Passing of objects to methods

Question

pl go thru this code...
public class A{
public static void main(String args[]) {
A aa = new A();
StringBuffer a = new StringBuffer("One");
StringBuffer b = new StringBuffer("Two");
aa.swap(a,b);
System.out.println("a is "+ a +" b is " + b);
void swap (StringBuffer a, StringBuffer b) {
a.append(" more");
b=a;
Output : a is One more b is Two
In java always parameters (objects) are passed by reference... then why doesnt the above make the references a and b to point to the same object in memory within the swap() method.

Answer 1

Hi.
This is a point of frequent confusion.
All parameter passing in Java is pass by value. In the case of parameters of reference type, it is the reference that is the value to be passed.
In your example, you call the swap() method using variables a and b as actual parameters. a and b both hold references to StringBuffer instances. These references (i.e. the values of the variables) are copied into the formal parameters a and b of your swap() method. This means that your assignment 'b=a' sets the value of the local variable b equal to the value of the local variable a (the values of b and a are references to StringBuffers). The fact that these variables hold references doesn't change the fact that it's just a local assignment.
Regards,
Lance
Lance Walton - [email protected]
Team In A Box - Software without Tragedy
http://www.teaminabox.co.uk