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plz explain about static references to objects

Suppose i have this code:
class A{
static A a;
and A a1 = new A();
and a1.a = new A();

and A a2 = new A();
and a2.a = new A();

now a1.a and a2.a should actually point to the same thing coz they belong to one class but since they can also be referenced by instance of class, they appear to point to different objects, so it means a1.a and a2.a is actually A.a so how can they point to different objects. Please explain this?
2 Solutions
A.a is a single, global variable, with class scope.

A static member is a single variable, not specific to an object instance. Instance variables, on the other hand, are stored in each object instance, each object has a copy.

no they would point to the same object, and the second assignment replaces the first.
Accessing a static via a object is undesirable though as it causes the confusion you are having, should always use A.a
Well let's see. The member variable 'a' belongs to the entire CLASS and NOT to any specific instance of that class.

Both a1 and a2 are two different instances/objects of type A. Since the reference 'a' is static, it belongs to the entire class. Therefore, both a1 and a2 will contain only one reference 'a' and specifying a1.a or a2.a does not make a difference. In fact, some IDEs throw up a warning saying a static member is being referenced in a non-static way if you specify a static reference this manner.

When you say a1.a = new A(), you are essentially making this static reference point to this new object that is created.

Then, when you say a2 = new A(); a2 already contains the reference 'a' which points to the object you created earlier in the previous statement (a1.a = new A() )

The final statement, a2.a = new A(), makes both a1.a and a2.a point to this new object being created.

Hence they will BOTH point to the SAME object and NOT a different one.
Ken ClementSoftware DeveloperCommented:
To expand a bit on nikhilmesnon's comment above:

The statement

    if (a1.a == a2.a)
        System.out.println ("same");
        System.out.println ("different");

Will ALWAYS result in the output "same", though you might get the warning about a static member referenced in a non-static way.
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