[Haskell-cafe] convert a list of booleans into Word*
Roel van Dijk
vandijk.roel at gmail.com
Wed Sep 30 06:20:01 EDT 2009
I wrote a few variants for fun. Probably equally inefficient. I
suggest you look at Data.Binary as Andrew suggested.
-- Your original function, but with a more generic type signature.
encodeBits :: Bits n => [Bool] -> n
encodeBits bs = go 0 0 bs
where
go n r [] = r
go n r (b:bs) = go (n+1) (if b then setBit r n else clearBit r n) bs
-- Combine the flags with their index and then set bits when appropriate.
encodeBits2 :: Bits n => [Bool] -> n
encodeBits2 = foldr (\(n, b) x -> setBitIf b x n) 0 . zip [0..]
where
setBitIf False x _ = x
setBitIf True x n = setBit x n
-- Shift the result left while constructing and only toggle the first bit.
encodeBits3 :: Bits n => [Bool] -> n
encodeBits3 bs = foldr (\b x -> setBitIf b (x `shiftL` 1)) 0 bs
where
setBitIf False x = x
setBitIf True x = setBit x 0
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