问题
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)
I got the following error:
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. The above error can be avoided by letting bessy
be Cow
instead of Animal
: val bessy:Cow = new Cow
.
Does this mean dynamic binding doesn't work here?
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"
}
And I had this, where x:Parent
gave Child as well:
scala> new Child().sig
res1: String = Child
val x:Parent = new Child()
x: Parent = Child@3a460b07
x.sig
res2: String = Child
回答1:
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.
If you allow the compiler to know that bessy
is a Cow
(val bessy = new Cow
), then she'll eat grass just fine.
来源:https://stackoverflow.com/questions/20754143/no-dynamic-binding-when-abstract-type-involved-in-scala