Haskell Show for String的实例是如何编写的？

我有一个关于定义类型类实例的基本问题。我以ShowTypeClass为例，只考虑类中的函数Show。像Bool这样的具体类型的Show实例很简单

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

但对于字符串，它不是：

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

产生一个可以理解的错误

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’

当然，以下情况是不允许的：

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

我可以定义一个新类型

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

但这不允许我做“测试”，就像我在Haskell做的那样

---

我缺少类型类的哪些基本功能？

Show类型类实际上有三个成员函数，

Show

、

showsPrec

和

showList

。在

Show Char

的实例中，

showList

函数被重载以输出引号，并将所有字母挤在一起而不使用分隔符：

发件人：

其中定义为：

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)

---

因此没有

Show String

实例，只是

Show Char

定义了如何在

[Char]

值上调用

Show

。

字符串格式是Char实例的一部分，通过showList方法。

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)