boolean isOdd(int num) {
  if(num == 1)
    return true;
  if(num > 0)
    return isEven(num - 1);
  else
    return isEven(num + 1);
}

boolean isEven(int num) {
  if(num > 0)
    return isOdd(num - 1);
  else
    return isOdd(num + 1);
}

[–] affiliate@lemmy.world 9 points 2 months ago (1 children)

the downside with this approach is that it will eventually terminate. the version in the original post has the advantage of giving me plenty of time to contemplate life’s many mysteries.

[–] cogman@lemmy.world 2 points 2 months ago

What can I say, I'm a performance nerd.

[–] Valmond@lemmy.world 1 points 2 months ago

isEeven(∞);

[–] Rednax@lemmy.world 0 points 2 months ago (1 children)

Why the complicated if statements to check the sign? Just let the number overflow. Would be functionaly the same, and result in much prettier code.

[–] cogman@lemmy.world 5 points 2 months ago

That's a platform dependent change. Overflow is undefined behavior. I'd rather have my code portable so it can run on my Univac 1101.