內部Haskell Monads

Question

我有以下代碼可以編譯並運行良好。 我試圖通過用case scanEmployee p of替換case newEmployee of來使其更緊湊，但是它不起作用。 從代碼中刪除newEmployee （和newTeam ）可能是一種簡單的方法，對嗎？

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 ...

Answer 1

您可以使用LambdaCase並使用>>=進行顯式表示，而不要使用do塊。 結果不會短很多：

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

Answer 2

您可以使用mapM和sequence來簡化代碼：

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

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

（請注意，這些類型簽名是簡化的，實際類型更通用。具體而言， mapM和sequence適用於任何monad（不僅是Either String ）和任何可遍歷的（不僅是([]) ））

並編寫一個簡單的解決方案：

scanTeam = fmap sequence . mapM scanEmployee