我可以将对象转换为具有接口的类吗？

Question

I have a object obj and I want to cast it to View that implements interface Tool .

class Whatever {
    View viewObj;
    Tool toolObj;
    ...
    public void setViewTool( View obj )
    {
        viewObj = obj;
        toolObj = (Tool) obj;
    }

    ...
}

I want to avoid creating two objects to handle the situation or using one that I have to cast each time I want to call an implemented method of View or Tool.

The same thing, if possible, when using two interfaces or more. I recall a language I could do something like this View<Tool, and, others, interfaces> .

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Example

I think the title and the first line of the question is clear. But, may it's not.

I have three tools to show in the app. One is made extending LinearLayout, and the others two, FrameLayout. These view objects has to implements Tool interface because there are methods called when it inits, finishes and when the user do some gestures. But, it's a view too, and when I need to show, it needs to call things like setLayout, addView, animate .

Just to clarify, it's works fine in the example above. But I have to set two variables to handle the same object, one for doing stuff related to the view aspect, and one to the Tool interface.

I can't say generic class doesn't work but the way I have tried, it doesn't. What I wish, was the possibility of create a class property as private (View implements Tool) myObj; .

Now, I almost sure it's not possible from this other answer: Java - Defining a member that extends class A and implements interface B

Answer 1

If your question is

I know I can cast an object (A) to another (B), but can I do it if the B class implement an interface (C) ?

Then the answer is yes. It does not change anything that the class B implement C or not. But as for any object cast you can do it only if the instance a of A is also an instance of B. You can test it like that :

A a = new A()
if(a instanceof B) {
    //a is an instance of B
}

If the instance a of A is also an instance of B it means that it implements C. So you can do

C c = (C) a;
B b = (B) a;