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); }bool isEven(int num) { return num == 0 || !isEven(num - (num > 0 ? 1 : -1)); }Damn that’s some solid optimization.
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
…a recursive is-even
wow
Man I love how horrible this is!
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.
In prod??
Listen up folks. This is why we do code reviews. This right here.
A few members of my team were reviewing codes but lots of PRs could be merged without tests or checks passing and only about 2 people before I joined understood what cicd is, no one else believed in its importance. They thought doing otherwise would “slow down the work precess and waste time, we know what we’re doing anyway!”.
I learned a lot from having to implement best practices and introduce tests in teams that don’t give a fuck or were never required to do it. I’m amazed at the industry standards and fully understand why job ads keep listing git as a requirement.
That’s something I would do
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.
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: ...Teach me
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
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;Obviously you couldn’t account for every number with only two lines.
Maybe you couldn’t
Eventually, it would wrap onto a second line, wouldn’t it?
When you run out of space, you just add a monitor to the right side. Honestly, it’s like you guys aren’t real developers.
Only if your line is not long enough…
The number of comments posting a better solution is funny and somewhat concerning.
Yeah, “just use modulo” - no shit, you must be some kind of master programmer
This is your brain on python:
def is_even (num): return num in [x*2 for x in range(sys.maxsize / 2)]That won’t work tho, you need to make it sys.maxsize//2 to coerce the output into int form
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)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)) > 0spoiler
i 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 230. but hey, at least it’s O(1).
Nice, how about bitwise & operator?
// n&1 is 1, then odd, else even return (!(n & 1));
Don’t use recursion. Each call will need to allocate a new stack frame which leads to a slower runtime than an iterative approach in pretty much any language.
TCO baby
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.
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.
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));Or modulo %
private bool IsEven(int number){ return number % 2 ? false : true; }
number % 2 == 0 and (number & 0b1) == 0Are the only sane ways to do this. No need to floor. Although If its C and you can’t modulo floats then (number/2 == floor(number/2))
If you are using floats, you really do not want to have an isEven function …
Take out the
elseand I’m inValid point.
Huh?
return number % 2 == 0
That’s the only sane solution.
Do note how I said “a decent” way, not “the best” way. Get that huh outta here.
Just do
npm install isEvenand don’t worry about it.Removed by mod
Extract an interface and let the consumer supply the implementation.
while (true){ if (number == 0) return true; if (number == 1) return false; number -= 2 }return !(number % 2)
Setting number to -1 might cause you to wait a while.
You know, shortly after posting this I did think about whether it’s still working if I just pass the underflow that will happen at some point or if I have to fix something in that case (like subtract 1 after the underflow). I deemed it “too complicated” and would just issue a warning that my code is only tested on positive numbers, but I think it will still work.
deleted by creator
Still some of YandereDev’s best code
Oh man, in js we have a package for this magic.
And it is so light, it only requires is-odd package!
left-pad PTSD intensifies
That one is bad, I use this one https://www.npmjs.com/package/is-is-is-even
Why even do that, just check if this outputs false https://www.npmjs.com/package/is-is-is-is-is-is-odd
Oh fuck, gonna refactor asap!
I always forget if is even requires is odd or the other way around.
Weekly Downloads: 293.319
Good job my young padawan, let me teach you about the modulo operator …
Actually the modulo operator is the wrong solution.
No its not the wrong solution! Premature optimization is a waste of time.
Using if or case are not a solution because they are way too verbose and very easy to introduce an error.
Modulo is a solution, and using bit-wise and is another faster solution.
It’s only the wrong solution if you’re writing something where every operation needs to be accounted for. Modulo is a great, easy, readable method otherwise.
Not too certain on C++, but I think this would be the cleanest implementation that still somewhat optimizes itself:
private bool IsEven(int number){ return !(number % 2) }You call it premature optimization. I call it obvious.
You use a flat head as a Phillip’s.
You can call it whatever you like, the fact of the matter remains - code readibility is more important than most optimizations you can ever hope to make.
Bad programmers optimize everything and produce code that is not understandabe and 0.001% “faster”
Optimization isn’t inherently better than Readable code.
What’s most important is what matters to your project.
If your project manages itself with limited resources, Optimization is better than Readable code, and it should be supplemented with comments.
If your project has a wealth of resources, readable code is probably the better option.
The “This is the ONLY WAY” mindset in coding is the only thing that I would argue is completely wrong. The “One-size fits all” mindset in development is short sighted, sure. but what makes it the worst, is when it becomes “One-Size fits all ONLY”.
“One-size fits all” means the project runs well on everything. “Optimized for one” means the project runs exceptionally on one Architecture/OS.
Basically, this situation is the highly unsatisfying “It’s your preference”.
But given the context of Yandere Dev and the Target Audience of Yandere Simulator, his code is perfect. It’s code that runs terribly on everything and the project is for no one.
Okay wow no need to get personal.
I call it making assumptions that may be incorrect, and do you know if the compiler will do the optimization anyway in this case?
What statement do you flag as assumption? Yes, I do. The modulo operator is only a subset of bit masks. It is more explicit to write:
if ( (variable &EVEN_MASK) == 0) …
To act upon even numbers then:
if ( (variable %2) == 0) …
How would you name the 2 in the above statement for more expressive power?
EVEN_MODULO_OP ? That may throw more people off imo.
You shouldn’t rename 2 at all. “Even” has a commonly understood meaning that is instantly recognizable from
(variable %2) == 0. The bitmask is an overgeneralization.I give up, I was wrong to even think about the modulo operator, you are clearly the master programmer. 🥇
This reminds me of a discussion about the ternary operator
? :, some people think its the one true way of writing code because its just so clear what it does. And I say please use it sparingly because if you start doing nested ternary operators its very hard to unpack what your code does, and I prefer readability over compact code, especially with today’s compilers.I feel you said about JavaScript as a whole.
https://realpython.com/python-modulo-operator/#how-to-check-if-a-number-is-even-or-odd
I just wonder why module is the wrong solution.
Plot twist: it’s generated code for the purpose of the joke.
Being yandere dev, that’s likely the actual code
He wouldn’t have the self-awareness to post it himself though, so this is likely a montage to put his name on it.
And they’re using the suggestions to efficiently generate even more code
