this post was submitted on 05 Aug 2023
537 points (92.8% liked)
Programmer Humor
19594 readers
1028 users here now
Welcome to Programmer Humor!
This is a place where you can post jokes, memes, humor, etc. related to programming!
For sharing awful code theres also Programming Horror.
Rules
- Keep content in english
- No advertisements
- Posts must be related to programming or programmer topics
founded 1 year ago
MODERATORS
you are viewing a single comment's thread
view the rest of the comments
view the rest of the comments
Doubles have a much higher max value than ints, so if the method were to convert all doubles to ints they would not work for double values above 2^31-1.
(It would work, but any value over 2^31-1 passed to such a function would get clamped to 2^31-1)
So why not return a long or whatever the 64 bit int equivalent is?
To avoid a type conversion that might not be expected. Integer math in Java differs from floating point math.
Math.floor(10.6) / Math.floor(4.6) = 2.5 (double)
If floor returned a long, then
Math.floor(10.6) / Math.floor(4.6) = 2 (long)
If your entire code section is working with doubles, you might not like finding Math.floor() unexpectedly creating a condition for integer division and messing up your calculation. (Have fun debugging this if you're not actively aware of this behavior).
Because even a long (64-bit int) is too small :)
A long can hold 2^64-1 = 1.84E19
A double can hold 1.79E308
Double does some black magic with an exponent, and can hold absolutely massive numbers!
Double also has some situations that it defines as "infinity", a concept that does not exist in long as far as I know (?)
But there's really no point in flooring a double outside of the range where integers can be represented accurately, is there.
what about using two ints
What about two
int64_tyeah that would be pretty effective. could also go to three just to be safe
Make it four, just to be even
A BigDecimal?