this post was submitted on 12 Sep 2025
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[–] humanspiral@lemmy.ca 2 points 1 hour ago* (last edited 1 hour ago)

easy fix... if infinity return false.

mathematical breakthrough bonus proof: all numbers are neither even nor odd.

[–] mercano@lemmy.world 3 points 3 hours ago (1 children)

When it fails, it at least points you to the site where everyone asks for help.

[–] orhtej2@eviltoast.org 2 points 3 hours ago

Closed as unclear

[–] FiskFisk33@startrek.website 13 points 9 hours ago (3 children)

https://www.npmjs.com/package/is-even

don't look at the weekly downloads if you are faint of heart.

[–] Decq@lemmy.world 7 points 6 hours ago (1 children)

To be fair in a dynamic typed language with dumb string to int coercions, I kinda get why such a would library exists. So it's more a symptom of terrible language design than modern dependency hell.

[–] bobo@lemmy.ml 7 points 5 hours ago* (last edited 5 hours ago) (1 children)

in a dynamic typed language with dumb string to int coercions, I kinda get why such a would library exists.

If string return nan, else % 2

So it's more a symptom of terrible language design than modern dependency hell.

Dependency chain: is-even depends on is-odd which depends on is-number

[–] Decq@lemmy.world 1 points 4 hours ago* (last edited 4 hours ago) (2 children)

If string return nan, else % 2

So now you return a number type if it's a string and a boolean if it's an integer. How does that make sense?

The is-even lib exists to sanitize input by throwing an exception which imho is better.

Edit: having looked at the code better. Apparently it still allows string coercion (boo). It only checks for non integer numbers.

[–] bobo@lemmy.ml 1 points 4 hours ago

Good point, but you can do if === true... and else if === false...

But definitely better to throw an error instead of nan.

[–] Hawk@lemmy.dbzer0.com 6 points 8 hours ago

If you really want to see some horror, follow the dependencies

https://10xengineersqualityprogramming.github.io/ https://www.npmjs.com/package/@falsejs/falsejs This is hilarious, has 262 of the best useless dependencies. In all seriousness though how does anyone ever audit a non package, it's dependency hell!

[–] FiskFisk33@startrek.website 5 points 9 hours ago* (last edited 9 hours ago) (1 children)
isEven(0) ->
    true;
isEven(Num) ->
    isOdd(Num-1).
isOdd(0) ->
    false;
isOdd(Num) ->
    isEven(Num-1).
[–] Knock_Knock_Lemmy_In@lemmy.world 3 points 7 hours ago* (last edited 7 hours ago) (1 children)

Hmm.

isEven(-2)...<out of stack error>

[–] killingspark@feddit.org 2 points 5 hours ago

Nah, tail recursion optimization can just reuse the same stack frame again and again. It's going to loop until it wraps around what ever integer width it has and then tells you if the biggest integer is even or odd. Or, if it's nice, it's going to complain about the wrap around

[–] cogman@lemmy.world 7 points 17 hours ago* (last edited 17 hours ago) (3 children)

Fixed

boolean isOdd(int num) {
  if(num == 1)
    return true;
  if(num > 0)
    return isEven(num - 1);
  else
    return isEven(num + 1);
}

boolean isEven(int num) {
  if(num > 0)
    return isOdd(num - 1);
  else
    return isOdd(num + 1);
}
[–] Valmond@lemmy.world 1 points 7 hours ago

isEeven(∞);

[–] affiliate@lemmy.world 11 points 17 hours ago (1 children)

the downside with this approach is that it will eventually terminate. the version in the original post has the advantage of giving me plenty of time to contemplate life’s many mysteries.

[–] cogman@lemmy.world 4 points 16 hours ago

What can I say, I'm a performance nerd.

[–] Rednax@lemmy.world 1 points 17 hours ago (1 children)

Why the complicated if statements to check the sign? Just let the number overflow. Would be functionaly the same, and result in much prettier code.

