Logical Question

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Revision as of 25 June 2013 at 20:47.
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Logical Question

Is there any kind of bit operation or arithmetical trick that would allow me to do something like integerWhichCouldOnlyHaveAValueOfZeroOrOne == 0 ? otherIntegerWhichCouldOnlyHaveAValueOfZeroOrOne == 0 ? 0 : 1 : 2 without so many expensive conditionals?

    Sheldor (talk)16:20, 25 June 2013

    Like this?

    integerWhichCouldOnlyHaveAValueOfZeroOrOne | otherIntegerWhichCouldOnlyHaveAValueOfZeroOrOne + integerWhichCouldOnlyHaveAValueOfZeroOrOne

      MN (talk)18:02, 25 June 2013

      Yes, I think that would work.

      Thanks

        Sheldor (talk)18:58, 25 June 2013
         

        If there is no possibility that integerWhichCouldOnlyHaveAValueOfZeroOrOne == 1 and otherIntegerWhichCouldOnlyHaveAValueOfZeroOrOne == 0, or if you don't care, then a simple addition would work.

          Skilgannon (talk)19:59, 25 June 2013

          There is a chance that the first and second integers would add up to 1 in two different scenarios. I do care to avoid this because I was planning to use this in VCS segmentation. It would not be prudent to represent two very different situations by the same segment.

            Sheldor (talk)20:29, 25 June 2013

            If you're using it for wall segmentation, it should be fine just adding. The chances of the closer one being triggered but not the further is quite rare (although it can happen, I admit).

              Skilgannon (talk)21:03, 25 June 2013

              If both wall checks be either 1 or 0 at the same time a vast majority of the time, what's the point of even having a second wall check?

                Sheldor (talk)21:41, 25 June 2013

                Hang on, is this two forward wall checks, or one forward and one reverse?

                I was thinking two forward, where obviously the check which extends further will also be triggered in 99% of cases where the one that extends less is triggered.

                If this is one forward, one reverse then I think they should actually be in different segments.

                  Skilgannon (talk)21:47, 25 June 2013