当使用.isInstanceOf[GenericType[SomeOtherType]]
,其中GenericType
和SomeOtherType
是任意类型的(合适的那种),Scala编译器给出了一个未检查警告由于类型擦除:为什么`Some(123).isInstanceOf [Option [List [String]]]`* *不会给出未经检查的警告?
scala> Some(123).isInstanceOf[Option[Int]]
<console>:8: warning: non variable type-argument Int in type Option[Int] is unchecked since it is eliminated by erasure
Some(123).isInstanceOf[Option[Int]]
^
res0: Boolean = true
scala> Some(123).isInstanceOf[Option[String]]
<console>:8: warning: non variable type-argument String in type Option[String] is unchecked since it is eliminated by erasure
Some(123).isInstanceOf[Option[String]]
^
res1: Boolean = true
然而,如果SomeOtherType
本身是一个通用类型(例如List[String]
),无警告发出:
scala> Some(123).isInstanceOf[Option[List[String]]]
res2: Boolean = true
scala> Some(123).isInstanceOf[Option[Option[Int]]]
res3: Boolean = true
scala> Some(123).isInstanceOf[Option[List[Int => String]]]
res4: Boolean = true
scala> Some(123).isInstanceOf[Option[(String, Double)]]
res5: Boolean = true
scala> Some(123).isInstanceOf[Option[String => Double]]
res6: Boolean = true
(记得,元组和=>
是Tuple2[]
和Function2[]
一般类型语法糖)
为什么没有发出警告? (所有这些都在斯卡拉REPL 2.9.1,与-unchecked
选项。)
大调查! – 2012-07-19 08:06:30
的确,工作很好! – pedrofurla 2012-07-19 14:46:52
用于引用编译器。 :-) – 2012-07-19 18:03:19