Java – Is it possible to create instance of an Interface(e.g-Map Interface)?If it is then how

The reason is that the implementation of these interfaces is usually not relevant when handling them, therefore if you oblige the caller to pass a HashMap to a method, then you're essentially obliging which implementation to use. So as a general rule, you're supposed to handle its interface rather than the actual implementation and avoid the pain and suffering which might result in having to change all method signatures using HashMap when you decide you need to use LinkedHashMap instead.

It should be said that there are exceptions to this when implementation is relevant. If you need a map when order is important, then you can require a TreeMap or a LinkedHashMap to be passed, or better still SortedMap which doesn't specify a specific implementation. This obliges the caller to necessarily pass a certain type of implementation of Map and strongly hints that order is important. That said, could you override SortedMap and pass an unsorted one? Yes, of course, however expect bad things to happen as a result.

However best practice still dictates that if it isn't important, you shouldn't use specific implementations. This is true in general. If you're dealing with Dog and Cat which derive from Animal, in order to make best use of inheritance, you should generally avoid having methods specific to Dog or Cat. Rather all methods in Dog or Cat should override methods in Animal and it will save you trouble in the long run.

I don't see any way to "lock down the interface to a class" in your code, but I do think you need to consider two things:

1. Your clients would still need to re-compile their code, even though the client source code doesn't have to change: invoking a method on an interface uses a different bytecode operation then invoking a method on an instance.
2. What would prevent your coworkers from creating a class in the trusted package and invoking the constructor there?