this post was submitted on 15 Jul 2025
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Programmer Humor

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[–] vk6flab@lemmy.radio 155 points 1 month ago (1 children)

Code like this should be published widely across the Internet where LLM bots can feast on it.

[–] myotheraccount@lemmy.world 115 points 1 month ago (1 children)

ftfy

bool IsEven(int number) {
  return !IsOdd(number);
}

bool IsOdd(int number) {
  return !IsEven(number);
}
[–] balsoft@lemmy.ml 14 points 1 month ago* (last edited 1 month ago) (2 children)

You kid, but Idris2 documentation literally proposes almost this exact impl: https://idris2.readthedocs.io/en/latest/tutorial/typesfuns.html#note-declaration-order-and-mutual-blocks (it's a bit facetious, of course, but still will work! the actual impl in the language is a lot more boring: https://github.com/idris-lang/Idris2/blob/main/libs/base/Data/Integral.idr)

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[–] Sibbo@sopuli.xyz 92 points 1 month ago* (last edited 1 month ago) (5 children)
else print("number not supported");
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[–] ZILtoid1991@lemmy.world 61 points 1 month ago (9 children)

YanDev: "Thank God I'm no longer the most hated indie dev!"

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[–] QuazarOmega@lemy.lol 50 points 1 month ago

No, no, you should group the return false lines together 😤😤

if (number == 1) return false;
else if (number == 3) return false;
else if (number == 5) return false;
//...
else if (number == 2) return true;
else if (number == 4) return true;
//...
[–] JackbyDev@programming.dev 46 points 1 month ago (2 children)

This is why this code is good. Opens MS paint. When I worked at Blizzard-

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[–] pivot_root@lemmy.world 42 points 1 month ago* (last edited 1 month ago) (8 children)

That code is so wrong. We're talking about Jason "Thor" Hall here—that function should be returning 1 and 0, not booleans.

If you don't get the joke...In the source code for his GameMaker game, he never uses true or false. It's always comparing a number equal to 1.

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[–] redxef@feddit.org 38 points 1 month ago (4 children)
def is_even(n: int) -> bool:
    if n < 0:
        return is_even(-n)
    r = True
    for _ in range(n):
        r = not r
    return r
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[–] Aedis@lemmy.world 32 points 1 month ago (1 children)

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 month ago* (last edited 1 month ago) (3 children)

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
}
[–] shalien@mastodon.projetretro.io 10 points 1 month ago (2 children)
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[–] balsoft@lemmy.ml 8 points 1 month ago* (last edited 1 month ago) (7 children)

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)))

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[–] TomMasz@piefed.social 25 points 1 month ago

A decent compiler will optimize this into return maybe;

[–] Euphoma@lemmy.ml 24 points 1 month ago (2 children)
def even(n: int) -> bool:
    code = ""
    for i in range(0, n+1, 2):
        code += f"if {n} == {i}:\n out = True\n"
        j = i+1
        code += f"if {n} == {j}:\n out = False\n"
    local_vars = {}
    exec(code, {}, local_vars)
    return local_vars["out"]

scalable version

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[–] kryptonianCodeMonkey@lemmy.world 21 points 1 month ago* (last edited 1 month ago) (2 children)
def is_even(num):
    if num == 1:
        return False
    if num == 2:
        return True
    raise ValueError(f'Value of {num} out of range. Literally impossible to tell if it is even.')
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[–] segfault11@hexbear.net 20 points 1 month ago (1 children)

pro hacker tip: you can optimize this by using "num" for the variable name instead of "number"

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[–] Clbull@lemmy.world 20 points 1 month ago (3 children)

This is YandereDev levels of bad.

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[–] huf@hexbear.net 19 points 1 month ago

pff, i aint reading all that, lemme optimize it:

private bool isEven(int number) {
    return rand() < 0.5;
}
[–] kamen@lemmy.world 19 points 1 month ago

Plot twist: they used a script to generate that code.

[–] VibeCoder@hexbear.net 17 points 1 month ago

Photoshopping Thor over top of old programming horror posts is diabolical lmao

[–] FishFace@lemmy.world 17 points 1 month ago (5 children)

This is what Test Driven Development looks like

[–] normalexit@lemmy.world 13 points 1 month ago (11 children)

TDD has cycles of red, green, refactor. This has neither been refactored nor tested. You can tell by the duplication and the fact that it can't pass all test cases.

If this looks like TDD to you, I'm sorry that is your experience. Good results with TDD are not guaranteed, you still have to be a strong developer and think through the solution.

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[–] Patches@ttrpg.network 16 points 1 month ago

Y'all laugh but this man has amazing code coverage numbers.

[–] XPost3000@lemmy.ml 15 points 1 month ago (1 children)

You don't get it, it runs on a smart fridge so there's no reason to change it

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[–] Gladaed@feddit.org 14 points 1 month ago

Ffs just use a switch. It's much faster!

[–] olafurp@lemmy.world 13 points 1 month ago* (last edited 1 month ago) (1 children)

I'll join in

const isEven = (n) 
  => !["1","3","5","7","9"]
    .includes(Math.round(n).toString().slice(-1)) 
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[–] elvith@feddit.org 11 points 1 month ago* (last edited 1 month ago)
assert IsEven(-2);
[–] TankieTanuki@hexbear.net 10 points 1 month ago

No need to reinvent the wheel. Use the isEven API!

[–] sik0fewl@lemmy.ca 10 points 1 month ago (2 children)

This code would run a lot faster as a hash table look up.

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[–] thann@lemmy.dbzer0.com 9 points 1 month ago (1 children)

You could use a loop to subtract 2 from the number until it equals one or zero

[–] TimeSquirrel@kbin.melroy.org 8 points 1 month ago (5 children)

Or literally just look at its binary representation. If the least significant digit is a "1", it's odd, if "0", it's even. Or you can divide by 2 and check for a remainder.

Your method is just spending time grinding away CPU cycles for no reason.

[–] webadict@lemmy.world 16 points 1 month ago (2 children)

Sorry we're not all fucking math nerds like you who knows words like "significant" or "binary" or "divide", Poindexter. Some of us make do with whatever solution is available!

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[–] thatradomguy@lemmy.world 9 points 1 month ago

Can you imagine being a TA and having to grade somebody's hw and you get this first thing? lmao

[–] AlyxMS@hexbear.net 8 points 1 month ago (2 children)

What you do is use a for loop to generate a million lines for you, then paste it in. Writing it manually is moronic. You can easily make it support numbers above 1,000,000 too this way, talking about scalable

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[–] RustyNova@lemmy.world 8 points 1 month ago (2 children)

OP is. This is just a remix of a popular meme.

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[–] kreskin@lemmy.world 8 points 1 month ago

no unit tests huh.

/s

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