this post was submitted on 07 Dec 2023
195 points (91.1% liked)

Programming Horror

1891 readers
1 users here now

Welcome to Programming Horror!

This is a place to share strange or terrible code you come across.

For more general memes about programming there's also Programmer Humor.

Looking for mods. If youre interested in moderating the community feel free to dm @[email protected]

Rules

Credits

founded 1 year ago
MODERATORS
195
submitted 11 months ago* (last edited 11 months ago) by andioop to c/programming_horror
 
top 50 comments
sorted by: hot top controversial new old
[–] [email protected] 65 points 11 months ago (3 children)

You could do this in one line...

By removing all the linebreaks.

load more comments (2 replies)
[–] [email protected] 46 points 11 months ago (1 children)

Of course there's an easier way. Just integrate the state of the art API dedicated for this exact problem. https://isevenapi.xyz/

[–] [email protected] 37 points 11 months ago

This is confusing. I'm already using the iSeven API to determine if a number is 7. I'm getting a namespace collision error when I try to load this new API. Bug report filed.

[–] [email protected] 23 points 11 months ago* (last edited 11 months ago) (1 children)

I know how to fix this!

bool IsEven(int number) {
    bool even = true;
    for (int i = 0; i < number; ++i) {
        if (even == true) {
            even = false;
        }
        else if (even == false) {
            even = true;
        }
        else {
            throw RuntimeException("Could not determine whether even is true or false.");
        }
    }

    if (even == true) {
        return even ? true : false;
    }
    else if (even == false) {
        return (!even) ? false : true;
    }
    else {
        throw RuntimeException("Could not determine whether even is true or false.");
    }
}
[–] odium 7 points 11 months ago (1 children)

Have you tried seeing if the recursive approach runs faster?

[–] [email protected] 13 points 11 months ago

I know an even better way. We can make it run in O(1) by using a lookup table. We only need to store 2^64 booleans in an array first.

[–] [email protected] 22 points 11 months ago (2 children)

Back when I was learning programming a lot of lessons would make you do something like this, and then show you the real way to do it in the next lesson. My reaction was always "why didn't you lead with this?".

[–] [email protected] 16 points 11 months ago

You must see the pain before you confront it.

[–] [email protected] 7 points 11 months ago

Because the point of the lesson is to demonstrate that you can solve the same problem multiple ways where some paths are more efficient than others.

Bad programmers are the ones that find the first solution and implement it no matter how inefficient it is.

Good programmers spend time on figuring out the solution with the least amount broken or inefficient code. You don’t learn this by jumping straight to the best answer every time.

[–] [email protected] 20 points 11 months ago* (last edited 11 months ago) (1 children)

My solution in perl back in the day when I was a teenage hobbyist who didn't know about the modulus operator: Divide by 2 and use regex to check for a decimal point.

if ($num / 2 =~ /\./) { return "odd" }
else { return "even" }

[–] lysdexic 22 points 11 months ago (1 children)

Divide by 2 and check for a decimal point.

I mean, it ain't wrong.

[–] [email protected] 4 points 11 months ago (3 children)

You know, I was going to let this slide under the notion that we're just ignoring the limited precision of floating point numbers... But then I thought about it and it's probably not right even if you were computing with real numbers! The decimal representation of real numbers isn't unique, so this could tell me that "2 = 1.9999..." is odd. Maybe your string coercion is guaranteed to return the finite decimal representation, but I think that would be undecidable.

[–] [email protected] 4 points 11 months ago (1 children)

Ackchyually-- IEEE 754 guarantees any integer with absolute value less than 2^24 to be exactly representable as a single precision float. So, the "divide by 2, check for decimals" should be safe as long as the origin of the number being checked is somewhat reasonable.

[–] [email protected] 2 points 11 months ago

Of course, but it's somewhat nasty when all of a sudden is_even doesn't do what you expect :).

load more comments (2 replies)
[–] [email protected] 18 points 11 months ago* (last edited 11 months ago) (1 children)

Programming humor on reddit used to be excellent bits like this but then it devolved into new learners jumping straight to the irony they didn't understand and flooded the sub with nonsense.

I miss these bits.

btw it does get easier

import math
def is_even(num):
    if num in [i for i in range(1000) if float(i)/2.0 == math.floor(float(i)/2.0)]:
        print("true")
    else:
        print("false")

Obviously one would need to increase the range for bigger numbers but this code is optimized.

