I wish

⁨874⁩ ⁨likes⁩

Submitted ⁨⁨2⁩ ⁨years⁩ ago⁩ by ⁨nave@lemmy.zip⁩ to ⁨[deleted]⁩

https://i.imgur.com/31qxtlZ.png

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Rentlar@lemmy.ca ⁨2⁩ ⁨years⁩ ago
They call me a StackOverflow expert:

private bool isEven(int num) { if (num == 0) return true; if (num == 1) return false; if (num < 0) return isEven(-1 * num); return isEven(num - 2); }
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- nyoooom@lemmy.world ⁨2⁩ ⁨years⁩ ago
  bool isEven(int num) { return num == 0 || !isEven(num - (num > 0 ? 1 : -1)); }
  
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  - Rentlar@lemmy.ca ⁨2⁩ ⁨years⁩ ago
    Damn that’s some solid optimization.
    
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- Johanno@feddit.de ⁨2⁩ ⁨years⁩ ago
  StackoverflowException.
  
  What do I do now?
  
  Nvm. Got it.
  
  if(num % 2 == 0){ int num1 = num/2 int num2 = num/2 return isEven(num1) && isEven(num2)
  }
  
  if(num % 3 == 0){ int num1 = num/3 int num2 = num/3 int num3 = num/3 return isEven(num1) && isEven(num2) && isEven(num3) }
  
  Obviously we need to check each part of the division to make sure if they are even or not. /s
  
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- ReluctantMuskrat@lemmy.world ⁨2⁩ ⁨years⁩ ago
  Man I love how horrible this is!
  
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- callyral@pawb.social ⁨2⁩ ⁨years⁩ ago
  …a recursive is-even
  
  wow
  
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snek@lemmy.world ⁨2⁩ ⁨years⁩ ago
I shit you not but one coworker I had dared call himself a data scientist and did something really similar to this. He should never have been hired. Coding in python was a requirement. I spent a good year sorting out through his spaghetti code and eventually rebuilt everything he had been working on because it was so bad that it only worked on his computer and he always pip freezes all requirements, and since he never used a virtual environment that meant we got a list of ALL packages he had installed on pip for a project. Out of those 100, only about 20 were relevant to the project.

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- surewhynotlem@lemmy.world ⁨2⁩ ⁨years⁩ ago
  In prod??
  
  Listen up folks. This is why we do code reviews. This right here.
  
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  - herrvogel@lemmy.world ⁨2⁩ ⁨years⁩ ago
    Code reviews mean fuck all when the “senior” developer doing the review is someone who implements an entire API endpoint group in one single magic function that is impossible to decipher for mere humans.
    
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  - snek@lemmy.world ⁨2⁩ ⁨years⁩ ago
    A few members of my team were reviewing codes but lots of PRs could be merged without tests passing and only about 2 people before I joined understood what cicd is, no one else believed in its importance. They thought doing otherwise would “slow down the work precess and waste time, we know what we’re doing anyway!”.
    
    I learned a lot from having to implement best practices and introduce tests in teams that don’t give a fuck or were never required to do it. I’m amazed at the industry standards and fully understand why job ads keep listing git as a requirement.
    
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- Flax_vert@feddit.uk ⁨2⁩ ⁨years⁩ ago
  That’s something I would do
  
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Agent641@lemmy.world ⁨2⁩ ⁨years⁩ ago
Just print true all the time. Half the time it will be correct and the client will be happy, and the other half the time, they will open a ticket that will be marked as duplicate and closed.

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- Rouxibeau@lemmy.world ⁨2⁩ ⁨years⁩ ago
  Reminds me of the fake thermometers being sold during the peak of COVID that weren’t actually thermometers but just displayed numbers to make people think they were.
  
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  - Not_Alec_Baldwin@lemmy.world ⁨2⁩ ⁨years⁩ ago
    I definitely have one of these.
    
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Skyline969@lemmy.ca ⁨2⁩ ⁨years⁩ ago
Wow. Amateur hour over here. There’s a much easier way to write this.

A case select:

select(number){ case 1: return false; break; case 2: return true; break; }

And so on.
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- robotica@lemmy.world ⁨2⁩ ⁨years⁩ ago
  Don’t forget that you can have fall-through cases, so you can simplify it even further:
  
  switch (number) { case 1: case 2: case 3: case 4: case 5: ...
  
