java.lang.ClassCastException on creating an Array of classes in java

Question

I want to create an array of classes and I have these:

1)works fine

Class<MyClass>[] array = (Class<MyClass>[]) new Class<?>[list.size()];

2)java.lang.ClassCastException

Class<MyClass>[] array = (Class<MyClass>[]) Array.newInstance(MyClass.class, list.size());

3)we have a generic method , works fine

public static <T> List<T> method(T arg , T[] array){
    array = (T[]) Array.newInstance(arg.getClass(), 1);
    ArrayList<T> list = new ArrayList<T>(Arrays.asList(array));
    array[0] = arg;
    return new ArrayList<T>(Arrays.asList(array));
}

Please explain why we get the exception in 2 ? As I understand we are doing the same thing in 3 , so what's the difference ?

UPDATE

I guess my confusion is because we have:

Type mismatch: cannot convert from List< Class < MyClass > > to List< MyClass >

List<MyClass> list = Utils.addToList(MyClass.class, null);

Works

List<Class<MyClass>> list = Utils.method(MyClass.class, null);

I think I'm missing something , but I'm not sure what...

So it's finally clear what exacly I had in mind:

List<Class<MyClass>> list = Utils.method(MyClass.class, null);

Class<MyClass>[] array2 = (Class<MyClass>[]) Array.newInstance(MyClass.class.getClass(), list.size());

List<Class<MyClass>> list2 = Utils.method(MyClass.class, list.toArray(array2));

Jon Skeet · Accepted Answer · 2015-03-12 10:27:02Z

4

Both snippets 2 and 3 are creating a MyClass[], not a Class<MyClass>[].

The element type is determined by the first argument to Array.newInstance. If you want an array where each element is a Class reference, you need to pass in Class.class.

Note that generics and arrays don't play terribly nicely together - which is why you get a warning even for your "works fine" code.

answered Mar 12, 2015 at 10:27

Jon Skeet

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5 Comments

Duncan Jones Over a year ago

... and I would guess it's more likely the OP wants MyClass[].

Jon Skeet Over a year ago

@Duncan: It's very unclear what the OP is really trying to do, to be honest. I'm just explaining their current code, and hoping that's enough to help...

Natasha Over a year ago

I think , I understand what you mean , but I'm still confused . I updates my question , could you look?

Jon Skeet Over a year ago

@KatrinnaL: Well that's because you're passing a Class<MyClass> reference as the value of arg, so T is inferred to be Class<MyClass>, and you end up with a List<MyClass>. It's still very unclear what you're really trying to achieve.

Natasha Over a year ago

Thanks, it helped me a lot to understand what I wanted to do :)

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