[–] cogman@lemmy.world 6 points 16 hours ago

That's a platform dependent change. Overflow is undefined behavior. I'd rather have my code portable so it can run on my Univac 1101.

[–] meme_historian@lemmy.dbzer0.com 45 points 1 day ago (1 children)
[–] Valmond@lemmy.world 2 points 7 hours ago

Oh, Python!

[–] cupcakezealot@piefed.blahaj.zone 45 points 1 day ago* (last edited 1 day ago) (5 children)
bool isOdd(int num) {  
	const oddNumbers = [];  
	for (let i = 1; i <= 10000000; i += 2) {  
  		oddNumbers.push(i);  
	}  
	if (oddNumbers.includes(num) {  
		return true;  
	}  
}  
[–] Malix@sopuli.xyz 16 points 1 day ago (1 children)
[–] dependencyinjection@discuss.tchncs.de 8 points 1 day ago (1 children)

this incident has been reported

[–] moseschrute@piefed.social 11 points 1 day ago* (last edited 1 day ago)

Maybe memo just to be safe, but LGTM!

[–] marcos@lemmy.world 4 points 1 day ago

You should make it oddNumbers.includes(num%10000000)...

[–] schema@lemmy.world 1 points 1 day ago

And if not, unicorns!

[–] a14o@feddit.org 48 points 1 day ago (1 children)

To fix this, add if(num == 255) return true; before line 10.

[–] CannonFodder@lemmy.world 19 points 1 day ago

Peak efficiency there.

But use 2147483647 to be safe.

[–] OpenStars@piefed.social 21 points 1 day ago (1 children)

Boss: don't spend any time on it, just vibe code a solution.

You: sure, I enjoy receiving a salary, what could go wrong?

[–] Aneb@lemmy.world 1 points 15 hours ago
[–] HappyFrog@lemmy.blahaj.zone 44 points 1 day ago (2 children)

Will this ever return? Won't it just overflow the stack?

[–] Valmond@lemmy.world 1 points 7 hours ago

Program it with template meta programming and cause a stack overflow when compiling 🤓😎

[–] sjmarf@sh.itjust.works 74 points 1 day ago (4 children)

Yep, this will cause a stack overflow.

[–] rovingnothing29@lemmy.world 28 points 1 day ago

A mod will appear in my office and claim my problem is a duplicate when it's not?

[–] Mad_Punda@feddit.org 24 points 1 day ago* (last edited 1 day ago)

Might very well be an endless loop because tail recursion can be optimized to reuse the stack frame. Depends on a lot of things of course.

[–] MonkderVierte@lemmy.zip 2 points 1 day ago* (last edited 1 day ago) (2 children)

~~Hm, stack overflow is basically a forkbomb in programming?~~ ok, bullshit.

[–] orhtej2@eviltoast.org 21 points 1 day ago

Forkbomb kills the entire system so not really.

With the stack overflow the runtime will gracefully terminate the program.

[–] calcopiritus@lemmy.world 11 points 1 day ago* (last edited 1 day ago) (1 children)

No.

A stack overflow is a symptom, not the illness. A fork bomb is an illness.

Software coming from the mathematical point of view, assummes it has infinite resources. However, a real computer has many resources that are finite.

CPU time is finite. Memory amount is finite. There is a finite number of network ports. And so on.

A stack overflow just means: "you have run out of this resource called 'the stack'". The stack is a region of the memory. Each thread of each process has 1 stack, and it is not infinite in size. This program will cause a stack overflow because it is infinitely recursive, and each function call will consume a bit of the stack.

A forkbomb is not the end of a finite resource. A fork bomb is a program that uses "forking" to rapidly consume system resources. A fork bomb might cause a stack overflow. Or an out of memory issue. Slow the computer a lot. Or if the OS has a hard limit for process amount, it might reach that limit.

[–] davidgro@lemmy.world 1 points 17 hours ago

A program such as the one in this post is a loop designed (intentionally or not) to run out of stack regardless of how much there is. I'd call that an illness rather than a symptom.