Is there a more efficient way than `round number length leng kotlinlang #getting-started

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getting-started

Title

a

Ayfri

03/11/2022, 4:34 AM

Is there a more efficient way than

round(number / length) * length

to round a number by a length ?

m

Michael de Kaste

03/11/2022, 8:44 AM

is it rounding for output only? Than I suggest using

"%.${length}f".format(number)

otherwise, mathmatically theres is not really much you can do to round more efficiently

k

Klitos Kyriacou

03/11/2022, 9:20 AM

The print formatting is not equivalent to the OP's rounding (which means to round to the nearest multiple of

length

). I think, if

length

is an integer, you can probably make it slightly more efficient using just integer arithmetic, but it's probably not worth the effort.

a

Ayfri

03/11/2022, 11:38 AM

Yes it's integer, but it's an operation that will be used at least 300 times per second, it's for a Kotlin/JS game

k

Klitos Kyriacou

03/11/2022, 12:32 PM

If it's for JS, I wouldn't bother, as there is no integer type in Javascript; only Number.

t

Tim Oltjenbruns

03/11/2022, 6:34 PM

If you can guarantee length is a multiple of 2, you can use bitwise operators to round your number to a multiple of length

@Ayfri for example if you want to round by 8, shift right 3 and then shift left 3 (8 = 2^3)

That why for example, Minecraft, uses a lot of 8's and 16's as their grid size, because it makes the frequent rounding to snap to a grid much more efficient.

technically, if you’re going to shift right/left 3 you can just use bitwise & to delete the last 3 bits

For example, in JS

const numberToRound = 15; 
const length = 8
const bitmask = Number.MAX_SAFE_INTEGER - length + 1;

console.log(numberToRound & bitmask);

But again, only works for multiples of 2

Make sure to test the performance though, I’m not sure if JavaScript’s representing all numbers in floating point makes this optimization useless or not. I know it works on platforms that have true integer types.

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