如何编写Show for String的实例？

Question

I have a basic question about definining instances of typeclasses. I am using the Show typeclass as an example, and I am considering only the function show within the class. The Show instance for a concrete type like Bool is simple

instance Show Bool where
  show x = {function of x here}

but for String it is not:

instance Show String where
  show x = {function of x here}

produces understandably an error

Illegal instance declaration for ‘Formatter String’
  (All instance types must be of the form (T t1 ... tn)
   where T is not a synonym.
   Use TypeSynonymInstances if you want to disable this.)
In the instance declaration for ‘Formatter String’

and of course the following is not allowed:

instance Show [Char] where
  show x = {function of x here}

I could define a newtype

newtype String2 = String2 String 
instance Formatter String2 where
  format (String2 x) = {function of x here}

which however does not allow me to do show "test", as I am able to do in Haskell.

What essential feature of typeclasses am I missing?

Answer 1

The Show typeclass actually has three member functions, show , showsPrec , and showList . In the instance for Show Char , the showList function is overloaded to output quote marks and shove all the letters together without delimiters:

From GHC.Show :

instance  Show Char  where
    showsPrec _ '\'' = showString "'\\''"
    showsPrec _ c    = showChar '\'' . showLitChar c . showChar '\''

    showList cs = showChar '"' . showLitString cs . showChar '"'

Where showLitString is defined as:

showLitString :: String -> ShowS
-- | Same as 'showLitChar', but for strings
-- It converts the string to a string using Haskell escape conventions
-- for non-printable characters. Does not add double-quotes around the
-- whole thing; the caller should do that.
-- The main difference from showLitChar (apart from the fact that the
-- argument is a string not a list) is that we must escape double-quotes
showLitString []         s = s
showLitString ('"' : cs) s = showString "\\\"" (showLitString cs s)
showLitString (c   : cs) s = showLitChar c (showLitString cs s)

So there is no Show String instance, it's simply that Show Char defines how to call show on [Char] values specifically.