Haskell Monads内部

Question

我有以下代码，可以很好地编译和运行。我试图通过用case scanEmployee p of替换case newEmployee of来使它更紧凑，但它不起作用。可能有一种简单的方法可以从代码中删除newEmployee (和newTeam)，对吧？

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mapM和sequence来简化代码

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

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

(请注意，这些类型签名是简化的，实际的类型更通用。具体来说，mapM和sequence适用于任何monad (不仅仅是Either String)和任何可遍历的(不仅仅是([])))

然后写一个简单的解决方案：

scanTeam = fmap sequence . mapM scanEmployee

Answer 2

您可以使用LambdaCase和显式的>>=，而不是使用do块。结果并不是很短：

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