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submitted 9 months ago* (last edited 9 months ago) by [email protected] to c/[email protected]
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[-] [email protected] 8 points 9 months ago* (last edited 9 months ago)

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

[-] [email protected] 21 points 9 months ago* (last edited 9 months ago)

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

[-] [email protected] 8 points 9 months ago

here is somewhat less:

return (number % 2) == 0;

[-] [email protected] 9 points 9 months ago
[-] [email protected] 8 points 9 months ago* (last edited 9 months ago)

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

return !(number & 0x1);
[-] [email protected] 5 points 9 months ago* (last edited 9 months ago)

oh shit yo

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

[-] [email protected] 4 points 9 months ago

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

return !(n&1);

[-] [email protected] 2 points 9 months ago

are the negative numbers all even?

[-] [email protected] 7 points 9 months ago
[-] [email protected] 4 points 9 months ago
#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.

[-] [email protected] 4 points 9 months ago

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

[-] [email protected] 3 points 9 months ago

Name doesn't check out.

[-] [email protected] 2 points 9 months ago

are u a wizard?

[-] [email protected] 2 points 9 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] 1 points 9 months ago

John carmak posting

this post was submitted on 07 Dec 2023
191 points (91.0% liked)

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