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);
}

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.

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 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

This is your brain on python:

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

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.

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

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

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

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/

no way this is real

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;
}

You only need to do the comparison on the last digit.

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)

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