定义空特性测试:
trait Test
复合型中使用的是什么:
scala> val a : Int with Test = 10.asInstanceOf[Int with Test]
a: Int with Test = 10
和带有复合类型参数的case类(如Unboxed Tagged Type):
scala> case class Foo(a: Int with Test)
error: type mismatch;
found : Double
required: AnyRef
Note: an implicit exists from scala.Double => java.lang.Double, but
methods inherited from Object are rendered ambiguous. This is to avoid
a blanket implicit which would convert any scala.Double to any AnyRef.
You may wish to use a type ascription: `x: java.lang.Double`.
但它完全适用于:
scala> case class Foo(a: List[Int] with Test)
defined class Foo
方法定义没问题:
scala> def foo(a: Int with Test) = ???
foo: (a: Int with Test)Nothing
Scala版 2.10.3
编译器行为是否正常?
答案 0 :(得分:5)
你遇到了Scala尝试统一原语和对象的情况之一。由于Scala中的Int
表示Java基元类型int
,因此不能将任何特征混合到其中。在执行asInstanceOf时,Scala编译器将Int
自动装箱到java.lang.Integer
:
scala> val a: Int with Test = 10.asInstanceOf[Int with Test]
a: Int with Test = 10
scala> a.getClass
res1: Class[_ <: Int] = class java.lang.Integer
但是,在声明类型时不会发生自动装箱,因此您必须手动执行:
scala> case class Foo(x: Integer with Test)
defined class Foo
但是在检查类型之前,编译器类型检查器不会自动装箱:
scala> Foo(a)
<console>:12: error: type mismatch;
found : Int with Test
required: Integer with Test
Foo(a)
^
因此,您必须将变量声明为Integer with Test
:
scala> val a: Integer with Test = 10.asInstanceOf[Integer with Test]
a: Integer with Test = 10
scala> Foo(a)
res3: Foo = Foo(10)
或在调用案例类时使用演员:
val a : Int with Test = 10.asInstanceOf[Int with Test]
scala> a: Int with Test = 10
scala> Foo(a.asInstanceOf[Integer with Test])
res0: Foo = Foo(10)
答案 1 :(得分:2)
@Travis Brown所说,这是已知的issue,已在scala 2.11.7中修复。
在ammonite Repl 2.0.4(scala 2.12.10 java 1.8.0_242)下运行
@case class Foo(a: Int with Test)
a: Int with Test = 10