Haskell 返回十六进制值列表

Question

I am trying to define the function "decToHex" which returns a list of chars that corresponds to the hexadecimal value of a given number. For example, 1128 = ['4','6','8']

This is my function

decToHex :: Int -> [Char]
list = ["0","1","2","3","4","5","6","7","8","9","A","B","C","D","E","F"]
listHexa = []
decToHex x
  | x < 16 = list !! x
  | otherwise = decToHex (x `div` 16) : ((list !! (x `mod` 16)) : listHexa)

But I get this error:

• Couldn't match type ‘[Char]’ with ‘Char’
  Expected type: [Char]
    Actual type: [[Char]]
• In the expression:
    decToHex (x `div` 16) : ((list !! (x `mod` 16)) : listHexa)

What am I missing?

Answer 1

Your list is of type [String] , because every letter there is a String . Instead, make it a list of Char by replacing double quotes with single ones.

Answer 2

Recall asymmetric operator : adds an element to the left of a list, while symmetric operator ++ concatenates two lists.

Hence, for an expression that starts with:

decToHex (x `div` 16) :  ... whatever ...

the left operand of : is already a string, that is a list of Char's. So the type of the whole expression has to be [[Char]] . But this conflicts with your type signature for function decToHex , which says it returns just a [Char] simple list.

Side note 1: things are less confusing when the list of hexadecimal digits is called hexDigits rather than plain list .

Side note 2: in a division, if you need both the quotient and the remainder, it is more readable and more efficient to use function divMod rather than div and mod separately.

We thus have the following code:

hexDigits  = ["0","1","2","3","4","5","6","7","8","9","A","B","C","D","E","F"]

decToHex :: Int -> [Char]
decToHex x
    | x < 0      =  '-' : decToHex (-x)
    | x < 16     =  hexDigits !! x
    | otherwise  =  let  (q, r) = divMod x 16
                    in  (decToHex q)  ++  (hexDigits !! r)

Side note 3:

Just like everything else, lists are immutable in Haskell. This implies that operator ++ has to work by duplicating its left operand . Hence, using ++ for recursion in such a way is very inefficient. You might want to find a way to use operator : as your main recursion engine instead. You will probably need a simpler hexDigits list like "0123456789ABCDEF" as mentioned in arrowd's answer.