Calling a Static Variable or Method

[Maxwell Xu]

Hello fellow programmers!

I am preparing for the OCA Java SE 8 Programmer I exam and to prepare for it I use the book by Jeanne Boyarsky and Scott Selikoff. However, I came upon some confusion. Here:

Calling a Static Variable or Method
Usually, accessing a static member is easy. You just put the classname before the method or variable and you are done. For example:

System.out.println(Koala.count);
Koala.main(new String[0]);

Both of these are nice and easy. There is one rule that is trickier. You can use an instance of the object to call a static method. The compiler checks for the type of the reference and uses that instead of the object—which is sneaky of Java. This code is perfectly legal:

5: Koala k = new Koala();
6: System.out.println(k.count); // k is a Koala
7: k = null;
8: System.out.println(k.count); // k is still a Koala

Believe it or not, this code outputs 0 twice. Line 6 sees that k is a Koala and count is a static variable, so it reads that static variable. Line 8 does the same thing. Java doesn’t care that k happens to be null. Since we are looking for a static, it doesn’t matter.

Above is a part of the book.

The problem is that I still don't get why the code outputs 0 twice. Please explain with detail.

Thanks,

Maxwell

[Paul Clapham]

I was going to say that line 6 outputs "0" and then line 8 outputs "0", which adds up to twice. But I see you posted something which already says that. So I'm not clear on what part of it is confusing you.

[Henry Wong]

Maxwell Xu wrote:
The problem is that I still don't get why the code outputs 0 twice. Please explain with detail.

The first zero is from line 6. The second zero is from line 8.

Henry

[Maxwell Xu]

Yes but wasn't Koala assigned to null?

[Henry Wong]

Maxwell Xu wrote:Yes but wasn't Koala assigned to null?

From the book that you quoted...

Book Quoted Maxwell Xu wrote:Line 8 does the same thing. Java doesn’t care that k happens to be null. Since we are looking for a static, it doesn’t matter.

Henry

[Mala Gupta]

Hi Maxwell -

Don't worry; you aren't alone. This is one of the questions, asked often by new programmers or new Java programmers.

Only one copy of a static variable exists, which is shared by all the instances of that class. When a class is loaded in memory by Java Virtual Machine (JVM), a class's static variables are also made accessible. So you don't need instances of a class to access its static members. You can use a class name to access its static variables.

Accessing static variables is confusing at times because Java allows usage of both - class name and an instance variable, to access a static variable.

For example, for the following class:

class MyClass {
    public static String aVar = "Comic";
}

You can access the static variable, by using the class name (that is, Book) or by using the name of its instance variable (refVar):

// Access using class name
System.out.println(MyClass.aVar);         

// Access using a non-null reference variable
MyClass refVar = new MyClass();
System.out.println(refVar.aVar);         

// Access using a null reference variable
// If a class is loaded in memory, its static variables are also accessible

refVar = null;
System.out.println(refVar.aVar);         

All calls to System.out.prinltn() in the preceding code, will output 'Comic'.



If you understand the above explanation, answer these question for me:

1) Why is the 'main' method in an executable class, defined as a static method?
2) Who calls this 'main' method?

Good Luck!


With much respect,
Mala