Rounding a number without lua math. thingy

Help and Feedback Code Review
#1

How do I improve this code

So I used my remaining brain cells to make this function that rounds numbers for a module, It works but has a high delay because of the for loop, And i don’t know if there is any faster way than this I need help pls

Note: It works because while I was doing for loops, I realized that the goal gets rounded by the index, Then created this thingy

Code

local function Abs(Number: number): number
	if Number > 0 then 
		return Number 
	else 
		return -Number 
	end
end

local function Round(Number : number): number
	local NearestNumber, LastNumber = Number, 0
	
	for Index=1, Number + 2, if Number > 0 then 1 else -1 do
		local NumDustance = Abs(Number - Index)
		
		if NumDustance < NearestNumber then 
			LastNumber = Index 
			NearestNumber = NumDustance 
		end
		
		continue
	end
	
	return LastNumber
end

print(Round(10.5)) -- 10
print(Round(2147483645)) -- YourPCWillCrashlol
#3

you could use strings i guess?

local n = X
local truncated = string.split(n,'.')[1]
#4

Why are you writing the rounding function in Lua when C/C++ one provided by math library works fine.

I am someone who has worked closely with Lua VM, so I can say this with some authority.

C/C++ code is always faster than extremely optimized Lua Code, so you should always go with the built in functions because:

  • Well Tested
  • Natively executes (close to hardware) so it’s fast.
  • Luau inlines library calls, so it’s almost equivalent of calling just one instruction.

Whatever your use case is, if you are wanting to make a custom one in Lua, I have a O(1) solution for you.

Because Luau uses double precision IEC-559 (aka IEEE-754) floats, as most do, and assuming your numbers are relatively small (the method is guaranteed to work for inputs between -251 and 251), the following efficient code will perform rounding, which is usually round to nearest, ties to even:

local function round(num)
  return num + (2^52 + 2^51) - (2^52 + 2^51)
end

But for larger numbers, you can go with another (less efficient, of course) algorithm that performs the same rounding but works for all numbers:

local function round(num)
  local ofs = 2^52
  if math.abs(num) > ofs then
    return num
  end
  return num < 0 and num - ofs + ofs or num + ofs - ofs
end
2 Likes