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Best way to get an accurate result?

 
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Flux.
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PostPosted: Fri Jul 19, 2024 1:20 pm    Post subject: Best way to get an accurate result? Reply with quote
Hello All,
Whats the best way to get an accurate result for the 1st equation?
Code:
local a = 8608480567731124086
local b = 4304240283865562043 -- half of a

print(math.tointeger(a / 2))
-- result = 4304240283865562112 fail expected b
print(math.tointeger(b * 2))
-- result = 8608480567731124086 success = a
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ParkourPenguin
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PostPosted: Fri Jul 19, 2024 1:53 pm    Post subject: This post has 1 review(s) Reply with quote
`a/2` converted the value to a double. Use `a//2` (integer division) instead
Code:
local a = 8608480567731124086
local b = 4304240283865562043

assert(a//2 == b)

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Flux.
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PostPosted: Fri Jul 19, 2024 2:00 pm    Post subject: Reply with quote
Ok that works, thanks for that quick reply.
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AylinCE
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PostPosted: Fri Jul 19, 2024 5:22 pm    Post subject: Reply with quote
What if there is no "b"? What if the results need to be extracted only from "a"?

Losses cannot be calculated because possible fractional division results are
passed through the "math.tointeger" or "math.floor" filter and rounded to an integer.

Here is a formula to calculate losses;

Code:
local a = 8608480567731124086

function divtointeger(...)
aa1 = tostring(...):gsub("%s","")
aa2,aa3,aa4 = string.match(aa1, "(%d+)(%p)(%d+)")
value = load('return '..aa1)()
a1 = math.tointeger(tonumber(value))
a3 = math.tointeger(tonumber(aa2) - a1)
a4 = math.tointeger(a3 - a1)
a5 = math.tointeger(a4 / 2)
a6 = math.tointeger(a1 + a5)
return a6
end

aa8 = divtointeger(a.." / 2")
print(aa8) -- 4304240283865562043

print(math.tointeger(aa8 * 2)) -- 8608480567731124086


-------------------------- Edit:
Ah this works @Parkour. Ok.

Code:
local a = 8608480567731124086

print(math.tointeger(a//2)) --> 4304240283865562043

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