Consider this is interface:
public interface Node {
Node next();
Node prev();
void setNext(Node next);
void setPrev(Node prev);
void setVal(int val);
int getVal();
}
I want know how I can instantiate this class, for example:
Node p = tail;
I would like to use it something like this.
you cant create instance for an interface , Instead make some class to implement this interface and create instance for that class
public class NodeImpl implements Node{
// your impl
}
then
NodeImpl tail = new NodeImpl();
Node p =tail;
Node tail = new NodeImpl();Node is an interface, not an interface class (there is no such thing). In a nutshell, you cannot instantiate an interface or an abstract class; you can only instantiate a concrete class.
To take your example, if tail is of type that's either Node or is a class/interface that implements/extends Node, then the assignment will just work.
If you need to create a new object, the code would look something like:
public interface Node { ... }
public class NodeImpl implements Node { ... }
Node p = new NodeImpl();
For the interface you can only do something like that for the initialization (anonymous class):
Node p = new Node() {
public Node next() {}
public Node prev() {}
public void setNext(Node next) {}
public void setPrev(Node prev) {}
public void setVal(int val) {}
public int getVal() {}
};
and implement all the methods. Otherwise you can create an example class:
class NodeImpl implements Node {
public Node next() {}
public Node prev() {}
public void setNext(Node next) {}
public void setPrev(Node prev) {}
public void setVal(int val) {}
public int getVal() {}
}
Node p = new NodeImpl();
It's better to create a class implementing this interface.
But, you still could write:
Node n = new Node() {
//and here you MUST implement all Node methods
};
Node p = new SomeClassThatImplementsNode();