[Haskell-cafe] int to bin, bin to int
Brent Yorgey
byorgey at gmail.com
Fri Sep 28 07:39:15 EDT 2007
On 9/27/07, PR Stanley <prstanley at ntlworld.com> 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
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