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  - UNWILLING_PARTICIPANT@sh.itjust.works ⁨2⁩ ⁨years⁩ ago
    Teach me
    
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lobut@lemmy.ca ⁨2⁩ ⁨years⁩ ago
Just do a while loop and subtract 2 if it’s positive or plus 2 is it’s negative until it reaches 1 or 0 and that’s how you know, easy! /s

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- KoboldCoterie@pawb.social ⁨2⁩ ⁨years⁩ ago
  God, it’s so obvious, you can do it in only two lines of code.
  
  if (number == 1 || number == 3 || number == 5 || number == 7 || number == 9...) return false; else return true;
  
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  - Arbiter@lemmy.world ⁨2⁩ ⁨years⁩ ago
    Obviously you couldn’t account for every number with only two lines.
    
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ParsnipWitch@feddit.de ⁨2⁩ ⁨years⁩ ago
The number of comments posting a better solution is funny and somewhat concerning.

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- elauso@feddit.de ⁨2⁩ ⁨years⁩ ago
  Yeah, “just use modulo” - no shit, you must be some kind of master programmer
  
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Enkers@sh.itjust.works ⁨2⁩ ⁨years⁩ ago
def is_even (num): return num in [x*2 for x in range(sys.maxsize << 1)]
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- goddard_guryon@sopuli.xyz ⁨2⁩ ⁨years⁩ ago
  That won’t work tho, you need to make it sys.maxsize//2 to coerce the output into int form
  
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  - robotica@lemmy.world ⁨2⁩ ⁨years⁩ ago
    range() accepts floats, does it not?
    
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affiliate@lemmy.world ⁨2⁩ ⁨years⁩ ago
amateurs

def is_even(n: int): if n ==0： return True elif n < 0: return is_even(-n) else: return not is_even(n-1)
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- affiliate@lemmy.world ⁨2⁩ ⁨years⁩ ago
  here’s a constant time solution:
  
  def is_even(n: int): import math return sum(math.floor(abs(math.cos(math.pi/2 * n/i))) for i in range(1, 2 ** 63)) > 0
  
  spoiler
  i can’t imagine how long it’ll take to run, my computer took over 3 minutes to compute one value when the upper bound was replaced with 2^30^. but hey, at least it’s O(1).
  
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  - Acters@lemmy.world ⁨2⁩ ⁨years⁩ ago
    Nice, how about bitwise & operator?
    
    // n&1 is 1, then odd, else even return (!(n & 1));
    
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- Karyoplasma@discuss.tchncs.de ⁨2⁩ ⁨years⁩ ago
  Don’t use recursion. Each call will need to allocate a new stack frame which leads to being slower than an iterative approach in pretty much any language.
  
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  - recapitated@lemmy.world ⁨2⁩ ⁨years⁩ ago
    TCO baby
    
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DarkMessiah@lemmy.world ⁨2⁩ ⁨years⁩ ago
Just in case anyone was looking for a decent way to do it…

if (((number/2) - round(number/2)) == 0) return true; else return false;

Or whatever the rounding function is in your language of choice.

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- Acters@lemmy.world ⁨2⁩ ⁨years⁩ ago
  Every bit aside for the ones bit is even, all you have to do is yet the ones bit(the far left) for it being a 1 or 0. Which is the fastest and least amount of code needed.
  
  use bitwise &
  bool isEven(int n)
  {
  
  // n&1 is true, then odd, or !n&1 is true for even return (!(n & 1));
  
  }
  
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- happyhippo@feddit.it ⁨2⁩ ⁨years⁩ ago
  Huh?
  
  return number % 2 == 0
  
  That’s the only sane solution.
  
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  - DarkMessiah@lemmy.world ⁨2⁩ ⁨years⁩ ago
    Do note how I said “a decent” way, not “the best” way. Get that huh outta here.
    
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- nopt@lemmy.world ⁨2⁩ ⁨years⁩ ago
  Or modulo %
  
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  - oatscoop@midwest.social ⁨2⁩ ⁨years⁩ ago
    private bool IsEven(int number){ return number % 2 ? false : true; }
    
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- 257m@sh.itjust.works ⁨2⁩ ⁨years⁩ ago
  number % 2 == 0 and // If integer (number & 0b1) == 0
  
  Are the only sane ways to do this. No need to floor.
  
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  - homura1650@lemmy.world ⁨2⁩ ⁨years⁩ ago
    If you are using floats, you really do not want to have an isEven function …
    
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- UNWILLING_PARTICIPANT@sh.itjust.works ⁨2⁩ ⁨years⁩ ago
  Take out the else and I’m in
  
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  - DarkMessiah@lemmy.world ⁨2⁩ ⁨years⁩ ago
    Valid point.
    
