java variable scope

[deva1]

public class example {
int i[] = {0};
public static void main(String args[]) {
int i[] = {1};
change_i(i);
System.out.println(i[0]);
}
public static void change_i(int i[]) {
i[0] = 2;
i[0] *= 2;
}
}


Can anyone tell how the scope of the array int i[] works? I thought if we declare something inside a method it is local to that method.Here int i [] declared inside example(class level) and then inside main method and change_i mehtod.If I run this program it prints 4 . why? Since it is declared inside main method i [] = {1}; I thought it will print 1.Hello Gurus I am bit confused,please help me to understand this?

Thanks

 

[sedj]

A primitive array (ie int[]) is actually a mutable object - so if you pass it to a method and alter the contents in that method - then it is modified.

Its all about how Java handles passing objects by *reference or value?*, and whether an object is mutabale.

Read up on the subject :

http://www.javaworld.com/javaworld/javaqa/2000-05/03-qa-0526-pass.html

http://java.sun.com/docs/books/tutorial/java/javaOO/arguments.html

http://www.jguru.com/faq/view.jsp?E....net/professional/java/myths/pass-by-ref.html

 

[stefanwagner]

You have a class-variable int i [], which is hidden in main, where you have a local i.
Your method gets a parameter i, which hides again the class-attribute.

- Write the class name in uppercase.
- Call

Example example = new Example ();

as first statement in main.
Now you have an example, and therefore an i.
- Remove the 'static'-keyword.
- call change_i without parameter.
- initialize i without telling again 'int array':

i = new int [1];
i[0]=1;

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