scala - 为什么删除后 "Banana with Cream"和 "Banana"的类型一样？如何修复？

Question

我想要根据标记特征(如 Cream)绑定(bind)不同的方法(在编译时!)。如何实现？

我下面的解决方案无法编译。

如何修复？

class Apple
class Banana
trait Cream

object HasOverloadedMethods
{
  def method(p:Apple)=println("Apple")
  def method(p:Banana)=println("Banana")
  def method(p:Banana with Cream)=println("Banana with Cream")
}

object Question extends App{
  HasOverloadedMethods.method(new Apple())
  HasOverloadedMethods.method(new Banana())
  HasOverloadedMethods.method(new Banana() with Cream)
}

错误:

double definition:
method method:(p: Banana with Cream)Unit and
method method:(p: Banana)Unit at line 9
have same type after erasure: (p: Banana)Unit
  def method(p:Banana with Cream)=println("Banana with Cream")
      ^

Answer 1

不幸的是，你不能那样修复它，因为 JVM 不知道混合类型(并且 method(x: Banana with Cream) 的字节码签名因此只是 method (x: 香蕉)).

您有几个选择，它们都有各自的缺点。

1. 只服用奶油。这会让你失去 Banana
2. 创建一个BananaWithCream 特征。这会使您的层次结构困惑。

3. 使用类型类:

   def method[T : MethImpl](x: T) = implicitly[MethImpl[T]].impl(x)
   
   trait MethImpl[T] {
     def impl(x: T): Unit
   }
   
   trait LowPrioMethImpl {
     implicit object BananaImpl extends MethImpl[Banana] {
       def impl(x: Banana) = println("Banana")
     }
   }
   
   object MethImpl extends LowPrioMethImpl {
     implicit object AppleImpl extends MethImpl[Apple] {
       def impl(x: Apple) = println("Apple")
     }
   
     implicit object BananaWithCreamImpl extends MethImpl[Banana with Cream] {
       def impl(x: Banana with Cream) = println("Banana with Cream")
     }
   }
   

   现在您可以:

   method(new Banana) // > Banana
   method(new Banana with Cream) // > Banana with Cream
   method(new Apple) // > Apple
   
   method("adsf") // error: Could not find implicit value ...
   

   缺点显然是此解决方案引入的困惑。