Scala中涉及抽象类型时没有动态绑定吗? [英] No dynamic binding when abstract type involved in Scala?

问题描述

当我在Martin Odersky的 Scala编程中尝试抽象类型的Animal/Food示例时，

When I was trying the Animal/Food example for abstract types in Martin Odersky's Programming in Scala,

class Food
abstract class Animal {
  type SuitableFood <: Food
  def eat(food:SuitableFood)
}
class Grass extends Food
class Cow extends Animal {
  type SuitableFood=Grass
  override def eat(food:SuitableFood) {}
}
val bessy:Animal = new Cow
bessy.eat(new Grass)

我遇到以下错误:

scala> <console>:13: error: type mismatch;
 found   : Grass
 required: bessy.SuitableFood
                  bessy.eat(new Grass)
                            ^

Martin最初的示例是bessy.eat(new Fish)，它肯定会失败，但是我没想到Grass也会失败.通过使bessy为Cow而不是Animal:val bessy:Cow = new Cow可以避免上述错误.

The original example by Martin was bessy.eat(new Fish), which would definitely fail, but I didn't expect it'd fail for Grass as well. The above error can be avoided by letting bessy be Cow instead of Animal: val bessy:Cow = new Cow.

这是否意味着动态绑定在这里不起作用?

Does this mean dynamic binding doesn't work here?

已 在Scala中进行常规继承的简单动态绑定:

Edited: Simple dynamic binding for regular inheritance in Scala:

abstract class Parent {
  def sig:String = "Parent"
}
class Child extends Parent {
  override def sig:String = "Child"
}

我有这个，x:Parent也给了 Child :

scala> new Child().sig
res1: String = Child

val x:Parent = new Child()
x: Parent = Child@3a460b07

x.sig
res2: String = Child

推荐答案

Scala是静态类型的.任意动物都不能吃草，而您刚刚尝试将草喂给任意动物.它恰好是一头母牛，但是您已经(使用: Animal声明)编译器可能仅假设它是动物.

Scala is statically typed. An arbitrary animal cannot eat grass, and you have just tried to feed grass to an arbitrary animal. It happens to be a cow, but you have stated (with : Animal) that the compiler may only assume that it is an animal.

如果让编译器知道bessy是Cow(val bessy = new Cow)，那么她会吃草的.

If you allow the compiler to know that bessy is a Cow (val bessy = new Cow), then she'll eat grass just fine.

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