Question

我有一个关于定义类型类实例的基本问题。我使用Show类型类作为示例，我只考虑类中的函数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}

但是对于String，它不是：

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

产生可以理解的错误

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’

当然不允许以下内容：

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}

然而，我不允许我做表演＆＃34;测试＆＃34;，因为我能够在Haskell中做。

我错过了类型类的哪些基本特征？

Answer 1

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推送到一起而没有分隔符：

来自GHC.Show：

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

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

showLitString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个实例，只是Show Char定义了如何专门针对show值调用[Char]。

如何编写Show for String的实例？

1 个答案: