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Using Haskell you can write it way more concise:
iseven :: Int -> Bool iseven 0 = True iseven 1 = False iseven 2 = True iseven 3 = False iseven 4 = True iseven 5 = False iseven 6 = True iseven 7 = False iseven 8 = True ...However, we can be way smarter by only defining the 2 base cases and then a recursive definition for all other numbers:
iseven :: Int -> Bool iseven 0 = True iseven 1 = False iseven n = iseven (n-2)It’s having a hard time with negative numbers, but honestly that’s quite a mood
Recursion is its own reward
so did someone draw this by hand or was it a filter
tbh it looks like an AI broke this down slightly & reconstructed it
a wise programmer knows to always ask the question “can i solve this problem in python using metaprogramming?” in this instance, the answer is yes:
def is_even(n: int): s = "def is_even_helper(number: int):\n" b = True for i in range(0, abs(n)+2): s += f"\tif (abs(number) == {i}): return {b}\n" b = not b exec(s) return locals().get("is_even_helper")(n)Gotta love how human readable Python always is!
Zero people in this post get the YanDev reference
so nobody actually really got the joke. very sad Moment.
It’s really just us… I’ve seen the basic programming joke a bunch of times, but people really aren’t understanding the YanDev/font embellishment. Sad indeed.
return trueis correct around half of the time
return Math.random() > 0.5would also be correct about half the time
Wouldn’t that only be correct about 25% of the time?
assert IsEven(2) == True assert IsEven(4) == True assert IsEven(6) == TrueAll checks pass. LGTM
import re def is_even(i: int) -> bool: return re.match(r"-?\d*[02468]$", str(i)) is not Nonei was gonna suggest the classic
re.match(r"^(..)\1*$", "0" * abs(i)) is not NoneCursed
“If it’s not an npm package it’s impossible”
- JS devs, probably
Can’t you just
If (number % 2 == 0){return true}
return number % 2 === 0Yeah, that’s even simpler
but what if
numberisn’t an integer, or even a number at all? This code, and the improved code shared by the other user, could cause major problems under those conditions. Really, what you would want, is to validate thatnumberis actually an integer before performing the modulo, and if it isn’t, you want to throw an exception, because something has gone wrong.That’s exactly what that NPM module does. And this is why it’s not a bad thing to use packages/modules for even very simple tasks, because they help to prevent us from making silly mistakes.
That would already cause an exception when calling the function because it has int number in the parameters
Javascript doesn’t have strongly-typed variables
yup, which is why I find the download stats truly horrifying
no
ok
That’s a lot of downloads
Just divide the number into its prime factors and then check if one of them is 2.
or divide the number by two and if the remainder is greater than
-(4^34)but less than
70 - (((23*3*4)/2)/2)then
trueWhat if the remainder is greater than the first, but not less than the latter?
Like, for example, 1?
Then you should return false, unless the remainder is also greater than or equal to the twenty second root of 4194304. Note, that I’ve only checked up to 4194304 to make sure this works, so if you need bigger numbers, you’ll have to validate on your own.
i hate to bring this up, but we also need a separate function for negative numbers
You can just bitwise AND those with …000000001 (for however many bits are in your number). If the result is 0, then the number is even, and if it’s 1, then the number is odd. This works for negative numbers because it discards the negative signing bit.
I remember coding actionscript in Flash and using modulo (%) to determine if a number was even or odd. It returns the remainder of the number divided by 2 and if it equals anything other than 0 then the number is odd.
Yeah. The joke is that this is the obvious solution always used in practise, but the programmer is that bad that they don’t know it and use some ridiculous alternative solutions instead.
I believe that’s the proper way to do it.
deleted by creator
if (!(number & 1))if (~number & 1)
Ask AI:
public static boolean isEven(int number) { // Handle negative numbers if (number < 0) { number = -number; // Convert to positive } // Subtract 2 until we reach 0 or 1 while (number > 1) { number -= 2; } // If we reach 0, it's even; if we reach 1, it's odd return number == 0; }Anything but using modulo I guess
And bit operations (:
This makes me happy that I don’t use genai
I’m not sure how fucked up their prompt is (or how unlucky they were). I just did 3 tries and every time it used modulo.
I’m assuming they asked it specifically to either not use modulo or to do a suboptimal way to make this joke.
When you sacrifice memory for an O(1) algorithm.
In this case still O(n)
Smh this is literally what switch statements are for
If number%2 == 0: return("Even") Else: return("odd")Not all ARM CPUs support mod operations. It’s better to use bit operations. Check if the last bit is set. If set it’s odd else it’s even.
I came here to post this
just check the least significant bit smh my head
I thought they were going to turn into Saddam Husseins.
