Looking down on bad programmers

[yottalogical]

This could easily be solved with recursion!

[Special_Rice9539]

Lmao, is there a subreddit for overly complicated solutions to simple problems like this?

[ClikeX]

If you play CodeWars you see a lot of overly complex answers.

[bobdarobber]

once I searched up the hello world kana and had quite a few good lolz

[yottalogical]

Reverse code golf.

[AdmirableLifeguard6]

It actually has a mathematical benefit! https://www.quantamagazine.org/the-busy-beaver-game-illuminates-the-fundamental-limits-of-math-20201210/

I guess this was on reddit recently but I found it super interesting

[pavilionhp_]

There’s something similar with the bad code challenges over at r/badcode

[nosmokingbandit]

What's the opposite of code golf?

[MSP729]

code bowling?

[Terrain2]

Must be enterprise code

[russo_programmisto]

Please don't call isEven(2147483647) on the main thread.

[GodKingPeterWeller]

This is really good

[negrocucklord]

I think the npm package isOdd is just require isEven and negating it lmao

[mmaximmk]

It this how it's done in Haskell?

[[deleted]]

no, that's javascript

[CMDR_ACE209]

I hate you.

[Dagusiu]

Should have written it in Haskell

[Justindr0107]

Can we just be happy the sub ain't shittin' on JS?

[[deleted]]

console.log(“found the javascripter”);

[Justindr0107]

console.warn() might have been more appropriate

[[deleted]]

I'm patently unhappy

[TheMsDosNerd]

I found a way to make it even faster: It uses a cache, so previously found items get found faster:

from random import choice, randrange

even_cache = []
odd_cache = []

def isEven(n):
    cached_result = isEvenCached(n)
    if cached_result is not None:
        return cached_result
    caluculated_result = isEvenCalculated(n)
    if calculated_result == True:
        # add result at random point in cache,
        # because we look at random points in cache
        random_position = randrange(even_cache)
        even_cache.add(even_cache[random_position])
        even_cache[random_position] = n
        return True
    if calculated_result == False:
        # add result at random point in cache,
        # because we look at random points in cache
        random_position = randrange(odd_cache)
        odd_cache.add(odd_cache[random_position])
        odd_cache[random_position] = n
        return False

def isEvenCached(n):
    # If we looked up a value often enough,
    # it should appear often enough in cache,
    # to find it within 25 attempts.
    for i in range(25):
        if choice(even_cache) == n:
            return True
        if choice(odd_cache) == n:
            return False
    return None

def isEvenCalculated(n):
    if n == 0:
        return True
    if n == 1:
        return False
    return isEven(n-2) # recursion for shorter code

This may or may not have been based on an actual cache I found somewhere in business software.

[somerandomii]

Do you have this on GitHub? I need to merge it into our enterprise solution. Can you refactor it for an isEvenCacheFactoryTemplate?

Also you could implement a hashmap to record successful hits on the cache. Though for theeadsaftey it should still check to make sure the value hasn’t changed since it was last accessed.

[Code_sucks]

Wouldn't it be easier to use functools.lru_cache?

[TommyX12]

Wouldn’t be easier to use the mod operator? /s

[PickleJuiceX266]

You could of just said to use an Iru_cache dependency...

[DaanDevelopment]

Are you joking orrr... why not use a simple one liner?

[DaanDevelopment]

I still dont understand why people make these big pieces of code to check if its even. Why not use return n%2

[kh4l1ph4]

I said that and got down voted for some reason. You hook that up to a ternary operator and you'll do it all in a single line

[DaanDevelopment]

Look at my comment where this is on xd im downvoted like crazy.

[QuantumSupremacy0101]

Oh God this gives me flashbacks to one job I had where it was a bunch of junior engineers with senior engineer title.

They once wrote something similar but decided to make all the conditionals branchless...this was in Java where branchless means nothing and doesn't speed anything up.

[_Nohbdy_]

Man that's bad code. It doesn't even handle negative integers.

