• fartripper@lemmy.ml
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      5 days ago

      or divide the number by two and if the remainder is greater than

      -(4^34)
      

      but less than

      70 - (((23*3*4)/2)/2)
      

      then

      true
      
      • superkret@feddit.org
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        5 days ago

        What if the remainder is greater than the first, but not less than the latter?

        Like, for example, 1?

        • prime_number_314159@lemmy.world
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          5 days ago

          Then you should return false, unless the remainder is also greater than or equal to the twenty second root of 4194304. Note, that I’ve only checked up to 4194304 to make sure this works, so if you need bigger numbers, you’ll have to validate on your own.

            • prime_number_314159@lemmy.world
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              4 days ago

              You can just bitwise AND those with …000000001 (for however many bits are in your number). If the result is 0, then the number is even, and if it’s 1, then the number is odd. This works for negative numbers because it discards the negative signing bit.

    • tipicaldik@lemmy.world
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      5 days ago

      I remember coding actionscript in Flash and using modulo (%) to determine if a number was even or odd. It returns the remainder of the number divided by 2 and if it equals anything other than 0 then the number is odd.

      • Korne127@lemmy.world
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        5 days ago

        Yeah. The joke is that this is the obvious solution always used in practise, but the programmer is that bad that they don’t know it and use some ridiculous alternative solutions instead.