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Thanks for your answers. Another example of unsound transformation on Boolean algebra. According to the LLVM documentation (http://llvm.org/docs/LangRef.html#undefined-values) it is unsafe to consider ' a & undef = undef ' and ' a | undef = undef ' but 'undef xor undef = undef' is safe. Now, given an expression ((a & (~b)) | ((~a) & b)) where a and b are

...ef ^ B = undef > > Thus the transform is correct. > > A is not undef but B is undef: > Similar reasoning follows. > > -- Sanjoy -------------- next part -------------- An HTML attachment was scrubbed... URL: <http://lists.llvm.org/pipermail/llvm-dev/attachments/20160914/2f61b9d5/attachment.html>