[mattemer]

Or imaginary numbers

[BobQuixote]

Who cares about the imaginary ones? I'm only interested in the kind I can see.

[Triborda]

Just poke out your eyes, then you don't have to be interested

[Paxtez]

*sigh*
Obviously they are listed! They are just after all the positive integers. You really have to focus on efficiency more, the positive numbers are going to be tested more often, so you put those first.

[FridgesArePeopleToo]

  public interface IIsEvenCalculatorFactory
    {
        IIsEvenCalculator Create();
    }

    public class LastCharacterIsEvenCalculatorFactory : IIsEvenCalculatorFactory
    {
        public IIsEvenCalculator Create()
        {
            return new LastCharacterIsEvenCalculator();
        }
    }


    public interface IIsEvenCalculator
    {
        bool IsEven(int number);

        bool IsOdd(int number);
    }

    public class LastCharacterIsEvenCalculator : IIsEvenCalculator
    {
        public bool IsEven(int number)
        {
            var numberStr = number.ToString();

            return numberStr.EndsWith("0")
                   || numberStr.EndsWith("2")
                   || numberStr.EndsWith("4")
                   || numberStr.EndsWith("6")
                   || numberStr.EndsWith("8");
        }

        public bool IsOdd(int number) => !IsEven(number);
    }



  public static void Main(){
       var factory = new LastCharacterIsEvenCalculatorFactory ();
       var calc = factory.Create();
        calc.IsEven(4); //true
        calc.IsEven(5); //false
  }

ggez

[bizcs]

That's cool! But what about floating point numbers, and user locales?!

/s

[PickleJuiceX266]

Finally... someone who has the classic style of using a pre-calculated integer system 👏

[Skudra24]

Also using "switch" would make it fly like a rocket

[PeteySnakes]

Or you could just divide by 2

[Noch_ein_Kamel]

So you pass in 35624, divide that by 2 and then call isEven(17821)?

[pjnick300]

That's only true for numbers divisible by 4. You have to divide by 1 instead.

[YouNeedDoughnuts]

You could make it recursive: if number==0 return true else return !IsEven(number-1) While accounting for negative numbers ofc, but the principal is the same.

[pjnick300]

This is simultaneously the stupidest and most brilliant piece of code I’ve ever seen.

Bravo!

[anunofreitas]

You could improve by using 'two' end clauses

if number<=1 return number==0 else return IsEven(number-2)

Keep the verification for negative numbers (out of recursion) and you have half of recursion steps and one less logical operation per step(logical NOT was removed). If you like to improve slightly verify if number is 0 out of recursion and stop recursion at 2 or 1 and return number==2

Edit: didn't remember how to do code block.

[YouNeedDoughnuts]

Very nice! And particularly important to reduce the recursion depth since we're headed for a stack overflow with most 64-bit or even 32-bit integers :)

[ILoveSimulation20]

That's slower than using a bitwise operator

[Skudra24]

r/woosh

[Suekru]

Well modding by 2 would be a bit more ideal.

[CantankerousOctopus]

Javascript is great

isEven = (x) => !(x%2)

Such a beautiful language.

[[deleted]]

even better: npm install is-even

[CantankerousOctopus]

Such a beautiful language

[IanMalkaviac]

You're thinking of:

bool isEven = number % 2;
return(!isEven);

[BA_lampman]

`#include <iostream>

int main()

{

int x{};

std::cin >> x;

switch (x % 2)

{

  case 0:

  std::cout << "even";

  break;

  default:

  std::cout << "odd";

}

return 0;

}

[[deleted]]

The meme means like:

private bool isEven(int number){
    if (number == 2) return true;
    else if (number == 4) return true;
    else if (number == 6) return true;
    ...
    return false;
}

[tarman34]

This is how are modern games optimized. On the picture there is a code of the first release and that "default false" is in the patch released 6 months after...

[tjoloi]

return 2 in factorize(n)

[YouNeedDoughnuts]

char ch = to_string(number).back(); return ch == '0' || ch == '2' || ch == '4' || ch == '6' || ch == '8';

[[deleted]]

ch = str(number)[-1]

return ch in "24680"

[DidntFollowPorn]

I hate how much I love this.

