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

On 9/27/07, 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

IntToBin diverges for inputs <= 0. You could get 0 "for free" with intToBin :: Int -> [Int] intToBin 0 = [] intToBin n = (intToBin (n`div`2)) ++ [n `mod` 2] And why not use [Bool] for the "Bin" type? Or data Bin = Zero | One Regards, Chris