Re: [Haskell-cafe] int to bin, bin to int
On 9/27/07, PR Stanley [EMAIL PROTECTED] 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 Others have already given many good suggestions, but I'll add something specifically about the order of the bits in the result. You have the generated list of bits in reading order, i.e. high-order bits first. But perhaps it would make more sense to have the low-order bits first? At least, it would be more efficient that way. Then you could do intToBin n = (n `mod` 2) : (intToBin (n `div` 2) The way you have it now, you will end up with something like [1] ++ [0] ++ [0] ++ [1] ++ ... which ends up inefficiently traversing the list multiple times. To undo, just (for example) binToInt xs = sum $ zipWith (*) xs (iterate (*2) 1). -Brent ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
[Haskell-cafe] int to bin, bin to int
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 ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] int to bin, bin to int
prstanley: 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. One of my favourites is: unroll :: Integer - [Word8] unroll = unfoldr step where step 0 = Nothing step i = Just (fromIntegral i .. 1, i `shiftR` 1) roll :: [Word8] - Integer roll = foldr unstep 0 where unstep b a = a `shiftL` 1 .|. fromIntegral b -- Don ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] int to bin, bin to int
On 9/27/07, PR Stanley [EMAIL PROTECTED] 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 ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] int to bin, bin to int
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 [EMAIL PROTECTED] 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 ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
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 [EMAIL PROTECTED] 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 ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
