I'm in a special circumstance where I would like to convert an integer into a bytes object of the smallest length possible. I currently use the following method to covert to bytes:
number = 9847
bytes = number.to_bytes(4, 'little')
However I would like to scale that the amount of bytes used down (the 4) to the smallest possible size. How can I achieve this?
You can get an exact conversion without needing to use log arithmetic yourself by querying the int for how many bits it needs for its value like so:
import math
def convert(i: int) -> bytes:
return i.to_bytes(math.ceil(i.bit_length() / 8))
I figured it out on my own! I use the following function to do the conversion to bytes for me now:
import math
def int_to_bytes(self, integer_in: int) -> bytes:
"""Convert an integer to bytes"""
# Calculates the least amount of bytes the integer can be fit into
length = math.ceil(math.log(integer_in)/math.log(256))
return integer_in.to_bytes(length, 'little')
This works because with exponents a = b^e is equivalent to e = log(a)/log(b)
In this case our problem is integer_in = 256^e, and we want to solve for e. This can be solved by rephrasing it to e = log(integer_in)/log(256). Lastly, we use math.ceil() to round up the answer to an integer.
