Converting number base

问题

Is there a platform function that will do the following?

convertBase :: (Num a, Num b) => Int -> Int -> [a] -> [b]

Convert a number from base 'a' to base 'b' where each list item is a digit in the number. for example:

convertBase 2 10 [1,1,0,1] = [1, 3]

I hope that makes sense, let me know if i can clear anything up

回答1:

Using the digits package from Hackage:

import Data.Digits (digits, unDigits)

convertBase :: Integral a => a -> a -> [a] -> [a]
convertBase from to = digits to . unDigits from

You can add a fromIntegral in there if you need the input and output types to be different. Also, the Integral constraint makes more sense than Num, since you probably don't want to deal with complex or floating-point digits.

回答2:

The closest thing in the haskell platform is from module Numeric:

readInt :: Num a => a -> (Char -> Bool) -> (Char -> Int) -> ReadS a
showIntAtBase :: Integral a => a -> (Int -> Char) -> a -> ShowS

fromBase :: Int -> String -> Int
fromBase base = fst . head . readInt base ((<base).digitToInt) digitToInt

toBase :: Int -> Int -> String
toBase base num = showIntAtBase base intToDigit num ""

fromBaseToBase :: Int -> Int -> String -> String
fromBaseToBase from to = toBase to . fromBase from

回答3:

Couple of ideas:

• use showIntAtBase or Text.printf to convert to a string, and convert back to a different base
• write it yourself -- easier when one base is always a multiple of the other

Here is a link that might help you: http://rosettacode.org/wiki/Non-decimal_radices/Convert#Haskell -- Non-decimal radices/Convert

来源：https://stackoverflow.com/questions/10028213/converting-number-base