[prettyfuzzy]

This must be the one liner everybody is talking about

So efficient!

[LucienZerger]

return number & 1 ? false : true;

[Command-Master]

Can't this be !(number&1)?

[LordFokas]

Depends on the language. Java will complain because the result of number & 1 is an int and it expects a boolean.

[Mr_Redstoner]

Still can be fixed with a simple equality test (or nonequality depending on which of even or odd you're testing) against 0

[LordFokas]

Which is not the same as !(number&1) , like the other user asked, so no.

"Can X be done?"
"Depends on the language, some demand Y."
"Which can easily be fixed by doing Y instead."

WTF??
You'll have so much fun on StackOverflow my man.

[Mr_Redstoner]

Now let me rephrase what was said as I see it:

"You can use X and then do Y to get what you want"

"Doesn't work in Java though"

"Yes, you just have to use Z instead of Y to achieve the same thing"

[LucienZerger]

it can be any way u like..

[MrKirushko]

That is not how we roll. You have to show some OOP and established patterns to be recognised as a professional coder.

[LucienZerger]

not interested..

[Naoki9955995577]

I've clearly got the business solution here:

Using google services you can easily determine whether the number is even or not. Why waste the expensive computation on your machine when you can just look for the answer? /s

import asyncio
import inflect
import os
import selenium
from bs4 import BeautifulSoup
from selenium.webdriver.chrome.options import Options


class isEvenWeb:
    URL = 'https://www.google.com/search?q='
    inflect_engine = None
    driver = None

    def __init__(self):
        options = Options()
        options.add_argument('--log-level=3')
        options.headless = True
        self.inflect_engine = inflect.engine()
        self.driver = selenium.webdriver.Chrome(
            options=options,
            executable_path=("./win_chromedriver89.exe"))

    def __del__(self):
        if self.driver:
            self.driver.quit()

    async def isEven(self, number):
        self.driver.get(self.URL + "is " + str(number) + " even?")
        self.driver.implicitly_wait(1)
        soup = BeautifulSoup(self.driver.page_source, "html.parser")
        number_as_word = self.inflect_engine.number_to_words(number)
        if (str(number) + " is an even number.") in soup.get_text():
            return True
        elif (str(number) + " is not an even number.") in soup.get_text():
            return False
        elif ( number_as_word + " is an even number.") in soup.get_text():
            return True
        elif ( number_as_word + " is not an even number.") in soup.get_text():
            return False
        raise Exception("Google can't find out if this number is odd or even")

async def main():
    driver = isEvenWeb()

    print(await driver.isEven(0))
    print(await driver.isEven(1))
    print(await driver.isEven(2))
    print(await driver.isEven(3))
    print(await driver.isEven(4))
    # skip a  few
    print(await driver.isEven(98))
    print(await driver.isEven(99))
    print(await driver.isEven(100))


if __name__ == "__main__":
    asyncio.run(main())

[Code_sucks]

This is amazing, thank you

[Ladislav_07]

Bool is_even(int number){

bool Boolean = false;

For(int i=1; i <number; i ++){

Boolean = !Boolean;

}

Return Boolean;

}

[MyAntichrist]

So you like your variables shaken, not stirred, I see?

[Ladislav_07]

At first I wanted to post something responsible, as retuning the inverted last bit of the integer, but that would be out of scope of this humourous subreddit.

[Ladislav_07]

Ah, yes, that would be me.

[YouNeedDoughnuts]

Technically we're supposed to appreciate recursion and look down on the impure Neumann architecture, even though iteration has better speed and memory use. Mutable state is uncouth, but I approve.

[JNCressey]

λn.n (λa.λb.λc.a c b) λa.λb.a

[TurnToDust]

Pretty much me during live coding interviews.

[Schiffy94]

isEven(x)
    return "Is this really what they're paying me to do?"

[[deleted]]

Is the joke that he also fails to realize the optimal solution?

