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Thread: Logical XOR in C/C++?

  1. #16
    Join Date
    May 2013
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    Re: Logical XOR in C/C++?

    Quote Originally Posted by NMTop40 View Post
    your best way to do it would be

    (!b1 ^ !b2)


    where b1 and b2 are expressions.
    This does not implement XOR. Try substituting b1<-0, b2<-0
    You can also see it in the xor karnaugh map which cannot be simplified

    simplest way is to use bitwise ^ or (!b1&&b2)||(b1&&!b2)

    cheers!

  2. #17
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    Re: Logical XOR in C/C++?

    Quote Originally Posted by juncode
    This does not implement XOR. Try substituting b1<-0, b2<-0
    If we use the rules from C: (!(-1) ^ !(-2)) = (0 ^ 0) = 0. This is correct.

    Quote Originally Posted by juncode
    You can also see it in the xor karnaugh map which cannot be simplified
    I do not see how that matters here since the proposed alternative expression is not a simplification.

    Quote Originally Posted by juncode
    simplest way is to use bitwise ^
    It was already pointed out that this would not work in general, back in post #4, some 12 years ago.
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  3. #18
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    May 2013
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    Re: Logical XOR in C/C++?

    Of course. I confused ^ with &&. Clearly !bn gives a result of type bool, which can be then correctly used with bitwise xor ^.

    Sorry and thanks for pointing that out!

  4. #19
    Join Date
    Apr 2000
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    Re: Logical XOR in C/C++?

    It depends how you define 'correctly'
    the result of 2 bools being bitwise xor'ed is an int. if you expect the result being of type bool, then you may not be getting the result you want. the "correctest" approach for logical xor behaviour is really using a != operator.
    with != either of the 2 operators is evaluated exactly once, and the result is a bool.


    Using a != will probably also better convey what the code is supposed to do, making it more readable. The only exception I can think of to this is probably trying to actually model logic gates in a circuit.

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