Array assignment -- need help

[Jayashree Mohan]

 
class Test2 {
  int[] i1 = {1}, i2 = {3};
  void m1() {
    m2(i1, i2);
    System.out.print(i1[0] + "," + i2[0]);
  }
  void m2(int[] i1, int[] i2) {
    int[] i3 = i1;
    this.i1 = i2;
    this.i2 = i3;
System.out.println(i1[0] + "," + i2[0]);
  }
  public static void main (String[] args) {
    new Test2().m1();
}}  

Why does it print 3,1???
Can somebody explain in detail....

[Ken Blair]

int[] i3 = i1;
this.i1 = i2;
this.i2 = i3;

Because you just swapped i1 and i2.
[ November 28, 2005: Message edited by: Ken Blair ]

[Miguel Antonioli]

Do you seriously ask this question? What do you expect this code produce?

[Jayashree Mohan]

 
class Test2 {
  static int[] i1 = {1}, i2 = {3};
  static void m1(int[] i1) {
    int[] i3 = i1; i1 = i2; i2 = i3;
System.out.println(i1[0] + "," + i2[0]); //first
  }
  public static void main (String[] args) {
    m1(i1);
    System.out.print(i1[0] + "," + i2[0]);//second
}}
 

but why does the second println statement print 1,1???

[Miguel Antonioli]

The second println should NOT print 1, 1. It should, however, print 3,1.

Notice in your m2() method, you swap the value of i1 and i2 using i3 as your temporary array.
[ November 28, 2005: Message edited by: Miguel Antonioli ]

[Jayashree Mohan]

yeah.. even i thought the same as we are passing the object reference in the methods.. but when I run the program,,, i get the answer 1,1...
I am not able to understand why???

[Ernest Friedman-Hill]

Originally posted by Jayashree Mohan:

I am not able to understand why???

In the first program, the method parameters to the "swapping" method have the same names as the member variables, but you use "this" to disambiguate the two. In the second program, there's only one parameter, and it has the same name as one member variable, but you don't use the class name to disambiguate, so while Test2.i2 gets reassigned, Test2.i1 does not. See?

[Ken Blair]

private Object o; // Object 1
 
public void m(Object o) { // Object 2
   o.toString();  // Prints Object 2
   this.o.toString(); // Prints Object 1
}

In the above code the parameter is said to "hide" the member because they have the same name. When a paramter hides a member you can refer to the member by prefixing it with the keyword "this". Consequently, in the above code while within m() "o" refers to the parameter (object 2) and the only way to access the member (object 1) is to refer to it by "this.o". If you didn't understand Ernest before reading this, try reading his reply again and it should make sense.