短路函数列表返回\/答案

【问题标题】：Short circuiting a list of functions returns \/短路函数列表返回\/
【发布时间】：2015-03-19 10:15:38
【问题描述】：

所以我有一个使用这个签名的昂贵方法

def func(param: Int): \/[String, Int]

我正在尝试遍历参数列表并返回\/[String, List[Int]]，但只要方法返回-\/，就停止循环。

我想出了这个：

scala> def func(i: Int) = {
     |   if( i > 1 ) { println{"!!"} ;"error".left[Int]}
     |   else i.right[String]
     | }
func: (i: Int)scalaz.\/[String,Int]

scala> val toList = (dis: scalaz.\/[String,Int]) => {dis.map(i => List(i))}
toList: scalaz.\/[String,Int] => scalaz.\/[String,List[Int]] = <function1>

scala> val composed = (func _) andThen toList
composed: Int => scalaz.\/[String,List[Int]] = <function1>

scala> (1 to 3).toList.foldLeftM(List[Int]().right[String]){(dis, i) =>
     |   for{
     |     a <- dis
     |     b <- composed(i)
     |   } yield a |+| b
     | }
<console>:17: error: no type parameters for method foldLeftM: (f: (scalaz.\/[String,List[Int]], Int) => G[scalaz.\/[String,List[Int]]])(implicit M: scalaz.Monad[G])G[scalaz.\/[String,List[Int]]] exist so that it can be applied to arguments ((scalaz.\/[String,List[Int]], Int) => scalaz.\/[String,List[Int]])
 --- because ---
argument expression's type is not compatible with formal parameter type;
 found   : (scalaz.\/[String,List[Int]], Int) => scalaz.\/[String,List[Int]]
 required: (scalaz.\/[String,List[Int]], Int) => ?G[scalaz.\/[String,List[Int]]]

              (1 to 2).toList.foldLeftM(List[Int]().right[String]){(dis, i) =>
                                       ^
<console>:17: error: type mismatch;
 found   : (scalaz.\/[String,List[Int]], Int) => scalaz.\/[String,List[Int]]
 required: (scalaz.\/[String,List[Int]], Int) => G[scalaz.\/[String,List[Int]]]
              (1 to 2).toList.foldLeftM(List[Int]().right[String]){(dis, i) =>
                                                                            ^

这里的G[_] 是什么，foldLeftM 中的正确结果类型是什么？

【问题讨论】：

标签： scala scalaz either

【解决方案1】：

解决了我犯的两个错误：

误解了初始/结转值的类型，应该是List[Int]而不是\/

必须显式声明foldLeftM的类型参数

scala> def func(i: Int) = {
 |   if( i > 1 ) { println{"!!"} ;"error".left[Int]}
 |   else i.right[String]
 | }
func: (i: Int)scalaz.\/[String,Int]

scala> val toList = (dis: scalaz.\/[String,Int]) => {dis.map(i => List(i))}
toList: scalaz.\/[String,Int] => scalaz.\/[String,List[Int]] = <function1>

scala> val composed = (func _) andThen toList
composed: Int => scalaz.\/[String,List[Int]] = <function1>

scala> val f = (dis: List[Int], i: Int) => composed(i).map(r => dis |+| r)
f: (List[Int], Int) => scalaz.\/[String,List[Int]] = <function2>

scala> (1 to 4).toList.foldLeftM[({type l[a] = String \/ a})#l, List[Int]](List[Int]())(f)
!!
res16: scalaz.\/[String,List[Int]] = -\/(error)

【讨论】：