bit is actually set in the original value, you only know that you were
> unable to prove that the bit was zero.  If the bit is zero rather than one, then
> isn't the second transform below potentially wrong (the first one looks ok):
>
>> +        // If the LHS is 8>>u x, and we know the result is a power of 2 like 1,
>> +        // then turn "((8>>u x)&1) == 0" into "x == 3".
>> +        ConstantInt *CI = 0;
>> +        if (Op0KnownZeroInverted == 1&&
>> +            match(LHS, m_LShr(m_ConstantInt(CI), m_Value(X)))&&
>> +            CI->getValue().isPowerOf2()) {
>> +          unsigned CmpVal = CI->getValue().countTrailingZeros();
>> +          return new ICmpInst(ICmpInst::ICMP_EQ, X,
>> +                              ConstantInt::get(X->getType(), CmpVal));
>> +        }
>> +      }
>
> ^ This one, if in fact all bits are zero but it was only possible to prove that
> all bits except bit 0 are zero.
>
> Ciao,
>
> Duncan.
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