复合类型的案例类伴随对象生成错误

Question

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`.

但它完美地适用于:

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版本2.10.3

这是正常的编译器行为吗？

Answer 1

你已经碰到了Scala尝试统一原语和对象的情况之一.由于Int在Scala中表示Java原始类型int,因此不能将任何特征混合到其中.在执行asInstanceOf时,Scala编译器会自动将其Int转换为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)
                  ^

所以你必须将你的变量声明为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case类时使用强制转换:

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)