Convert Bits to Int8 Haskell

Question

Trying to convert a list of Bits (0,1) into an Int8 or something similar so that I am not wasting a byte from ByteString on only 1 bit

For example, I may have a list like [0,1,0,0,0,1,1,1,1,0], which as a ByteString represents each of those as a Byte instead of a Bit.

Answer 1

One solution would be to just fold over the list:

import Data.Bits (Bits, shiftL, (.|.))

pack :: (Num b, Foldable t, Bits b) => t b -> b
pack a = foldl go 0 a
    where go acc i = (acc `shiftL` 1) .|. i

and you get:

\> pack [0,1,0,0,0,1,1,1,1,0]
286