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);
}

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.

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.

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)

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

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

Still some of YandereDev’s best code

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

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.

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";
}```

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