Scala

Question

根据协方差定义：

Q[+B] means that Q can take any class, but if A is a subclass of B, then Q[A] is considered to be a subclass of Q[B].

让我们看下面的例子：

trait SomeA
trait SomeB extends SomeA
trait SomeC extends SomeB

case class List1[+B](elements: B*) 

val a = List1[SomeA](new SomeA{},new SomeB{})
val b = List1[SomeB](new SomeB{},new SomeC{})

一切正常，但我不明白为什么 List1[SomeB] 是 List1[SomeA] 的子类，或者换句话说为什么 b 是 a 的子类？

Answer 1

现在，List1[SomeB] 是 List1[SomeA] 的子类，这意味着您可以将第一个放在需要的地方。

scala> case class List1[+B](elements: B*)

scala> val a = List1[SomeA](new SomeA{},new SomeB{})
a: List1[SomeA] = List1(WrappedArray($anon@3e48e859, $anon@31ddd4a4))

scala> val b = List1[SomeB](new SomeB{},new SomeC{})
b: List1[SomeB] = List1(WrappedArray($anon@5e8c34a0, $anon@7c1c5936))

scala> val c: List1[SomeA] = b
c: List1[SomeA] = List1(WrappedArray($anon@5e8c34a0, $anon@7c1c5936))

scala> val c: List1[SomeA] = a
c: List1[SomeA] = List1(WrappedArray($anon@3e48e859, $anon@31ddd4a4))

如果它是不变的，那将是不可能的，请参阅：

scala> case class List1[B](elements: B*)
defined class List1

scala> val c: List1[SomeA] = b
<console>:16: error: type mismatch;
 found   : List1[SomeB]
 required: List1[SomeA]
Note: SomeB <: SomeA, but class List1 is invariant in type B.
You may wish to define B as +B instead. (SLS 4.5)
       val c: List1[SomeA] = b
                             ^

scala> val c: List1[SomeA] = a
c: List1[SomeA] = List1(WrappedArray($anon@45acdd11, $anon@3f0d6038))

Answer 2

由于 List1[SomeB] 中的所有元素都是 SomeA 的子类型（因为 SomeB 扩展了 SomeA），您可以简单地传递 List1[SomeB] 其中 [=预计 15=]。

Scala - 协方差

Scala - Covariance

covariance

scala-generics