Question

在2.8.1 / 2.9.0.1 REPL中尝试以下内容，第一个给出错误。

val l = List(Vector(1,2), List(3,4,5))
error: type mismatch;
 found   : scala.collection.immutable.Vector[Int]
 required: scala.collection.immutable.Seq[Int]{def companion:     scala.collection.generic.GenericCompanion[scala.collection.immutable.Seq[Any]]; protected def thisCollection: Seq[Int]{def companion: scala.collection.generic.GenericCompanion[Seq[Any]]}; def dropRight(n: Int): scala.collection.immutable.Seq[Int]{def companion: scala.collection.generic.GenericCompanion[scala.collection.immutable.Seq[Any]]}; def takeRight(n: Int): scala.collection.immutable.Seq[Int]{def companion: scala.collection.generic.GenericCompanion[scala.collection.immutable.Seq[Any]]}; def slice(start: Int,end: Int): scala.collection.immutable.Seq[Int]{def companion: scala.collection.generic.GenericCompanion[scala.collection.immutable....
      val l = List(Vector(1,2), List(3,4,5))
                         ^
:5: error: type mismatch;
 found   : List[Int]
 required: scala.collection.immutable.Seq[Int]{def companion:     scala.collection.generic.GenericCompanion[scala.collection.immutable.Seq[Any]]; protected def thisCollection: Seq[Int]{def companion: scala.collection.generic.GenericCompanion[Seq[Any]]}; def dropRight(n: Int): scala.collection.immutable.Seq[Int]{def companion: scala.collection.generic.GenericCompanion[scala.collection.immutable.Seq[Any]]}; def takeRight(n: Int): scala.collection.immutable.Seq[Int]{def companion: scala.collection.generic.GenericCompanion[scala.collection.immutable.Seq[Any]]}; def slice(start: Int,end: Int): scala.collection.immutable.Seq[Int]{def companion: scala.collection.generic.GenericCompanion[scala.collection.immutable.Seq[Any]]}; def take(n: Int):...
       val l = List(Vector(1,2), List(3,4,5))
                                    ^

虽然这成功了：

val l = List[Seq[Int]](Vector(1,2), List(3,4,5))
//evaluates fine to List[Seq[Int]] = List(Vector(1, 2), List(3, 4, 5))

scala在第一种情况下试图推断的类型是什么？它是具有结构类型的Seq吗？为什么它不能统一Vector和List？这是否有一些缺失的功能，还是这样（需要明确的类型def）以防止自己在脚下射击？

Answer 1

这显然是类型推理器中的一个错误，现在已经固定 2.9.1 ：

Welcome to Scala version 2.9.1.final (Java HotSpot(TM) Server VM, Java 1.6.0_18).
Type in expressions to have them evaluated.
Type :help for more information.

scala> import collection.immutable._
import collection.immutable._

scala>  List(Vector(1, 2, 3), List(4, 5))
res0: List[scala.collection.immutable.Seq[Int]] = List(Vector(1, 2, 3), List(4, 5))

Answer 2

根据daniel的回答here，Scala不使用Hindley-Milner类型推理，而是进行局部类型推断，从左向右移动。

在你的第一个声明中，第一个列表成员是Vector[Int]，所以Scala说“好吧我有List[Vector[Int]]但是当它到达第二个列表元素时，List[Int]，它无法将此与Vector[Int]统一起来。泛型必须是推理器的问题，因为包含数字和字符串的列表可以正确推断为List[Any]。

相关内容：向量和列表可以在==运算符的2.9.0.1中互操作。

scala> List[Int](1,2,3) == Vector[Int](1,2,3)
res2: Boolean = true

scala> List[Int](1,2,3) == Vector[Int](1,12,3)
res3: Boolean = false

Scala - 为什么无法推断包含Vector和List的列表的Seq列表？

2 个答案: