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method..

xie li
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Joined: Nov 30, 2005
Posts: 54
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
Ranch Hand

Joined: Dec 20, 2005
Posts: 114

Answer: A. 0


Devisri, SCJP 5.0, SCWCD 5.0
"Dream is Not what you see in sleep. Dream is that which never lets you sleep" - Abdul Kalam
Cheenu Subramanian
Ranch Hand

Joined: Aug 15, 2005
Posts: 40
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
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Joined: Oct 16, 2005
Posts: 146
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
Ranch Hand

Joined: Nov 30, 2005
Posts: 54
thank you
 
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