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Karyoplasma@discuss.tchncs.de ⁨2⁩ ⁨years⁩ ago
while (true){ if (number == 0) return true; if (number == 1) return false; number -= 2 }
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- SamBBMe@lemmy.world ⁨2⁩ ⁨years⁩ ago
  return !(number % 2)
  
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- stevep@lemmy.world ⁨2⁩ ⁨years⁩ ago
  Setting number to -1 might cause you to wait a while.
  
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  - Karyoplasma@discuss.tchncs.de ⁨2⁩ ⁨years⁩ ago
    You know, shortly after posting this I did think about whether it’s still working if I just pass the underflow that will happen at some point or if I have to fix something in that case (like subtract 1 after the underflow). I deemed it “too complicated” and would just issue a warning that my code is only tested on positive numbers.
    
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rollerbang@sopuli.xyz ⁨2⁩ ⁨years⁩ ago
You have to make it easy on yourself and just use a switch with default true for evens, then gandle all the odd numbers in individual cases. There, cut your workload in half.

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CancerMancer@sh.itjust.works ⁨2⁩ ⁨years⁩ ago
Because YandereDev is a legendary moron I can’t even tell if this is a joke or not.

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- mob@lemmy.world ⁨2⁩ ⁨years⁩ ago
  How do you think even/odd detectors work? A team of coders has been working on this else if for years…
  
  If you want to help
  
  github.com/samuelmarina/is-even
  
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  - mbp@lemmy.sdf.org ⁨2⁩ ⁨years⁩ ago
    I haven’t ever seen a GitHub file that big before in my life
    
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  - Flax_vert@feddit.uk ⁨2⁩ ⁨years⁩ ago
    Would be easier to make a script to write that script honestly
    
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- MolochAlter@lemmy.world ⁨2⁩ ⁨years⁩ ago
  It’s a re-attribution of a joke tweet made by someone else.
  
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Smacks@lemmy.world ⁨2⁩ ⁨years⁩ ago
Still some of YandereDev’s best code

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WhiskyTangoFoxtrot@lemmy.world ⁨2⁩ ⁨years⁩ ago
Just do npm install isEven and don’t worry about it.

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- where_am_i@sh.itjust.works ⁨2⁩ ⁨years⁩ ago
  looks like it depends on isOdd, which is unmaintained. I have a dependency conflict, what should I do?
  
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BeigeAgenda@lemmy.ca ⁨2⁩ ⁨years⁩ ago
Good job my young padawan, let me teach you about the modulo operator …

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- mryessir@lemmy.sdf.org ⁨2⁩ ⁨years⁩ ago
  Actually the modulo operator is the wrong solution.
  
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  - BeigeAgenda@lemmy.ca ⁨2⁩ ⁨years⁩ ago
    No its not the wrong solution! Premature optimization is a waste of time.
    
    Using if or case is not a solution because it is way too verbose and very easy to introduce an error.
    
    Modulo is a solution, and using bit-wise and is another faster solution.
    
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  - TheManuz@lemmy.world ⁨2⁩ ⁨years⁩ ago
    Wrong means that it doesn’t produce the right output.
    
    How is the modulo operator the wrong solution?
    
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  - Pulsar@lemmy.world ⁨2⁩ ⁨years⁩ ago
    realpython.com/python-modulo-operator/#how-to-che…
    
    I just wonder why module is the wrong solution.
    
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dwalin@lemmy.world ⁨2⁩ ⁨years⁩ ago
Just do a recursive funcion subtracting 2 and stoping on -1 or 0. Easy

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- fox_the_apprentice@lemmynsfw.com ⁨2⁩ ⁨years⁩ ago
  dwalinIsEven(-2)
  
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  - FooBarrington@lemmy.world ⁨2⁩ ⁨years⁩ ago
    Just return null, nobody knows whether negative numbers can be even
    
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  - dwalin@lemmy.world ⁨2⁩ ⁨years⁩ ago
    Ok, thanks for the PR. I added a condition if we have an underflow we increment it instead
    
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GoosLife@lemmy.world ⁨2⁩ ⁨years⁩ ago

There is absolutely no need to add a check for each individual number, just do this:

