parsing - 函数定义中的相互引用

Question

一些和许多定义here具有相互引用的内部表达式。我觉得这很难理解。

some :: f a -> f [a]
some v = some_v
  where
    many_v = some_v <|> pure []
    some_v = (:) <$> v <*> many_v

-- | Zero or more.
many :: f a -> f [a]
many v = many_v
  where
    many_v = some_v <|> pure []
    some_v = (:) <$> v <*> many_v

many_v 和 some_v 的类型签名是什么？

如何评估以下内容(使用parsec)？

Prelude Text.Parsec> parse (many (oneOf "abc")) mempty "abc"
Right "abc"

Answer 1

这两个函数都有像

这样的签名

many_v :: (Alternative f) => f [a]
some_v :: (Alternative f) => f [a]

用可以简化的定义替换绑定(bind)

some v = some_v
  where
    some_v = (:) <$> v <*> (some_v <|> pure [])

many v = many_v
  where
    many_v = ( (:) <$> v <*> many_v) <|> pure []

<小时/>

some v = (:) <$> v <*> (some v <|> pure [])
many v = ((:) <$> v <*> many v) <|> pure []

<小时/>

通常您将这些函数简单地定义为

many v = some v <|> pure []
some v = (:) <$> v <*> many v

在大多数项目中，递归函数都是定义的工作包装函数，以便编译器能够进行更多优化