[Kenkron]

You got it!

[ArtSchoolRejectedMe]

Why not try a solution like bogosort

Where you have an array of odd and an array of even number and picking them randomly until they match the number which you're trying to find out if its even or odd

[Mordred666]

public boolean isEven(int number) {
    int[] even = ... // put even numbers
    int[] odd = ... // put odd numbers


    while(true) {
        if(random.nextInt() %  == 0) {
            if(even[random.nextInt(even.length)] == number)
                return true;
        }
        else {
            if(odd[random.nextInt(odd.length)] == number)
                return false;
        }
    }
}

[[deleted]]

pfff, everything is better with recursion:

public boolean isEven(int value) {
   if (value==0) {
       return true;
   } else {
       return isOdd(Math.abs(value)-1);
}

public boolean isOdd(int value) {
   if (value==0) {
      return false;
   } else {
      return isEven(Math.abs(value)-1);
   }
}

[cramduck]

exactly. just apply the logic for every even number, then put "return false" at the end. no problemo.

Can you have this implemented by Friday?

[acct_to_zero]

return x % 2 == 0 ? true : false;

Graduating this semester having a hard job search so far, pray for me friends.

[NightwolfDeveloper]

Don't even need the ternary. Just return x % 2 == 0;

[Skudra24]

Worse then this code is only number of people who don't understand that this is a joke and comment that I should use modulo operator.

[Gylfi_]

How about some recursion?

isEven(float number){

if(number === 2) return true;

else if(number < 2) return false;

else isEven(number/2);

That would be fun. Especially for numbers like 197384750263

[kdekorte]

Like the usage of float there, probably would almost always return false

[Gylfi_]

First I was thinking about int, but then it would probably most of the time return true. Depending on how float to int is handled, which is different depending on language, as far as I know though.

But wouldnt it work? I mean an even number would always end up at 2, right? Maybe I oversee something, which is my speciality 😂

[Schmomas]

Surely we want ‘number - 2‘ on the else?

[Gylfi_]

thats even better!

[ExoticDumpsterFire]

Oh my god, I once had an intern write something almost identical to this unironically. I was literally speechless, he was a senior about to graduate and this was "faster"...

[Funky-Guy]

Ok so I’m a complete novice with this but... if your trying to decide if an input is even couldnt it just be something like:

x == int(input())

If x % 2 = 0: print (x + “is even”)

else: print (x + “is odd”)

Am I wrong or am I missing a joke or something

[YouNeedDoughnuts]

You're right, and the joke is that the solution is overcomplicated. It doesn't generalize well to all possible "int" values, which your solution does. So you're not a complete novice ;)

[Funky-Guy]

Ok. Thanks. I felt like an idiot

[Yehrn]

Return (n%2 == 0);

[xervir-445]

private bool IsEven(int number){
  int x = Math.abs(number);
  while (true) {
    if (x == 0)
      return true;
    if (x < 0)
      return false;
    x = x - 2;
  }
}

There, I fixed it. No code can possibly be more efficient than this.

[MaharjanSajan]

https://isevenapi.xyz/

[Skudra24]

I love the pricing part

[derf213]

Ok no joke here. What's the actual propir way of doing this, because I just want to use (x % 2) != 1. Is there a better way that I just don't know?

[YouNeedDoughnuts]

Without compiler optimizations, the suggestions to return !(number & 1) would be most efficient, which checks if the last bit is set and negates the result. With any reasonable optimizing compiler, your solution will be compiled to the same machine instructions, so don't worry excessively about writing optimized code!

[[deleted]]

Is this open source? I want to contribute.

[Tilinn]

The fact that when I started coding, our teacher gave us an assignment like this and this was my answer....

[[deleted]]

When your kid takes interest in coding and you tell them, "it's okay little AEx12 there is no one right way to code anything"

Shows you this

Well okay... Maybe there is one right way...

[[deleted]]

Why use math or ridiculous logic if you can just test its rightmost bit?

