How do I get the closes power of 2 from a number?

[Katterkatt1 (DUCKIE)]

So I need to get the nearest power2 like (2, 4, 8, 16, 32) etc.
What I mean by that is for example:

function getClosestPow2(n)
	--return some wizard math IDK	
end

local floatNumber = 15.0
local powerOf2Number = getClosestPow2(floatNumber)

print(powerOf2Number)

Output:

16 --as it is the closest power of 2 to 15.0

[0786ideal (ideal)]

return math.floor(x/2)*2 -- returns closest multiplication

[Rescriptedd (rescripted)]

does closest multiplication and closest exponent mean the same thing? no.

[Katterkatt1 (DUCKIE)]

@0786ideal

return math.floor(x/2)*2 -- returns closest multiplication

Not to sound rude, but this isn’t really what I asked about as it just returns the number n - 1

[Negativize]

Use the log function base 2. This question is vague because I’m not sure if you are wanting the closest power of 2 or the closest power of 2 that doesn’t exceed the result. I am going to assume the former. If you want the latter, simply get rid of the extra addition.

function closestpow(n)
  return math.floor(math.log(n, 2) + 0.5)
end

print(closestpow(56)) 
-- 6 (2^6 is 64)

EDIT: I misunderstood your question, but this function can be used to get what you want. Just raise 2 to the power of the output of closestpow:

function closestpow(n)
  return math.floor(math.log(n, 2) + 0.5)
end

print(math.pow(2, closestpow(15)))
-- 16

[Katterkatt1 (DUCKIE)]

@Negativize

function closestpow(n)
  return math.floor(math.log(n, 2) + 0.5)
end

print(math.pow(2, closestpow(15)))
-- 16

thank you soo much
I would never have figured something like this out on my own!

[Negativize]

If you want a reason as to why, it’s because the logarithm base 2 of a number returns what you would need to raise 2 by to get that number.

For example, log2(9) is roughly 3.17, which means 2^3.17 is roughly 9. Simply put, if you take the logarithm base 2 of your input and round it to get n, you can also find the closest integer value of 2^n, which is what the above function does.

In fact, you can generalize your solution to any base by adding an additional parameter to the function:

function closestpow(n, b)
  return math.pow(b, (math.floor(math.log(n, b) + 0.5)))
end

print(closestpow(166, 3))
-- Closest power of 3 to 166
-- 243

print(closestpow(987, 10))
-- Closest power of 10 to 987
-- 1000

[0786ideal (ideal)]

I wanted to show you overall math how to do multiplication, you could use it