How to generate number with infinite range?

[Ongyj888 (duccblox)]

I need to generate numbers but the range is infinite.

I have tried this:

local num = math.random(1, math.huge)
print(num)

But it throwed this error:

ServerScriptService.Script:6: invalid argument #2 to ‘random’ (interval is empty)

[DasKairo (Cairo)]

It isnt really Possible, and there are limitations to the pseudorandom number generation.

math.random() has an Integer Limit of (2^31) - 1 or (2147483647), Around 2.1 Billion, which the exact number for the 32-Bit Integer Limit.

Random.new() has an Integer Limit of (2^63) - 1 or (9223372036854776000), which is 9.2 Quintillion, or in other words, the 64-Bit Integer Limit.

These numbers are Important because the are the exact limits that these pseudorandom number generators can handle, and going over these limits will result in a Integer Overflow, Typically the error occurs because the number given in math.random() is going over the 32-Bit Integer Limit, that It becomes a Negative Number, The Second Argument Cannot be Negative, so it throws the error.

[skipper987 (skip)]

Edit: This post goes into some detail on the mechanics of the method below.

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Complementing the detailed response by @DasKairo , an implementation of Random.new() would look something like this:

rand = Random.new()
num = rand:NextInteger(1, 1e90)
print(num)

[Ongyj888 (duccblox)]

So basically if I wanted to generate the number “without range”

Will this work?

local num = math.random(-2147483647, 2147483647)
or
local ran = Random.new()
ran:NextNumber(-9223372036854776000, 9223372036854776000)

[DasKairo (Cairo)]

It should, as they arent going past any limits, but if you just want infinite, use math.huge