CODEWITHSUNDEEP
javagenerics
Nejc
Nejc

Reputation: 700

Declaring a field extending some type

I am having some problems with my syntax. I am looking all over and could not find the correct thing (I actually do not know, what would be the phrase to search). So here is the problem. For simplicity, lets assume I have 4 classes:

class A:

public abstract class A{
}

class B:

public class B extends A{
    //some methods and fields
}

class C:

public class C extends A{
    //some other methods
}

class D:

public class D{
    protected ? value;
}

Here I have a problem, because I do not know what to put instead of ? in class D, if I want to define ? as some sub-class of A. I know an option is just to write A instead of ?, but I think that this could be more specified, for example something like <T extends A>.

Is this even possible?

-edit-

From answers below, I saw that approach would be, to use raw types. But nevertheless, can anyone confirm that a construct like this, does not exist in java? That you could create a field of generic type T (extending some class), but the class containing this field does not have a generic parameter T?

Upvotes: 0

Views: 107

Answers (3)

Mik378
Mik378

Reputation: 22191

Use a generic class as follows:

public class D<T extends A> {
    protected T value;
}

MyClass isn't dependent of T as you pretend in your comment above. You can instantiate an instance without precising wildcard. Therefore, T would still be remplaced by A at compile-time This concept is called: Raw types

For more information: http://docs.oracle.com/javase/tutorial/java/generics/rawTypes.html

Upvotes: 1

Kumar Vivek Mitra
Kumar Vivek Mitra

Reputation: 33544

- Use Type Parameter,

Eg:

public class MyClass<T extends A>{

   T value;

}

Upvotes: 1

Subhrajyoti Majumder
Subhrajyoti Majumder

Reputation: 41240

I think you should specific to this value Type at run time .

public class D<T extends A>{
    protected T value;
}

D<B> d =new D<B>();
D<A> d1 = new D<A>();

Upvotes: 1

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