Question

查看此REPL会话（我为了便于阅读而对其进行了整理）：

scala> val x = 1 to 10
x: Range.Inclusive = Range(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)

scala> val y = x.toSeq
y: Range = Range(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)

scala> x eq y
res14: Boolean = true

scala> util.Random.shuffle(y)
<console>:10: error: Cannot construct a collection of type scala.collection.AbstractSeq[Int] with elements of type Int based on a collection of type scala.collection.AbstractSeq[Int].
              util.Random.shuffle(y)
                                 ^

scala> util.Random.shuffle(x)
res16: scala.collection.immutable.IndexedSeq[Int] = Vector(8, 3, 4, 2, 10, 9, 7, 5, 6, 1)

首先，无论类型是否不同，这都应该有效。 问题是＆＃39;为什么？＆＃39;

Answer 1

这是SI-6948，a bug由基本的scala破坏引起。

这是一个nice long commit message，还有一些额外的解释。

Answer 2

出于某种原因，shuffle的类型推断会为Inclusive而不是Range产生不同的结果。

toSeq导致Range的原因是它的定义无辜地缩小了类型：

override def toSeq = this

有一个未解决的问题来推断重写方法的结果类型。

显示REPL没有说谎：

scala> import util.Random.shuffle
import util.Random.shuffle

scala> val x = 1 to 10
x: scala.collection.immutable.Range.Inclusive = Range(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)

scala> val y = x.toSeq
y: scala.collection.immutable.Range = Range(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)

scala> val z: Range = x
z: Range = Range(1, 2, 3, 4, 5, 6, 7, 8, 9, 10)

scala> shuffle(x)
res0: scala.collection.immutable.IndexedSeq[Int] = Vector(1, 5, 2, 10, 9, 6, 3, 7, 4, 8)

scala> shuffle(y)
<console>:11: error: Cannot construct a collection of type scala.collection.AbstractSeq[Int] with elements of type Int based on a collection of type scala.collection.AbstractSeq[Int].
              shuffle(y)
                     ^

剪切以查看推断的内容以及使用的隐式内容：

scala> :replay -Xprint:typer

Replaying: shuffle(x)
        private[this] val res0: scala.collection.immutable.IndexedSeq[Int] = scala.util.Random.shuffle[Int, scala.collection.immutable.IndexedSeq]($line5.$read.$iw.$iw.x)(immutable.this.IndexedSeq.canBuildFrom[Int]);

scala> shuffle(y)
        private[this] val <res1: error>: <error> = scala.util.Random.shuffle[Int, scala.collection.AbstractSeq]($line6.$read.$iw.$iw.y)();

而不是你所希望的：

scala> shuffle[Int, collection.immutable.IndexedSeq](z)
res3: scala.collection.immutable.IndexedSeq[Int] = Vector(6, 5, 3, 8, 4, 1, 2, 10, 7, 9)

使用-Ytyper-debug，有一个额外的类型参数A似乎让它继续进行Inclusive，但我不知道它来自何处。

|    |    |    |    |-- x BYVALmode-EXPRmode-POLYmode (site: value res3  in $iw) 
|    |    |    |    |    \-> scala.collection.immutable.Range.Inclusive
|    |    |    |    solving for (T: ?T, CC: ?CC)
|    |    |    |    solving for (A: ?A)
|    |    |    |    [adapt] [A]=> scala.collection.generic.CanBuildFrom[scala.collect... adapted to [A]=> scala.collection.generic.CanBuildFrom[scala.collect... based on pt scala.collection.generic.CanBuildFrom[scala.collection.immutable.IndexedSeq[Int],Int,scala.collection.immutable.IndexedSeq[Int]]
|    |    |    |    |-- [T, CC[X] <: TraversableOnce[X]](xs: CC[T])(implicit bf: ... EXPRmode (site: value res3  in $iw) 
|    |    |    |    |    \-> scala.collection.immutable.IndexedSeq[Int]
|    |    |    |    [adapt] [T, CC[X] <: TraversableOnce[X]](xs: CC[T])(implicit bf: ... adapted to [T, CC[X] <: TraversableOnce[X]](xs: CC[T])(implicit bf: ...
|    |    |    |    \-> scala.collection.immutable.IndexedSeq[Int]

是错误还是行为？

说清楚：

scala> import language.higherKinds, collection.TraversableOnce, collection.generic.CanBuildFrom
import language.higherKinds
import collection.TraversableOnce
import collection.generic.CanBuildFrom

scala> def f[T, CC[X] <: TraversableOnce[X]](xs: CC[T])(implicit cbf: CanBuildFrom[CC[T],T,CC[T]]): CC[T] = null.asInstanceOf[CC[T]]
f: [T, CC[X] <: scala.collection.TraversableOnce[X]](xs: CC[T])(implicit cbf: scala.collection.generic.CanBuildFrom[CC[T],T,CC[T]])CC[T]

scala> f(1 to 10)
res0: scala.collection.immutable.IndexedSeq[Int] = null

scala> f(1 until 10)
<console>:12: error: Cannot construct a collection of type scala.collection.AbstractSeq[Int] with elements of type Int based on a collection of type scala.collection.AbstractSeq[Int].
              f(1 until 10)
               ^

斯卡拉的洗牌范围很奇怪

2 个答案: