How is the instance of Show for String written?
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?
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.
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