• lobut
    link
    fedilink
    arrow-up
    11
    ·
    edit-2
    1 day ago

    Because the abs(3) == 3 is true and that isn’t even.

    An even number of flips would be true and an odd number of flips would be false which works out.

    I was thinking a bitwise & or converting it to a string and testing if the right most character is 0, 2, 4, 6, 8 would be panic mode solutions too.

    • Kraiden@kbin.earth
      link
      fedilink
      arrow-up
      2
      ·
      23 hours ago

      you might be able to do it with a bitwise op? My track record tonight is not great so I’m not going to comment. Have a look at @ImplyingImplactions comment for a loopless solution

      • UID_Zero@infosec.pub
        link
        fedilink
        English
        arrow-up
        2
        ·
        20 hours ago

        Bitwise and with 0x1. If result is 0, it’s even. Least significant bit is always 1 for odd numbers.