The

Question

我得到了以下代码

 newtype Parser a = Parser { parse :: String -> Maybe (a,String) }

   instance Applicative Parser where
   pure a = Parser $ \s -> Just (a,s)
   f <*> a = Parser $ \s ->
     case parse f s of
       Just (g,s') -> parse (fmap g a) s'
       Nothing -> Nothing

    instance Alternative Parser where
      empty = Parser $ \s -> Nothing
      l <|> r = Parser $ \s -> parse l s <|> parse r s

    satisfy :: (Char -> Bool) -> Parser Char
    satisfy p = Parser f
       where f [] = Nothing
             f (x:xs) = if p x then Just (x,xs) else Nothing

    ws :: Parser ()
    ws = pure () <* many (satisfy isSpace)

我知道ws解析器会删除前导空格。但是我需要一个解释，说明它是如何完成的。我不了解<*语法（我知道<*>）。可以帮助我理解语法以及它为什么起作用吗？

Answer 1

没有恐惧！当我们经过小步骤并查看它将我们引向何方时，这很容易。下次，请尝试自己做。

[(<*) = liftA2 const和(<*) = liftA2 const，所以

liftA2 f x = (<*>) (fmap f x)

（（带有适用性表示法；以了解其用途。在以上链接中也列举了适用性法律。）

所以它执行liftA2 f x = (<*>) (fmap f x)的操作，除了它返回的计算值始终为

ws :: Parser ()
ws = pure () <* many (satisfy isSpace) 
     = {- (<*) = liftA2 const -}
     liftA2 const (pure ()) (many $ satisfy isSpace) 
     = {- liftA2 f x = (<*>) (fmap f x) -}
     fmap const (pure ()) <*> many (satisfy isSpace)
     = {- by Applicative laws -}
     (pure const <*> pure ()) <*> many (satisfy isSpace)
     = {- by homomorphism law of Applicatives -}
     pure (const ()) <*> many (satisfy isSpace)
     = {- with Applicative Do -}
     do { f <- pure (const ())
        ; r <- many (satisfy isSpace)
        ; pure (f r) }
     =
     do { r <- many (satisfy isSpace)
        ; pure (const () r) }
     =
     do { _ <- many (satisfy isSpace)
        ; pure () }
     =
     do { many (satisfy isSpace)
        ; pure () }
     = {- Monad Comprehension illustration -}
     [ () | _ <- many (satisfy isSpace) ]

。

在纯函数中，many (satisfy isSpace)将完全忽略其第二个参数。我们可以将其称为()，它不在乎。但是Applicative Functor表示计算，并且f a = ()在组合之前将众所周知的计算进行组合，因此无论以后是否使用其计算值，每个函数都将得到[[performed。