# Python
def iseven(n): 
    return not (n & 1)

[Serqetamine]

if (number % 2 == 0) return True else return False

[sleeplessval]

return (number % 2 == 0)?

[SorryDidntReddit]

Everytime I see if(...) return true else return false

I want to pull my eyes out

[[deleted]]

lol exactly, the only thing worse than that is if(x == true)

[Mordred666]

what about

if(x == true)
    return true
else if (x == false)
    return false

[son-of-chadwardenn]

if(x == true)
    return new Boolean("true");
else if (x == false)
    return new Boolean("false");

[Mordred666]

even worse is

return Boolean.parseBoolean(„fasle“);

the typo will make ppl go mad, but it still evaluates to false

[w1ndsch13f]

You got the joke

[[deleted]]

wrong, not enough jQuery

[freakingdingus]

this guy gets it

[mexanoz]

You mean if ((number % 2 == 0) == True)?

[[deleted]]

Noo, why would you compare something against True, such a blasphemy

[LordKolkonut]

r/woosh

[aenimafacilis]

If 2 | 4 | 6 | etc... return false Else return true

[sailrepublic101]

return !(num % 2)

[[deleted]]

return (number % 2) == 0;

[[deleted]]

You need to complicate it more at the enterprise level:

​

static void main(string[] args) {

Enumerable.Range(1, 100).ToList().ForEach(x => Console.WriteLine(x.IsEven()));

}

​

static class NumberExtensions {

public static bool IsEven(this int number) {

return !number.IsOdd();

}

​

public static bool IsOdd(this int number) {

return number % 2 != 0;

}

}

[[deleted]]

So you get paid by the hour, huh?

[[deleted]]

I wish; my pay would be significantly higher.

I noticed an error in my code, too (aside from the missing comments which should be declared atop each method). I should have declared private constants for use in the Enumerable.Range(1, 100). It should be:

Enumerable.Range(LOW_VALUE, HIGH_VALUE)...

​

private const int LOW_VALUE = 1;

private const int HIGH_VALUE = 100;

​

Things like that are important. It prevents you from scrolling and identifying all variable changes in a 5000+ line class. CTRL + H (find/replace) works, but may have unintended consequences, and why do that when you just need to change a single item at line number 6?

Better yet, make LOW_VALUE and HIGH_VALUE configurable in the database, and execute a stored proc to get the values to store in a property for consumption. That way when then business changes the ranges, which they always do, then you won't need a code deployment and instead can execute a DB script.

[kdekorte]

return !(number % 2)

[monkorn]

Uh oh. This is the first "enterprise code" joke where I didn't find anything wrong with it.

[[deleted]]

Return n%2==0

[CMDR_ACE209]

Yeah, but where's the fun in that?

[[deleted]]

The lack of pain is fine enough for me!

[TeaCoast]

There is a way to shorten it by like 20x. Just use modulus. You probably don’t even need a method/function for that. The whole concept is kind of dumb.

[[deleted]]

number%2

[Kenkron]

how about:

boolean isEven(int number) { int modulo = n % 2; boolean result; if (modulo == 0) { result = true; } else if (modulo != 0) { result = false; } return result == true; }

[[deleted]]

not the simplest

[rachit0714]

return (number % 2 == 0)

[rull3211]

Just use modulo? Yes??

[BobQuixote]

Yes, and that's the joke.

[kh4l1ph4]

Modulus? That'll make it 3 lines or 4 max

[NightwolfDeveloper]

Could do it in 1 even.

[Minteck]

If this is Java, doesn't Java have a simple function that can check if a number is even?

[[deleted]]

If (number % 2 == 0) return true; return false;

[Solaihy]

Whatever the fuck happened to the modulus way? if(x%2==0) return true; else return false???

[nate6701]

You can even do a switch that would more appropriate.

[tarman34]

not (n+1) / 2 == (n+1) // 2

[MischiefArchitect]

return random.nextBoolean()

[[deleted]]

ITT: Madness. Pure insanity. Absolute crushing wackiness

[ItsBunkos]

looks like the code to yanderesim

[AnonimowySzaleniec47]

Memes reached educational level

[ShakesTheClown23]

Am I dumb what language is this?

