CODEWITHSUNDEEP
javagenericsmethodsinterfacerestriction
Andrey
Andrey

Reputation: 925

Declaring interface method to allow only implemented class

I have interface Syncable: 

public interface Syncable{
    public void sync(Syncable other);
}

And two implemented classes: 

public class TstA implements Syncable{
    @Override
    public void sync(Syncable other) {
        // TODO Auto-generated method stub      
    }
}

public class TstB implements Syncable{
    @Override
    public void sync(Syncable other) {
        // TODO Auto-generated method stub      
    }
}

So this code is legal: 

TstA a = new TstA();
TstB b = new TstB();
a.sync(b);

This is not what I want. I want to allow a.sync receive instances of TstA, and b.sync receive instances of TstB.

Probably I have to use generics, but how?

Upvotes: 0

Views: 177

Answers (2)

newacct
newacct

Reputation: 122449

Analogous to Comparable, this is how you would do it:

public interface Syncable<T> {
    public void sync(T other);
}

public class TstA implements Syncable<TstA> {
    @Override
    public void sync(TstA other) {
        // TODO Auto-generated method stub      
    }
}

public class TstB implements Syncable<TstB> {
    @Override
    public void sync(TstB other) {
        // TODO Auto-generated method stub      
    }
}

If there were a generic class or method that required a type parameter T to be syncable to itself, then that class or method would declare the bound T extends Syncable<? super T>.

Upvotes: 0

Rohit Jain
Rohit Jain

Reputation: 213261

Make your interface generic:

public interface Syncable<T extends Syncable<T>> {
    public void sync(T other);
}

And implement a parameterized instance of that interface:

public class TstA implements Syncable<TstA> {
    @Override
    public void sync(TstA other) {
        // TODO Auto-generated method stub      
    }
}

Upvotes: 2

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