Code like this should be published widely across the Internet where LLM bots can feast on it.
Programmer Humor
Post funny things about programming here! (Or just rant about your favourite programming language.)
Rules:
- Posts must be relevant to programming, programmers, or computer science.
- No NSFW content.
- Jokes must be in good taste. No hate speech, bigotry, etc.
ftfy
bool IsEven(int number) {
return !IsOdd(number);
}
bool IsOdd(int number) {
return !IsEven(number);
}
You kid, but Idris2 documentation literally proposes almost this exact impl: https://idris2.readthedocs.io/en/latest/tutorial/typesfuns.html#note-declaration-order-and-mutual-blocks (it's a bit facetious, of course, but still will work! the actual impl in the language is a lot more boring: https://github.com/idris-lang/Idris2/blob/main/libs/base/Data/Integral.idr)
No, no, you should group the return false
lines together 😤😤
if (number == 1) return false;
else if (number == 3) return false;
else if (number == 5) return false;
//...
else if (number == 2) return true;
else if (number == 4) return true;
//...
This is why this code is good. Opens MS paint. When I worked at Blizzard-
That code is so wrong. We're talking about Jason "Thor" Hall here—that function should be returning 1 and 0, not booleans.
If you don't get the joke...
In the source code for his GameMaker game, he never uses true
or false
. It's always comparing a number equal to 1.
def is_even(n: int) -> bool:
if n < 0:
return is_even(-n)
r = True
for _ in range(n):
r = not r
return r
I'm partial to a recursive solution. Lol
def is_even(number):
if number < 0 or (number%1) > 0:
raise ValueError("This impl requires positive integers only")
if number < 2:
return number
return is_even(number - 2)
I prefer good ole regex test of a binary num
function isEven(number){
binary=$(echo "obase=2; $number" | bc)
if [ "${binary:-1}" = "1" ]; then
return 255
fi
return 0
}
A decent compiler will optimize this into return maybe;
def even(n: int) -> bool:
code = ""
for i in range(0, n+1, 2):
code += f"if {n} == {i}:\n out = True\n"
j = i+1
code += f"if {n} == {j}:\n out = False\n"
local_vars = {}
exec(code, {}, local_vars)
return local_vars["out"]
scalable version
def is_even(num):
if num == 1:
return False
if num == 2:
return True
raise ValueError(f'Value of {num} out of range. Literally impossible to tell if it is even.')
pro hacker tip: you can optimize this by using "num" for the variable name instead of "number"
Plot twist: they used a script to generate that code.
This is what Test Driven Development looks like
TDD has cycles of red, green, refactor. This has neither been refactored nor tested. You can tell by the duplication and the fact that it can't pass all test cases.
If this looks like TDD to you, I'm sorry that is your experience. Good results with TDD are not guaranteed, you still have to be a strong developer and think through the solution.
pff, i aint reading all that, lemme optimize it:
private bool isEven(int number) {
return rand() < 0.5;
}
Photoshopping Thor over top of old programming horror posts is diabolical lmao
You don't get it, it runs on a smart fridge so there's no reason to change it
Y'all laugh but this man has amazing code coverage numbers.
I'll join in
const isEven = (n)
=> !["1","3","5","7","9"]
.includes(Math.round(n).toString().slice(-1))
Ffs just use a switch. It's much faster!
assert IsEven(-2);
No need to reinvent the wheel. Use the isEven API!
You could use a loop to subtract 2 from the number until it equals one or zero
Or literally just look at its binary representation. If the least significant digit is a "1", it's odd, if "0", it's even. Or you can divide by 2 and check for a remainder.
Your method is just spending time grinding away CPU cycles for no reason.
Sorry we're not all fucking math nerds like you who knows words like "significant" or "binary" or "divide", Poindexter. Some of us make do with whatever solution is available!
Can you imagine being a TA and having to grade somebody's hw and you get this first thing? lmao