• Acters@lemmy.world
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    1 year ago

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

    use bitwise &

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