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内部Haskell Monads

[英]Inside Haskell Monads

I have the following code that compiles and runs fine. 我有以下代码可以编译并运行良好。 I tried to make it more compact by replacing case newEmployee of with case scanEmployee p of , but it didn't work. 我试图通过用case scanEmployee p of替换case newEmployee of来使其更紧凑,但是它不起作用。 There's probably an easy way to remove newEmployee (and newTeam ) from the code right? 从代码中删除newEmployee (和newTeam )可能是一种简单的方法,对吗?

module Main( main ) where
import Control.Monad.State

data Employee  = EmployeeSW  Int Int | EmployeeHW Int String deriving ( Show )
data Employee' = EmployeeSW'     Int | EmployeeHW'    String deriving ( Show )

scanTeam :: [Employee] -> State (Int,Int) (Either String [Employee'])
scanTeam [    ] = return (Right [])
scanTeam (p:ps) = do
    newEmployee <- scanEmployee p
    case newEmployee of
        Left errorMsg -> return (Left errorMsg)
        Right e -> do
            newTeam <- scanTeam ps
            case newTeam of
                Right n -> return (Right (e:n))
                Left errorMsg -> return (Left errorMsg)

scanEmployee :: Employee -> State (Int,Int) (Either String Employee')
-- actual code for scanEmployee omitted ...

You could use LambdaCase and be explicit with >>= instead of using do blocks. 您可以使用LambdaCase并使用>>=进行显式表示,而不要使用do块。 The result is not much shorter: 结果不会短很多:

scanEmployee p >>= \case
    Left errorMsg -> return (Left errorMsg)
    Right e       -> do ...

You can simplify your code a bit with mapM and sequence : 您可以使用mapMsequence来简化代码:

mapM scanEmployee :: [Employee] -> State (Int, Int) [Either String Employee')

sequence :: [ Either String a ] -> Either String [ a ]

(Note that these type signatures are simplifications and the actual types are more general. Specifically mapM and sequence work for any monad (not just Either String ) and any traversable (not just ([]) )) (请注意,这些类型签名是简化的,实际类型更通用。具体而言, mapMsequence适用于任何monad(不仅是Either String )和任何可遍历的(不仅是([]) ))

And write a simple solution: 并编写一个简单的解决方案:

scanTeam = fmap sequence . mapM scanEmployee

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