这是scalac bug吗?

时间:2015-01-13 01:58:50

标签: scala pattern-matching classcastexception partialfunction

class X[A](val value: A){
  def unapply[B <: A](x: X[B]) = true
}

object Main extends App {
  val int = new X(1)
  val string = new X("a")
  val pf: PartialFunction[Any, Int] = { case o @ int() => o.value }
  println(pf(string) + 1)
}
java.lang.ClassCastException: java.lang.String cannot be cast to java.lang.Integer
    at scala.runtime.BoxesRunTime.unboxToInt(BoxesRunTime.java:105)

我认为此代码应该抛出MatchError而不是ClassCastException。或者 scalac应警告此代码。

Scala 2.11.5

编辑:

Scala 2.9.3警告如下。但不是2.10,2.11

Main.scala:8: warning: non variable type-argument B in type pattern X[B] is unchecked since it is eliminated by erasure
  val pf: PartialFunction[Any, Int] = { case o @ int() => o.value }
                                                    ^

我从一开始就理解类型擦除。

换句话说,pf.isDefinedAt(string)返回true,pf.apply(string)抛出ClassCastException

Welcome to Scala version 2.11.5 (Java HotSpot(TM) 64-Bit Server VM, Java 1.7.0_67).
Type in expressions to have them evaluated.
Type :help for more information.

scala> :paste
// Entering paste mode (ctrl-D to finish)

class X[A](val value: A){
  def unapply[B <: A](x: X[B]) = true
}

val int = new X(1)
val string = new X("a")
val pf: PartialFunction[Any, Int] = { case o @ int() => o.value }

// Exiting paste mode, now interpreting.

defined class X
int: X[Int] = X@4f3cb3fc
string: X[String] = X@1a4c8e08
pf: PartialFunction[Any,Int] = <function1>

scala> pf.isDefinedAt(string)
res0: Boolean = true

scala> pf.apply(string)
java.lang.ClassCastException: java.lang.String cannot be cast to java.lang.Integer
  at scala.runtime.BoxesRunTime.unboxToInt(BoxesRunTime.java:105)
  ... 33 elided

2 个答案:

答案 0 :(得分:0)

他们竭尽全力:

scala> import reflect._
import reflect._

scala> class X[A: ClassTag](val v: A) { def unapply[B <: A](x: X[B]) = x.v match {
     |   case _: A => true
     |   case _    => false } }
defined class X

scala> val int = new X(1)
int: X[Int] = X@2a5ca609

scala> val s = new X("a")
s: X[String] = X@4667ae56

scala> val pf: PartialFunction[Any, Int] = { case o @ int() => o.v }
pf: PartialFunction[Any,Int] = <function1>

scala> pf(s)
scala.MatchError: X@4667ae56 (of class X)
  at scala.PartialFunction$$anon$1.apply(PartialFunction.scala:253)
  at scala.PartialFunction$$anon$1.apply(PartialFunction.scala:251)
  at $anonfun$1.applyOrElse(<console>:12)
  at scala.runtime.AbstractPartialFunction.apply(AbstractPartialFunction.scala:36)
  ... 33 elided

scala> pf(int)
res1: Int = 1

但我想你的意思是,如果他们静静地抱怨这个,为什么他们不为我插入一个测试实例呢?至少他们什么时候可以?

scala> val pf: PartialFunction[X[_], Int] = { case o @ int() => o.v }
<console>:12: error: inferred type arguments [_$1] do not conform to method unapply's type parameter bounds [B <: Int]
       val pf: PartialFunction[X[_], Int] = { case o @ int() => o.v }
                                                       ^
<console>:12: error: type mismatch;
 found   : _1
 required: Int
       val pf: PartialFunction[X[_], Int] = { case o @ int() => o.v }
                                                                  ^

答案 1 :(得分:0)

使用Any是代码中的一大气味。当你丢弃你的类型信息时,你就不会对编译器造成错误。

首先,我将Any替换为X[C]

def pf[C]: PartialFunction[X[C], Int] = { case o @ int() => o.value }

这会产生以下编译错误:

[error] Foo.scala:12: inferred type arguments [C] do not conform to method unapply's type parameter bounds [B <: Int]
[error]   def pf[C]: PartialFunction[X[C], Int] = { case o @ int() => o.value }
[error]                                                      ^

如果要将类型检查延迟到运行时,则需要在某处捕获类型信息。我们可以在X中执行此操作,如下所示:

import scala.reflect.runtime.universe._

class X[A: TypeTag](val value: A) {
  val typeTag = typeOf[A]
  def unapply[B](x: X[B]): Boolean =
    x.typeTag <:< this.typeTag
}

现在我们必须告诉编译器,如果匹配成功C实际上是Int

def pf[C]: PartialFunction[X[C], Int] = { case o @ int() => o.value.asInstanceOf[Int] }
println(pf(int) + 1)  
println(pf(string) + 1)

这似乎有效:

[info] Running Main 
2
[error] (run-main-7) scala.MatchError: X@1c34aaaa (of class X)
scala.MatchError: X@1c34aaaa (of class X)