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submitted 1 week ago by sirico@feddit.uk to c/programmerhumor@lemmy.ml

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[-] Aedis@lemmy.world 33 points 1 week ago

I'm partial to a recursive solution. Lol

def is_even(number):
    if number < 0 or (number%1) > 0:
        raise ValueError("This impl requires positive integers only") 
    if number < 2:
        return number
    return is_even(number - 2)

[-] tetris11@lemmy.ml 18 points 1 week ago* (last edited 1 week ago)

I prefer good ole regex test of a binary num

function isEven(number){
   binary=$(echo "obase=2; $number" | bc)
   if [ "${binary:-1}" = "1" ]; then
         return 255
   fi
   return 0
}

[-] balsoft@lemmy.ml 8 points 1 week ago* (last edited 1 week ago)

Amateur! I can read and understand that almost right away. Now I present a better solution:

even() ((($1+1)&1))

~~(I mean, it's funny cause it's unreadable, but I suspect this is also one of the most efficient bash implementations possible)~~

(Actually the obvious one is a slight bit faster. But this impl for odd is the fastest one as far as I can tell odd() (($1&1)))

[-] Aedis@lemmy.world 2 points 1 week ago

I'm waiting for a code golf style solution now.

[-] balsoft@lemmy.ml 2 points 1 week ago

I don't think there's much to codegolf. The "obvious" solution (even() (($1%2))) is both shorter and faster. Don't think it can be optimized much more.

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this post was submitted on 15 Jul 2025

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