[英]Haskell - Attempting Monad instance - syntax issues
I have the following definitions: 我有以下定义:
newtype Parser a = Parser { parse :: String -> [(a,String)] }
instance Functor Parser where
fmap g (Parser pa) = Parser { parse = \s -> [(g a,s') | (a,s') <- pa s] }
instance Applicative Parser where
pure x = Parser { parse = \s -> [(x,s)] }
(<*>) (Parser pg) (Parser pa) = Parser { parse = \s -> [(g a,s'') | (g,s') <- pg s, (a, s'') <- pa s'] }
instance Monad Parser where
return = pure
(>>=) (Parser pa) g = Parser { parse = \s -> [(b,s'') | [(a, s')] <- (pa s), [(b,s'')] <- (g a) s'] }
The trouble is with the bind operator implementation. 问题在于绑定运算符的实现。 I am not sure why I get type error there. 我不确定为什么会在那里输入类型错误。 I am attempting to define the bind operator on the same lines as (<*>) 我正在尝试在与(<*>)相同的行上定义绑定操作符
The error is: 错误是:
parser.hs:17:58: error:
• Couldn't match expected type ‘(a, String)’
with actual type ‘[(a, t0)]’
• In the pattern: [(a, s')]
In a stmt of a list comprehension: [(a, s')] <- (pa s)
In the expression:
[(b, s'') | [(a, s')] <- (pa s), [(b, s'')] <- (g a) s']
• Relevant bindings include
g :: a -> Parser b (bound at parser.hs:17:20)
pa :: String -> [(a, String)] (bound at parser.hs:17:16)
(>>=) :: Parser a -> (a -> Parser b) -> Parser b
(bound at parser.hs:17:2)
|
17 | (>>=) (Parser pa) g = Parser { parse = \s -> [(b,s'') | [(a, s')] <- (pa s), [(b,s'')] <- (g a) s'] }
| ^^^^^^^^^
parser.hs:17:92: error:
• Couldn't match expected type ‘t0 -> [[(b, String)]]’
with actual type ‘Parser b’
• The function ‘g’ is applied to two arguments,
but its type ‘a -> Parser b’ has only one
In the expression: (g a) s'
In a stmt of a list comprehension: [(b, s'')] <- (g a) s'
• Relevant bindings include
s' :: t0 (bound at parser.hs:17:63)
g :: a -> Parser b (bound at parser.hs:17:20)
(>>=) :: Parser a -> (a -> Parser b) -> Parser b
(bound at parser.hs:17:2)
|
17 | (>>=) (Parser pa) g = Parser { parse = \s -> [(b,s'') | [(a, s')] <- (pa s), [(b,s'')] <- (g a) s'] }
Update: For the sake of completeness, based on the answers. 更新:为了完整起见,基于答案。 Here is what worked: 这是起作用的:
instance Monad Parser where
return = pure
(>>=) (Parser pa) g = Parser { parse = \s -> [(b,s'') | (a, s') <- (pa s), (b,s'') <- parse (g a) s'] }
A guard pattern like [(a, s')] <- (pa s)
means you're expecting pa s
to return a list of lists with each exactly one pair of a
and s
in them, whereas it actually returns just a list with any number of those pairs in them. 像[(a, s')] <- (pa s)
类的保护模式意味着您期望pa s
返回一个列表列表,每个列表中正好有一对a
和s
,而实际上它仅返回一个列表与其中任何数量的对。
Basically, drop the brackets (and also from the other binding) and then it should work. 基本上,放下括号(以及从其他绑定中放下),然后它应该起作用。
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