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
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.
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
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
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.