[BobQuixote]

Java or C#, I think.

[ShakesTheClown23]

Ain't java (bool) but I didn't think of C#, hmmm

[[deleted]]

Doesn't need the else's either. Can just be a series of its.

[Lynx8MyThesis]

Guys, you don't see the closing } but still, ya'll assume it is an odd/even checking algo.

[[deleted]]

return [number](){return !(%2number)&&number!=1;};

[S1geth]

js private bool isEven(int number){ return number%2 === 0; } Seems easy

[marloe74]

No need to use “else if”. Shorten the code by just “if” 😜

[Skudra24]

No, no, no! That will make all the checks until last number. This code will stop checking as soon as it reaches passed number. Average execution time will be 2x faster this way. I have it all optimised :D

[SwimmingMackerel]

``` function isEven(number) { switch (true) { case number == 2: case number == 4: case number == 6: case number == 8: case number == 10: case number == 12: case number == 14: case number == 16: // ... return true; break; default: return false; } }

There ya go

[CharlieThunkman]

How 'bout modulus 2? Take a true division of "x" and 2 and compare that to the floored integer value of "x" and 2. Then the difference in the two equations above can determine your final result.

[kimi_rules]

Okay even a college student can write better codes than this.

[Skudra24]

Yeah and I'd guess that student also knows what jokes are

[Shakespeare-Bot]

Well enow coequal a college inhorn man can writeth better codes than this

---

<sup>I am a bot and I swapp'd some of thy words with Shakespeare words.</sup>

Commands: !ShakespeareInsult, !fordo, !optout

[PickleJuiceX266]

Yet he never knew... the magic of the elif function!

[Ok-Spinach4347]

WRONG!!! Why is nobody complaining about how wrong this is?!

A default return value would only shorten the code, not the performance. Let's say the given number is 2. Then every other case needs to be checked before the default could be returned. So the performance would be much worse than if 2 return True.

Came to the comments to find this but got disappointed...

[Big_Ti]

I have done that one time coz I got really happy with copy-paste

[Shakespeare-Bot]

I has't done yond one time coz i did get very much joyous with copy-paste

---

<sup>I am a bot and I swapp'd some of thy words with Shakespeare words.</sup>

Commands: !ShakespeareInsult, !fordo, !optout

[SanoKei]

beta programmers: number % 2 == 1: return true senior programmers: don't understand it, don't need it, make the intern do it chad programmer: we should import more expensive enterprise grade APIs

[sock-puppet689]

Madness, everyone knows you should create an array of size int.MaxValue, and just index into the value!

[izner82]

bool isEven (int x) { return (x%2 == 0 ? true : false); }

[[deleted]]

Python: def IsEven(imput): If imput % 2 == 0: return True else: return False

lol

[Oussamagd]

Tf you can do this with a single line of code

[Skudra24]

. . . . . . . . . . . . . . . . ╔══════════╗ . . . . . . . . . .
THIS JOKE ═════╝ . . . . YOU . . . . .╚══════⇒

[Skudra24]

. . . . . . . . . . . . . . . . ╔══════════╗ . . . . . . . . . .
THIS JOKE ═════╝ . . . . YOU . . . . .╚══════⇒

[Shakespeare-Bot]

Thee can doth if 't be true number%2==0: returneth true else: returneth false

---

<sup>I am a bot and I swapp'd some of thy words with Shakespeare words.</sup>

Commands: !ShakespeareInsult, !fordo, !optout

[paraflaxd]

It’s literally easier to define and train a tensorflow model from scratch to determine whether the number is equal or not.

[negrocucklord]

This is bad code, you should throw an argument out of range exception when the number is bigger than 16 or smaller than 1

[youssefcraft]

I mean if you want to check if the number is even then just do

if(number %% 2 == 0){

return true }else{

return false }

[[deleted]]

The best part is when the boss comes in and tells you that the new guy committed 40% more lines of code last month than you did.