Re: [Haskell-cafe] int to bin, bin to int

I might be inclined to use data Bin = Zero | One (or at least type Bin = Bool) to let the type system guarantee that you'll only ever have binary digits in your [Bin], not any old integer. Using [Int] is an abstraction leak, inviting people to abuse the representation behind your back. Rodrigo Queiro wrote:

If you don't like explicit recursion (or points):

intToBin = map (`mod` 2) . takeWhile (>0) . iterate (`div` 2)

binToInt = foldl' (\n d -> n*2+d) 0 or even: binToInt = foldl' ((+).(*2)) 0

On 27/09/2007, PR Stanley wrote:

...

Hi intToBin :: Int -> [Int] intToBin 1 = [1] intToBin n = (intToBin (n`div`2)) ++ [n `mod` 2]

binToInt :: [Integer] -> Integer binToInt [] = 0 binToInt (x:xs) = (x*2^(length xs)) + (binToInt xs) Any comments and/or criticisms on the above definitions would be appreciated. Thanks , Paul

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