求教等式为啥成立

[eote]

newtype Parser a = Parser (String -> [(a, String)])

apply :: Parser a -> String -> [(a, String)]
apply (Parser p) s = p s

instance Monad Parser where 
 return x = Parser (\s -> [(x, s)])
 p >>= q  = Parser (\s -> [(y, s'') | (x, s') <- apply p s, (y, s'') <- apply (q x) s'])


fail = Parser (\s -> [])

guard :: Bool -> Parser ()
guard True  = return ()
guard False = fail


sat :: (Char -> Bool) -> Parser Char
sat p = do { c <- getc; guard (p c); return c }


-- 以上是背景
-- 以下是问题

-- 为啥 guard (p c) >> return c = fail >> return c = fail
-- p c == True 很好理解 guard (p c)结果被弃掉了
-- p c == False 为啥没走到 return c

[iamzuoxinyu]

https://en.wikibooks.org/wiki/Haskell/do_notation#the_fail_method

[eote]

@iamzuoxinyu url 说的是 pattern failure 吧

[leomayleomay]

Prelude 里面的 guard 貌似需要 context 是 Alternative https://hackage.haskell.org/package/base-4.16.1.0/docs/Control-Monad.html#v:guard