Question

我正在通过learning scalaz和Learn You A Haskell For Greater Good工作，并想知道如何将filterM示例从LYAHFGG翻译为Scala。

fst $ runWriter $ filterM keepSmall [9,1,5,2,10,3]

将keepSmall定义为

keepSmall :: Int -> Writer [String] Bool  
keepSmall x  
    | x < 4 = do  
        tell ["Keeping " ++ show x]  
        return True  
    | otherwise = do  
        tell [show x ++ " is too large, throwing it away"]  
        return False

我天真的方法以编译错误结束，我不知道如何解决这个问题！

    val keepSmall: (Int => WriterT[Id, Vector[String], Boolean]) = (x: Int) => 
      if (x < 4) for {
        _ <- Vector("Keeping " + x.shows).tell
      } yield true
      else for {
        _ <- Vector(x.shows + " is too large, throwing it away").tell
      } yield false

println(List(9,1,5,2,10,3) filterM keepSmall)

编译错误：

 Error:(182, 32) no type parameters for method filterM: (p: Int => M[Boolean])(implicit evidence$4: scalaz.Applicative[M])M[List[Int]] exist so that it can be applied to arguments (Int => scalaz.WriterT[scalaz.Scalaz.Id,Vector[String],Boolean])
 --- because ---
argument expression's type is not compatible with formal parameter type;
 found   : Int => scalaz.WriterT[scalaz.Scalaz.Id,Vector[String],Boolean]
 required: Int => ?M[Boolean]
    println(List(9,1,5,2,10,3) filterM keepSmall)
                               ^

和

Error:(182, 40) type mismatch;
 found   : Int => scalaz.WriterT[scalaz.Scalaz.Id,Vector[String],Boolean]
 required: Int => M[Boolean]
    println(List(9,1,5,2,10,3) filterM keepSmall)
                                       ^

Answer 1

问题是由于Scala无法真正知道如何将具有三个 hole 的类型放入filterM期望的参数中，Boolean只有一个洞填充val keepSmall: (Int => ({type L[T] = WriterT[Id, Vector[String], T]})#L) = ...。

你可以使用像这样的奇怪的type-lambda语法来解决你的问题（未经过测试，可能无效）：

type MyWriter[T] = WriterT[Id, Vector[String], T]
val keepSmall: (Int => MyWriter[Boolean]) = ...

或者（更容易）通过引入类型别名如下：

filterM

这将确保chmod a+x yourfile.py所期望的参数类型与您提供的参数类型相匹配。

Scalaz作家Monad和filterM

1 个答案: