Scalaz作家Monad和filterM

Question

我正在学习学习黄曲柳和给你一个更好的哈斯克尔，并想知道如何将filterM示例从LYAHFGG转换到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

这个问题是因为Scala不能真正知道如何在filterM所期望的参数中加入一个有三个洞的类型，这个参数只有一个填充了Boolean的洞。

您可以使用这样的奇怪类型-lambda语法来解决您的问题(没有测试，可能不起作用)：

val keepSmall: (Int => ({type L[T] = WriterT[Id, Vector[String], T]})#L) = ...

或者(更容易)引入一个类型别名，如下所示：

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

这将确保filterM所期望的参数类型与所提供的参数类型相匹配。