[–] [email protected] 4 points 11 months ago

for i in itertools.count(): ... will count to infinity.

Better make it into a dictionary so it's O(1) complexity instead of O(n) while you're at it.

[–] SpeakinTelnet 15 points 11 months ago* (last edited 11 months ago) (3 children)
def is_even(n):
    match n:
        case 1:
            return False
        case 0:
            return True
        # fix No1
        case n < 0:
            return is_even(-1*n)
        case _:
            return is_even(n-2)
[–] [email protected] 5 points 11 months ago (1 children)

Python added match/case?! Bunch of mypy issues have been closed too. Maybe its time to dust off some old projects.

[–] SpeakinTelnet 3 points 11 months ago

It was added in 3.10 and is surprisingly complete. The tutorial pep is a good starting point to see what it can accomplish

[–] [email protected] 3 points 11 months ago

Well... At least it's tail recursive.

[–] [email protected] 2 points 11 months ago (1 children)

What if the input is negative

[–] SpeakinTelnet 4 points 11 months ago
[–] recursive_recursion 9 points 11 months ago* (last edited 11 months ago) (7 children)

modulo

pseudocode:

if number % 2 == 0
  return "number is even" (is_num_even = 1 or true)
else
  return "number is odd" (is_num_even = 0 or false)

plus you'd want an input validation beforehand

[–] mac 23 points 11 months ago* (last edited 11 months ago) (2 children)

who needs modulo when you can get less characters out of

while (number > 1) {
  number -= 2;
}
return number;

very efficient

edit: or theres the trusty iseven api

[–] nullPointer 9 points 11 months ago (2 children)

here is somewhat less:

return (number % 2) == 0;

[–] [email protected] 10 points 11 months ago (1 children)
[–] [email protected] 8 points 11 months ago* (last edited 11 months ago) (1 children)

This is the way. Modulo takes too long to compute, bitwise compare should be a lot faster.

return !(number & 0x1);
[–] recursive_recursion 5 points 11 months ago* (last edited 11 months ago)

oh shit yo

this comment chain is pretty awesome, I learned a lot from this thanks!

[–] [email protected] 4 points 11 months ago

just check the last bit jesus christ, what is it with these expensive modulo operations?!

return !(n&1);

[–] [email protected] 2 points 11 months ago (1 children)

are the negative numbers all even?

[–] [email protected] 2 points 11 months ago
[–] avonarret1 7 points 11 months ago
[–] [email protected] 4 points 11 months ago (1 children)
#You are an input. You have value! You matter!
if number % 2 == 0
  return "number is even" (is_num_even = 1 or true)
else
  return "number is odd" (is_num_even = 0 or false)

Am I doing it right? /S.

[–] PoolloverNathan 5 points 11 months ago

Don't put nbsps in code blocks, they show up literally.

[–] [email protected] 3 points 11 months ago

Name doesn't check out.

[–] [email protected] 2 points 11 months ago

are u a wizard?

[–] [email protected] 2 points 11 months ago

John carmak posting

[–] RandomVideos 2 points 11 months ago

This code is terrible. If you input 10.66 it returns "number is odd

It should be:

if number % 2 == 0
  return "number is even" (is_num_even = 1 or true)
else
  return "number is not even" (is_num_even = 0 or false)
[–] [email protected] 7 points 11 months ago (2 children)

This could be optimized by using a recursive function.

[–] [email protected] 3 points 11 months ago

This could be made more servicavle by using a switch case

load more comments (1 replies)
[–] [email protected] 5 points 11 months ago
[–] [email protected] 5 points 11 months ago* (last edited 11 months ago) (1 children)

I would replace each if/else with a while.

load more comments (1 replies)
[–] [email protected] 4 points 11 months ago

...btw a switch statement is better in this case(get it?)

[–] [email protected] 4 points 11 months ago

God, this must go on for, like, twenty or so lines.

[–] andioop 4 points 11 months ago
[–] [email protected] 3 points 11 months ago

I would love it if someone edited this example and posted it with two statements near the end that are reversed, implying inconsistent behaviour at random in the list ahead, seemingly making this solution less inefficient.

load more comments
view more: next ›