CODEWITHSUNDEEP
javagenericsinheritancenested-generics
james.cookie
james.cookie

Reputation: 455

How do I extend a builder class twice using Java generics?

I have a builder class that I want to extend, here is the simple version:

class A {
    public A withSomeAStuff() {
        return this;
    }
}
A a = new A().withSomeAStuff();

When I extend it, I know that I can do this without any problem:

class AA<T extends AA> {
    public T withSomeAStuff() {
        return (T) this;
    }
}
class BB extends AA<BB> {
    public BB withSomeBStuff() {
        return this;
    }
}
AA aa = new AA().withSomeAStuff();
BB bb = new BB().withSomeAStuff().withSomeBStuff();

But now I want to extend it further with another class, so I try this:

class AAA<T extends AAA> {
    public T withSomeAStuff() {
        return (T) this;
    }
}
class BBB<T extends BBB> extends AAA<T> {
    public T withSomeBStuff() {
        return (T) this;
    }
}
class CCC extends BBB<CCC> {
    public CCC withSomeCStuff() {
        return this;
    }
}
AAA aaa = new AAA().withSomeAStuff();
BBB bbb = new BBB().withSomeAStuff().withSomeBStuff(); //breaks here!
CCC ccc = new CCC().withSomeAStuff().withSomeBStuff().withSomeCStuff();

My new CCC class works fine, but my BBB class is broken and I cannot figure out why.

What do I need to do to fix it?

Upvotes: 2

Views: 163

Answers (2)

k5_
k5_

Reputation: 5558

Never ignore raw type warnings: What is a raw type and why shouldn't we use it?

I added a method self() so you only have a single unchecked cast in your code.

class AAA<T extends AAA<T>> {
    public T withSomeAStuff() {
        return self();
    }

    @SuppressWarnings("unchecked")
    protected T self() {
        return (T) this;
    }
}

class BBB<T extends BBB<T>> extends AAA<T> {
    public T withSomeBStuff() {
        return self();
    }
}

class CCC extends BBB<CCC> {
    public CCC withSomeCStuff() {
        return this;
    }
}


public static void main(String[] args) {
    AAA<?> aaa = new AAA<>().withSomeAStuff();
    BBB<?> bbb = new BBB<>().withSomeAStuff().withSomeBStuff(); 
    CCC ccc = new CCC().withSomeAStuff().withSomeBStuff().withSomeCStuff();
}

Upvotes: 1

Codebender
Codebender

Reputation: 14471

When you introduce generics in the type declaration, then use it while creating objects of the type too.

    AAA<AAA> aaa = new AAA<>().withSomeAStuff();
    BBB<BBB> bbb = new BBB<>().withSomeAStuff().withSomeBStuff(); //Does not break anymore.
    CCC ccc = new CCC().withSomeAStuff().withSomeBStuff().withSomeCStuff();

Note: While this will solve your compiler error, this is not a foolproof way and guaranteed to work in every case. You will get compiler warnings to confirm that.

For example you could do,

BBB<CCC> bbb = new BBB<CCC>().withSomeAStuff().withSomeBStuff();

and get a shock during runtime.

Upvotes: 2

Related Questions