有关声明可比较对象列表的Java问题

Question

It is a pretty simple one. I want to declare a list of objects, but I want make sure all objects implemented the Comparable interface. What should I do? I try to write

   List<Comparable> heap = new ArrayList<Comparable>();

But compiler gives me a warning. What is the proper way to write it?

Thanks

Follow up: I thought I had finished the question before I post it but apparently I didn't. So let me finish it up.

My purpose is to have a list of objects that: 1. Implements the Comparable interface 2. Are all with the same type

Can I make it

   List<Comparable<?>> heap = new ArrayList<Comparable<?>>()

No, I can't. The reason is because I need to retrieve the elements from the list and compare them. like:

   if( heap.get(0).compareTo(heap.get(1)) > 0)

If I use wildcard List>, the complier will give me an error. Saying heap.get(0) cannot match heap.get(1) So I need to know the correct way to declare it.

I find somebody asking me what the warning is.... that surprises me.....well, the warning is: Comparable is a raw type. References to generic type Comparable should be parameterized

Answer 1

It's because you're using the raw type Comparable . You can try giving Comparable a wildcard type parameter:

List<Comparable<?>> heap = new ArrayList<Comparable<?>>();

---

If you want the objects to all be of the same type, you can do something along the lines of

public static <T extends Comparable<? super T>> List<T> getList() {
    ...
}

and return your list from getList() .

Answer 2

由于所有对象都是同一类型（例如fe FooType ），因此只需使用List<FooType> 。

Answer 3

Perhaps you could write a method that only takes a type that extends Comparable and returns a List of that type:

public <T extends Comparable> List<T> getComparableList(T t) {
    List<T> heap = new ArrayList<T>();
    return heap;
}