method..

[xie li]

1 public class Test {
2 public static void add3 (Integer i) {
3 int val = i.intValue();
4 val += 3;
5 i = new Integer(val);
6 }
7
8 public static void main(String args[]) {
9 Integer i = new Integer(0);
10 add3(i);
11 System.out.println(i.intValue()); 12 } 13 } What is the result?

A. 0

B. 3

C. Compilation fails.

D. An exception is thrown at runtime.

[Devi Sri]

Answer: A. 0

[Cheenu Subramanian]

When u pass arguments to a method, only a copy of the reference /value is passed. so changes to the reference/value will not reflect in the calling method. But changes to the values referred by the reference will be reflected. Hope thats not too confusing

[vipul patel]

Xie,

Just remember that everything is passed by value. There is nothing like pointers.

As K&B says, and I love it. "Java is not C++". saying such thing by yourself will really(!) convience yourself to think java code seperately than C++ code.

By the way, output is 0. so option A is right.

[xie li]

thank you