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.

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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...

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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.

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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;
    }
}

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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)