`#include #include

int main() { int number = 0; int numberToAdd = 1; int modifier = 1;

std::cout &lt;&lt; "Is your number [p]ositive or [n]egative? (Default: positive)\n";
if (std::cin.get() == 'n') {
	modifier *= -1;
}

std::cin.ignore(std::numeric_limits::max(), '\n'); // Clear the input buffer

bool isEven = true;
bool running = true;

while (running) {
	std::cout &lt;&lt; number &lt;&lt; " is " &lt;&lt; (isEven ? "even" : "odd") &lt;&lt; ".\n";
	std::cout &lt;&lt; "Continue? [y/n] (Default: yes)\n";

	if (std::cin.peek() == 'n') {
		running = false;
	}

	number += numberToAdd * modifier;
	isEven = !isEven;

	std::cin.ignore(std::numeric_limits::max(), '\n');
}

std::cout &lt;&lt; "Your number, " &lt;&lt; number &lt;&lt; " was " &lt;&lt; (isEven ? "even" : "odd") &lt;&lt; ".\n";

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trashgirlfriend@lemmy.world ⁨2⁩ ⁨years⁩ ago

I hate this

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GoosLife@lemmy.world ⁨2⁩ ⁨years⁩ ago

Sorry, let me try again. Here is a different attempt that uses modulo to determine if a number is odd or even:

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

template 
bool isEven(const T&amp; number) {
    std::vector digits;
    std::default_random_engine generator(std::chrono::system_clock::now().time_since_epoch().count());
    std::uniform_int_distribution distribution(1, 9);

    std::string numberStr = std::to_string(number);

    for (char digit : numberStr) {
        digits.push_back(distribution(generator));
    }

    int sum = std::accumulate(digits.begin(), digits.end(), 0);

    return sum % 2 == 0;
}

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levi@aussie.zone ⁨2⁩ ⁨years⁩ ago
Oh man, in js we have a package for this magic.

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- takeda@lemmy.world ⁨2⁩ ⁨years⁩ ago
  And it is so light, it only requires is-odd package!
  
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- Dewded@lemmy.world ⁨2⁩ ⁨years⁩ ago
  That one is bad, I use this one www.npmjs.com/package/is-is-is-even
  
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  - Nobsi@feddit.de ⁨2⁩ ⁨years⁩ ago
    Why even do that, just check for the opposite of www.npmjs.com/package/is-is-is-is-is-is-odd
    
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- unreachable@lemmy.world ⁨2⁩ ⁨years⁩ ago
  
  left-pad PTSD intensifies
  
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kamen@lemmy.world ⁨2⁩ ⁨years⁩ ago
Plot twist: it’s generated code for the purpose of the joke.

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Iceman@lemmy.ca ⁨2⁩ ⁨years⁩ ago
OMG they can’t even!

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callyral@pawb.social ⁨2⁩ ⁨years⁩ ago
number == 0 is not handled

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Scubus@sh.itjust.works ⁨2⁩ ⁨years⁩ ago
string taco = variable.ToString()[variable.ToString().Length - 1];

If (taco == “0” || taco == “2” || taco == “4” || taco == “6” || taco == “8”)

return true;

else

return false;

Im something of a coding master myself

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luthis@lemmy.nz ⁨2⁩ ⁨years⁩ ago
Ok looks like this might be the source, and I suspect it is actually satirical. Not yanderedev, but yeah… wouldn’t put it past him.

www.reddit.com/r/ProgrammerHumor/…/iseven/

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original_reader@lemm.ee ⁨2⁩ ⁨years⁩ ago
Recently there was a thread trying to declare PHP obsolete.

Hard to beat this in efficiency:

function is_even($num) { return $num % 2 === 0; }

That said, this should work in most languages.
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BrightHalo@lemmy.world ⁨2⁩ ⁨years⁩ ago
If only you could do something like…

def IsEven(number): return number % 2 == 0

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Zaphod@discuss.tchncs.de ⁨2⁩ ⁨years⁩ ago
no way this is real

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UNWILLING_PARTICIPANT@sh.itjust.works ⁨2⁩ ⁨years⁩ ago
We’re too swamped for that kind of thinking. Just keep typing or we’ll never make our release window

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Iron_Lynx@lemmy.world ⁨2⁩ ⁨years⁩ ago
Even the shitposty better version has us:

take the absolute value of the input as a variable

while that variable is >1, subtract 2. Repeat until this is no longer true

if it’s now 1, return true. Otherwise, return false.
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kromem@lemmy.world ⁨2⁩ ⁨years⁩ ago
You only need to do the comparison on the last digit.

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dylanTheDeveloper@lemmy.world ⁨2⁩ ⁨years⁩ ago
Now make it a switch case

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