Force parameterized type as constructor parameter to be correct

I have been searching for quite a while on this, and nothing really comes close to what I need.

Example code:

public class MyQueue<E extends Delayed & Serializable> extends DelayQueue<E> {
    private Class<E> mClass;
    public MyQueue(Class<E> type) {
        super();
        mClass = type;
    }
}

MyQueue is created like: MyQueue q<MyObj> = new MyQueue<MyObj>(MyObj.class);.

My question: How can I write the constructor in such a way, that the parameter "type" is of the proper parameterized type "E extends Delayed & Serializable"?

I hope I have explained myself allright.

Thank you in advance for any assistance.

Edit: From the answers and remarks I was at first unable to select a proper answer. So for the first time I shall try to enhance my question with what I wanted in the first place, and what I am ending up with now. Eg answering in the original post what I have found.

Obviously what I need to know, is the class of the parameterized type E on construction time. For transparancy, the queue is using the E class name (being MyObj) to pass to a backing store. After a lot of reading I came to understand that I could not get at this information easier because of type erasure. I was forced to pass the class not only as a parameterized type, but also as a constructor variable.

I was afraid I could, by accident, call the constructor as such (MyObj and OthrObj both implement Delayed & Serializable):

MyQueue q<MyObj> = new MyQueue<MyObj>(OthrObj.Class);

My better question should have been: How can I write the constructor in such a way, forcing the constructor variable E to match the parameterized type E?

It would have made so much more sense, and it probably would have been easier to understand what I have written. Technically most of you are right in some sense, and after testing many of the variations, I c开发者_运维问答ame to the conclusion I have already done so giggles sorry seh and Tnem are completly right and both deserve the credits.

What you have written is what you have specified, the type parameter will be enforced to be of type Delayed & Serializable. I can't see what is wrong here...

You have already written 'the constructor in such a way, that the parameter "type" is of the proper parameterized type "E extends Delayed & Serializable"'.

The type restriction from the class definition itself

public class MyQueue<E extends Delayed & Serializable> ...

carries through. You don't repeat the upper bounds for the type in the constructor definition

public MyQueue(Class<E> type) {

and it's not necessary. E.g. an instantiation like

MyQueue<Long> myQueue = new MyQueue<Long>(Long.class);

results in a compiler error. - Just try it.

I think what you want is like:

public class MyQueue<E extends Delayed & Serializable> extends DelayQueue<E> {
    private E mClass;
    public MyQueue(E type) {
        super();
        mClass = type;
    }
}

public MyQueue(Class<E> type) {
    super();
    mClass = type;
    type.asSubclass(Delayed.class);
    type.asSubclass(Serializable.class);//both of these throw when it can't happen
}

this explicitly check whether the class passed is correct even when accounting type erasure (and throws classcastexception when it cant happen)