[英]No dynamic binding when abstract type involved in Scala?
When I was trying the Animal/Food example for abstract types in Martin Odersky's Programming in Scala , 当我在Martin Odersky 在Scala的Programming中尝试抽象类型的Animal / Food示例时,
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)
I got the following error: 我收到以下错误:
scala> <console>:13: error: type mismatch;
found : Grass
required: bessy.SuitableFood
bessy.eat(new Grass)
^
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. 马丁最初的例子是
bessy.eat(new Fish)
,它肯定会失败,但是我没想到Grass
也会失败。 The above error can be avoided by letting bessy
be Cow
instead of Animal
: val bessy:Cow = new Cow
. 可以通过将
bessy
设为Cow
而不是Animal
来避免上述错误: val bessy:Cow = new Cow
。
Does this mean dynamic binding doesn't work here? 这是否意味着动态绑定在这里不起作用?
Edited: Simple dynamic binding for regular inheritance in Scala: 编辑: Scala中用于常规继承的简单动态绑定:
abstract class Parent {
def sig:String = "Parent"
}
class Child extends Parent {
override def sig:String = "Child"
}
And I had this, where x:Parent
gave Child as well: 我有这个,
x:Parent
给了Child :
scala> new Child().sig
res1: String = Child
val x:Parent = new Child()
x: Parent = Child@3a460b07
x.sig
res2: String = Child
Scala is statically typed. Scala是静态类型的。 An arbitrary animal cannot eat grass, and you have just tried to feed grass to an arbitrary animal.
任意动物都不能吃草,而您只是试图将草喂给任意动物。 It happens to be a cow, but you have stated (with
: Animal
) that the compiler may only assume that it is an animal. 它碰巧是一头牛,但是您已经(使用
: Animal
)声明过,编译器可能仅假设它是动物。
If you allow the compiler to know that bessy
is a Cow
( val bessy = new Cow
), then she'll eat grass just fine. 如果让编译器知道
bessy
是一头Cow
( val bessy = new Cow
),那么她会吃草的。
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