• gerryflap@feddit.nl
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    5 hours ago

    Using Haskell you can write it way more concise:

    iseven :: Int -> Bool
    iseven 0 = True
    iseven 1 = False
    iseven 2 = True
    iseven 3 = False
    iseven 4 = True
    iseven 5 = False
    iseven 6 = True
    iseven 7 = False
    iseven 8 = True
    ...
    

    However, we can be way smarter by only defining the 2 base cases and then a recursive definition for all other numbers:

    iseven :: Int -> Bool
    iseven 0 = True
    iseven 1 = False
    iseven n = iseven (n-2)
    

    It’s having a hard time with negative numbers, but honestly that’s quite a mood

  • affiliate@lemmy.world
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    8 hours ago

    a wise programmer knows to always ask the question “can i solve this problem in python using metaprogramming?” in this instance, the answer is yes:

    def is_even(n: int):
        s = "def is_even_helper(number: int):\n"
        b = True
        for i in range(0, abs(n)+2):
            s += f"\tif (abs(number) == {i}): return {b}\n"
            b = not b
        exec(s)
        return locals().get("is_even_helper")(n)
    
      • Rai@lemmy.dbzer0.com
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        4 hours ago

        It’s really just us… I’ve seen the basic programming joke a bunch of times, but people really aren’t understanding the YanDev/font embellishment. Sad indeed.

    • Fiona@discuss.tchncs.de
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      6 hours ago

      And that isn’t even the worst thing about it…

      The implementation looks like this:

      function isEven(i) {
        return !isOdd(i);
      };
      

      And yes, is-odd is a dependency that in turn depends on is-number

      • drake@lemmy.sdf.org
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        6 hours ago

        but what if number isn’t an integer, or even a number at all? This code, and the improved code shared by the other user, could cause major problems under those conditions. Really, what you would want, is to validate that number is actually an integer before performing the modulo, and if it isn’t, you want to throw an exception, because something has gone wrong.

        That’s exactly what that NPM module does. And this is why it’s not a bad thing to use packages/modules for even very simple tasks, because they help to prevent us from making silly mistakes.

  • bob_lemon@feddit.org
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    16 hours ago
    import re
    
    def is_even(i: int) -> bool:
        return re.match(r"-?\d*[02468]$", str(i)) is not None
    
    • fartripper@lemmy.ml
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      17 hours 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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        16 hours 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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          15 hours 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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              12 hours 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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      17 hours 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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        16 hours 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.

  • lnxtx@feddit.nl
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    17 hours ago

    Ask AI:

    public static boolean isEven(int number) {
        // Handle negative numbers
        if (number < 0) {
            number = -number; // Convert to positive
        }
        
        // Subtract 2 until we reach 0 or 1
        while (number > 1) {
            number -= 2;
        }
        
        // If we reach 0, it's even; if we reach 1, it's odd
        return number == 0;
    }
    
    • istdaslol@feddit.org
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      8 hours ago

      Not all ARM CPUs support mod operations. It’s better to use bit operations. Check if the last bit is set. If set it’s odd else it’s even.