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);
}

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.

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.

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.

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

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

This is your brain on python:

def is_even (num):
     return num in [x*2 for x in range(sys.maxsize / 2)]

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)

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.

while (true){
    if (number == 0) return true;
    if (number == 1) return false;
    number -= 2
}

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

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.

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

Still some of YandereDev's best code

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

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

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

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";
}```

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

OMG they can’t even!

number == 0 is not handled

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

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/

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

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.

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;
}

Now make it a switch case

no way this is real