this post was submitted on 01 Nov 2023
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[–] [email protected] 107 points 1 year ago* (last edited 1 year ago) (4 children)

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);
}
[–] [email protected] 35 points 1 year ago (1 children)
bool isEven(int num) {
 return num == 0 || !isEven(num - (num > 0 ? 1 : -1));
}
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[–] [email protected] 17 points 1 year ago* (last edited 1 year 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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[–] [email protected] 87 points 1 year ago* (last edited 1 year ago) (2 children)

I shit you not but one coworker I had dared call himself a data scientist and did something really similar to this but in Python and in production code. 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.

[–] [email protected] 48 points 1 year ago (2 children)

In prod??

Listen up folks. This is why we do code reviews. This right here.

[–] [email protected] 16 points 1 year ago* (last edited 1 year 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 thousand-something lines magic function that is impossible to decipher for mere humans.

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[–] [email protected] 80 points 1 year ago* (last edited 1 year ago) (1 children)

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.

[–] [email protected] 24 points 1 year ago (1 children)

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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[–] [email protected] 69 points 1 year ago* (last edited 1 year ago) (1 children)

Wow. Amateur hour over here. There's a much easier way to write this.

A case select:

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

And so on.

[–] [email protected] 13 points 1 year ago* (last edited 1 year ago) (1 children)

Don't forget that you can have fall-through cases, so you can simplify it even further:

switch (number) {
    case 1:
    case 3:
    case 5:
    case 7:
    case 9:
      ...
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[–] [email protected] 46 points 1 year ago (1 children)

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

[–] [email protected] 48 points 1 year ago (5 children)

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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[–] [email protected] 42 points 1 year ago* (last edited 1 year ago) (3 children)

This is your brain on python:

def is_even (num):
     return num in [x*2 for x in range(sys.maxsize / 2)]
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[–] [email protected] 41 points 1 year ago (1 children)

The number of comments posting a better solution is funny and somewhat concerning.

[–] [email protected] 24 points 1 year ago

Yeah, "just use modulo" - no shit, you must be some kind of master programmer

[–] [email protected] 40 points 1 year ago (4 children)

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)
[–] [email protected] 28 points 1 year ago (1 children)

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

spoileri 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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[–] [email protected] 33 points 1 year ago* (last edited 1 year ago)

You have to make it easy on yourself and just use a switch with default true for evens, then handle all the odd numbers in individual cases. There, cut your workload in half.

[–] [email protected] 29 points 1 year ago (4 children)
while (true){
    if (number == 0) return true;
    if (number == 1) return false;
    number -= 2
}
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[–] [email protected] 28 points 1 year ago (2 children)

Because YandereDev is a legendary moron I can't even tell if this is a joke or not.

[–] [email protected] 16 points 1 year ago (3 children)

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

https://github.com/samuelmarina/is-even

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[–] [email protected] 27 points 1 year ago* (last edited 1 year ago) (13 children)

Just in case anyone was looking for a decent way to do it...

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

return false;

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

EDIT: removed unnecessary else.

[–] [email protected] 20 points 1 year ago

Modulo operator my dude.

[–] [email protected] 16 points 1 year ago* (last edited 1 year ago)

Every bit aside for the ones bit is even. All you have to do is get the ones bit(the far right) for it being a 1 or 0. Which is the fastest and least amount of code needed.

use bitwise &

// n&1 is true, then odd, or !n&1 is true for even  

 return (!(n & 1));  
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[–] [email protected] 26 points 1 year ago (1 children)

Just do npm install isEven and don't worry about it.

[–] [email protected] 15 points 1 year ago (1 children)

looks like it depends on isOdd, which is unmaintained. I have a dependency conflict, what should I do?

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[–] [email protected] 26 points 1 year ago

Still some of YandereDev's best code

[–] [email protected] 25 points 1 year ago (13 children)

Oh man, in js we have a package for this magic.

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[–] [email protected] 24 points 1 year ago (22 children)

Good job my young padawan, let me teach you about the modulo operator ...

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[–] [email protected] 23 points 1 year ago (3 children)

Plot twist: it's generated code for the purpose of the joke.

[–] [email protected] 33 points 1 year ago (1 children)

Being yandere dev, that's likely the actual code

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[–] [email protected] 23 points 1 year ago* (last edited 1 year ago) (1 children)

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 << "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 << number << " is " << (isEven ? "even" : "odd") << ".\n";
		std::cout << "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 << "Your number, " << number << " was " << (isEven ? "even" : "odd") << ".\n";
}```
[–] [email protected] 12 points 1 year ago (2 children)
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[–] [email protected] 21 points 1 year ago (3 children)

Just do a recursive funcion subtracting 2 and stoping on -1 or 0. Easy

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[–] [email protected] 19 points 1 year ago

OMG they can’t even!

[–] [email protected] 18 points 1 year ago (1 children)

number == 0 is not handled

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[–] [email protected] 16 points 1 year ago (2 children)

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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[–] [email protected] 15 points 1 year ago (1 children)

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.

https://www.reddit.com/r/ProgrammerHumor/comments/i15h4d/iseven/

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[–] [email protected] 14 points 1 year ago (1 children)

We're too swamped for that kind of thinking. Just keep typing or we'll never make our release window

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[–] [email protected] 13 points 1 year ago (1 children)

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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[–] [email protected] 13 points 1 year ago* (last edited 1 year ago) (3 children)
int is_even(int n)
{
    int result = -1;
    char number[8]; //should be enough
    sprintf(number, "%d", n);

    // check the number
    // TODO: handle negative numbers
    for (char *p=number; *p; p++)
    {
        if (*p=='0' || *p=='2' || *p=='4' || *p=='6' || *p=='8')
            result = 1;
        else if (*p=='1' || *p=='3' || *p=='5' || *p=='7' || *p=='9')
            result = 0;
        else {
           fprintf(stderr, "Your number is wrong!\n");
           exit(1); 
        }
    }
    return result;
}
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[–] [email protected] 12 points 1 year ago (5 children)
[–] [email protected] 15 points 1 year ago

From what I've heard about yanderedev... it's possible this is actually real.

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[–] [email protected] 12 points 1 year ago

Now make it a switch case

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