Specializing method arguments in subclasses in Java

Question

Consider the following situation:

public abstract class AnimalFeed{
}
public class FishFeed extends AnimalFeed{
}
public class BirdFeed extends AnimalFeed{
}

public abstract class Animal{

public void eat(AnimalFeed somethingToEat)

}

Now I would like to define a class "Bird" extending "Animal" being sure that when the bird eats, it eats only BirdFeed.

One solution would be to specify a sort of contract, in which the caller of "eat" must pass an instance of the appropriate feed

public class Bird extends Animal{

@Override 
public void eat(AnimalFeed somethingToEat){

    BirdFeed somethingGoodForABird

    if(somethingToEat.instanceOf(BirdFeed)){
    somethingGoodForABird = (BirdFeed) somethingGoodForABird
    }else{
    //throws error, complaining the caller didn't feed the bird properly
    }
}
}

Is it acceptable to delegate the responsibility of the parameter to the caller? How to force the caller to pass a specialization of the parameter? Are there alternative design solutions?

Andy Turner · Accepted Answer · 2016-11-25 15:42:35Z

13

You'd need to add a type variable to the class:

public abstract class Animal<F extends AnimalFeed> {
  public abstract void eat(F somethingToEat);
}

Then you can declare your subclasses as wanting a particular type of AnimalFeed:

public class Bird extends Animal<BirdFeed> {
  public void eat(BirdFeed somethingToEat) {}
}

public class Fish extends Animal<FishFeed> {
  public void eat(FishFeed somethingToEat) {}
}

answered Nov 25, 2016 at 15:42

Andy Turner

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Comments

GhostCat · Accepted Answer · 2016-11-25 18:44:45Z

4

What you are asking for doesn't make sense from an theoretical point of view.

Restricting a method parameter violates the Liskov Substitution Principle.

The idea there: any occurance (usage) of some base class object must be able to deal with some sub class object, too.

A more simple example: when your base interface goes:

void foo(Number n)

then you must not do

@Override
void foo(Integer i)

in a subclass. Because all of a sudden, a caller

someObject.foo(someNumber)

would run into ugly ugly problems when someObject is of that derived class; which only accepts Integers, but not Numbers.

In other words: good OO design is much more than just writting down A extends B. You have to follow such rules; or you end up with systems are already broken on a conceptual point!

And for the record: it is theoretically valid to widen method parameters (in general, but in Java); and it is also ok to restrict the return types of methods (because these changes can not break client code; and that even works in Java).

Long story short: the answer here is too change your design; for example by using Generics and dependent interfaces to somehow create a relationship between the Animal and the Feed class hierarchy.

edited Nov 25, 2016 at 18:44

answered Nov 25, 2016 at 15:37

GhostCat

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4 Comments

Andy Turner Over a year ago

"it is very much valid to widen method parameters" In general, but not in Java: you can't widen parameter types because of the way that Java method resolution and overloading works. See stackoverflow.com/questions/18337056/… (although I'm sure there are better Qs and As than that).

Andy Turner Over a year ago

I agree that wider parameters doesn't violate LSP. I'm just saying that wider arguments aren't supported in java (whereas narrower return types are; return type isn't part of the method signature used to determine which method to invoke).

GhostCat Over a year ago

Got it; updated my answer. Although I am wondering why your answers seems to fly six times better than mine ;-)

Paramvir Singh Karwal Over a year ago

An example that the user could directly relate and use with no or little modifications make them fly better :)

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