CODEWITHSUNDEEP
javagenericsabstract-class
Cem
Cem

Reputation: 743

Java Abstract Class Implementing an Interface with Generics

I am trying to define an abstract class implementing Comparable. When I define the class with following definition:

public abstract class MyClass implements Comparable <MyClass>

subclasses have to implement compareTo(MyClass object). Instead, I want every subclass to implement compareTo(SubClass object), accepting an object of its own type. When I try to define the abstract class with something like:

public abstract class MyClass implements Comparable <? extends MyClass>

It complains that "A supertype may not specify any wildcard."

Is there a solution?

Upvotes: 74

Views: 36118

Answers (7)

David Levy
David Levy

Reputation: 587

Found another solution:

  1. Define an interface on the fields which make up the comaprable (e.g ComparableFoo)
  2. Implement the interface on the parent class
  3. Implement Comparable on the parent class.
  4. Write your implementation.

Solution should look like this:

public abstract class MyClass implements ComparableFoo,Comparable<ComparableFoo> {
    public int compareTo(ComparableFoo o) {
    // your implementation
    }
}

This solution implies that more things might implement ComparableFoo - this is likely not the case but then you're coding to an interface and the generics expression is simple.

Upvotes: 1

Caroline Even
Caroline Even

Reputation: 121

I know you said you want "compareTo(SubClass object), accepting an object of its own type", but I still suggest declaring the abstract class like this:

public abstract class MyClass implements Comparable <Object>

and do an instanceof check when overriding compareTo in MySubClass:

@Override
public int compareTo(Object o) {
    if (o instanceof MySubClass)) {
        ...
    }
    else throw new IllegalArgumentException(...)
}

similarly to 'equals' or 'clone'

Upvotes: 0

seh
seh

Reputation: 15259

Apart from the mechanical difficulties you're encountering declaring the signatures, the goal doesn't make much sense. You're trying to establish a covariant comparison function, which breaks the whole idea of establishing an interface that derived classes can tailor.

If you define some subclass SubClass such that its instances can only be compared to other SubClass instances, then how does SubClass satisfy the contract defined by MyClass? Recall that MyClass is saying that it and any types derived from it can be compared against other MyClass instances. You're trying to make that not true for SubClass, which means that SubClass does not satisfy MyClass's contract: You cannot substitute SubClass for MyClass, because SubClass's requirements are stricter.

This problem centers on covariance and contravariance, and how they allow function signatures to change through type derivation. You can relax a requirement on an argument's type—accepting a wider type than the supertype's signature demands—and you can strengthen a requirement on a return type—promising to return a narrower type than the supertype's signature. Each of these freedoms still allows perfect substitution of the derived type for the supertype; a caller can't tell the difference when using the derived type through the supertype's interface, but a caller using the derived type concretely can take advantage of these freedoms.

Willi's answer teaches something about generic declarations, but I urge you to reconsider your goal before accepting the technique at the expense of semantics.

Upvotes: 21

newacct
newacct

Reputation: 122449

public abstract class MyClass<T> implements Comparable<T> {

}

public class SubClass extends MyClass<SubClass> {

    @Override
    public int compareTo(SubClass o) {
        // TODO Auto-generated method stub
        return 0;
    }

}

Upvotes: 1

irreputable
irreputable

Reputation: 45443

see Java's own example:

public abstract class Enum<E extends Enum<E>> implements Comparable<E>
    public final int compareTo(E o)

on seh's comment: usually the argument is correct. but generics makes type relations more complicated. a SubClass may not be a subtype of MyClass in Willi's solution....

SubClassA is a subtype of MyClass<SubClassA>, but not a subtype of MyClass<SubClassB>

type MyClass<X> defines a contract for compareTo(X) which all of its subtypes must honor. there is no problem there.

Upvotes: 3

whiskeysierra
whiskeysierra

Reputation: 5120

It's a little too verbose in my opinion, but works:

public abstract class MyClass<T extends MyClass<T>> implements Comparable<T> {

}

public class SubClass extends MyClass<SubClass> {

    @Override
    public int compareTo(SubClass o) {
        // TODO Auto-generated method stub
        return 0;
    }

}

Upvotes: 53

zevra0
zevra0

Reputation: 271

I'm not sure that you need the capture:

First, add the compareTo to the abstract class...

public abstract class MyClass implements Comparable <MyClass> {

@Override
public int compareTo(MyClass c) {
...
}    
}

Then add the implementations...

public class MyClass1 extends MyClass {
...
}

public class MyClass2 extends MyClass {
...
}

Calling compare will call the super type method...

MyClass1 c1 = new MyClass1();
MyClass2 c2 = new MyClass2();

c1.compareTo(c2);

Upvotes: 1

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