Comment on I wish

DarkMessiah@lemmy.world ⁨1⁩ ⁨year⁩ ago

• Acters@lemmy.world ⁨1⁩ ⁨year⁩ ago

  Every bit aside for the ones bit is even, all you have to do is yet the ones bit(the far left) for it being a 1 or 0. Which is the fastest and least amount of code needed.

  use bitwise &
  bool isEven(int n)
  {

  // n&1 is true, then odd, or !n&1 is true for even  
  
  return (!(n & 1));  
  

  }

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• happyhippo@feddit.it ⁨1⁩ ⁨year⁩ ago

  Huh?

  return number % 2 == 0

  That’s the only sane solution.

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  • DarkMessiah@lemmy.world ⁨1⁩ ⁨year⁩ ago

    Do note how I said “a decent” way, not “the best” way. Get that huh outta here.

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• nopt@lemmy.world ⁨1⁩ ⁨year⁩ ago

  Or modulo %

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  • oatscoop@midwest.social ⁨1⁩ ⁨year⁩ ago

     private bool IsEven(int number){
        return number % 2 ? false : true;
     }
    

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• 257m@sh.itjust.works ⁨1⁩ ⁨year⁩ ago

  number % 2 == 0
  and
  // If integer
  (number & 0b1) == 0
  

  Are the only sane ways to do this. No need to floor.

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  • homura1650@lemmy.world ⁨1⁩ ⁨year⁩ ago

    If you are using floats, you really do not want to have an isEven function …

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    • 257m@sh.itjust.works ⁨1⁩ ⁨year⁩ ago

      Whats the alternative a macro? An inline function is perfectly fine for checking if a nunber is even. Compiler will probably optimize it to a single and instruction.

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      • homura1650@lemmy.world ⁨1⁩ ⁨year⁩ ago

        No. The alternative is to not use a float. Testing if a float is even simply does not make sense.

        Even testing two floats for equality rarely makes sense.

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• UNWILLING_PARTICIPANT@sh.itjust.works ⁨1⁩ ⁨year⁩ ago

  Take out the else and I’m in

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  • DarkMessiah@lemmy.world ⁨1⁩ ⁨year⁩ ago

    Valid point.

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