passing string object in to a method...

[Ganeshkumar cheekati]

This is from Master exam

1. class Mystery {
2. public static void main(String args[]) {
3. Changer c = new Changer();
4. c.method(args);
5. System.out.println(args[0] + " " + args[1]);
6. }

7. static class Changer
{
8. void method(String s[])
{
9. String temp = s[0];
10. s[0] = s[1];
11. s[1] = temp;
12. }
13. }
14. }


java Mystery Mighty Mouse



output is: Mouse Mighty.


can anyone explain me clearly why i got this output?
[ November 01, 2008: Message edited by: Ganeshkumar cheekati ]

[Jesper de Jong]

Ganeshkumar, when you copy a question from a book or mock exam, we require that you quote your sources. So, please tell us where you copied this from.

[M Srilatha]

 1. class Mystery {
2. public static void main(String args[]) {
3. Changer c = new Changer(); //creates an object of Changer(static nested           
                              //class)
//Before calling method(): args[0] = "Mighty" ; args[1] = "Mouse"
4. c.method(args); // copy of the reference variable args is passed to
                   // method(). As args is a reference to a object, the
                   // changes made to that object inside method() will be 
                   // reflected. 
//After the method() is executed the contents of args will be
//args[0]="Mouse"; args[1] = "Mighty"
5. System.out.println(args[0] + " " + args[1]);
6. }
 
7. static class Changer
{
8. void method(String s[])
{
//Here s and args in main() method refers to the same object on the Heap.
9. String temp = s[0];
10. s[0] = s[1];
11. s[1] = temp;
12. }
13. }
14. }
   

Hope this is clear!

[Ganeshkumar cheekati]

can you explain me what is happening at line 9,10,11.. in void method(String[] s)?
[ November 01, 2008: Message edited by: Ganeshkumar cheekati ]

[Ravikanth kolli]

9. String temp = s[0];  // here the string value stored in s[0] is copied    into temp variable. 
10. s[0] = s[1];     // The value of s[1] is assigned to s[0]
11. s[1] = temp;     // The value of s[1] is assigned the value of temp
 

Here we are actually working on the string values rather than the references. the reference to the String array object "args" is passed to the "method". The reference variable "s" in the method is pointing to the same location as "args".

Am i clear?

[Ganeshkumar cheekati]

ok i got it.

i am treating that Mystery which is class name as args[0].oh shit i have mistaken.

now i am clear...