[Haskell-beginners] Convert bits to bytes

[Alec Benzer]

Thanks - I thought there would be some api for it but I see it's pretty
simple to do on your own.

I came up with this (dealing with non-multiple of 8 numbers of bits)

bitsToBytes :: [Bool] -> [Word8]
bitsToBytes [] = []
bitsToBytes bits = map bitsToByte (getChunks bits)
  where bitsToByte = foldl' (\byte bit -> byte*2 + if bit then 1 else 0) 0

getChunks :: [Bool] -> [[Bool]]
getChunks [] = []
getChunks xs
  | length xs < 8 = getChunks $ take 8 (xs ++ repeat False)
  | otherwise =
    let (these,rest) = splitAt 8 xs
    in these:getChunks rest


On Thu, Sep 2, 2010 at 3:18 AM, David Virebayre
<dav.vire+haskell at gmail.com<dav.vire%2Bhaskell at gmail.com>
> wrote:

> I could have been a bit more verbose.
>
> The way I see it, since you have an arbitrary long list of 'bits' that
> you want to convert into bytes, the first thing to do is to group this
> list into sublists of 8 bits.
>
> That's what chunk does: it splits the list at the 8th element, and
> recursively does it for the rest of the list, until the list is empty.
>
> One problem with that is that if the length of the list isn't a
> multiple of 8, then the last byte might be incorrect.
>
> > chunk :: [Bool] -> [[ Bool ]]
> > chunk [] = []
> > chunk l = a : chunk b
> >  where (a,b)  = splitAt 8 l
>
> This one converts a list of 'bits' into a number. The head of the list
> is assumed to be the most significant bit :
>
> > conv1 = foldl' (\a b -> a*2 + if b then 1 else 0) 0
>
> if we want the head of the list to be the least significant bit, then
> you can convert with foldr :
>
> > conv1' = foldr (\b a -> a*2 + if b then 1 else 0) 0
>
> Now converting the whole list is just a matter converting the whole
> list in groups, then converting each group :
>
> > convlist = map conv1 . chunk
> >
> > test = convlist (replicate 8 True ++ replicate 8 False :: [Bool] )
>
>
> David.
>
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