About Exception Handling.....

[chen ke-chien]

Can somebody explain the flow of the following code ?
why it result in 4 and 5....
thx...^_^
class Question1 {
public static void main(String[] args) {
for(int i=0;i<10;++i) {
try {
try {
if(i%3==0) throw new Exception("E0");
System.out.println(i);
}catch (Exception inner) {
i*=2;
if(i%3==0) throw new Exception("E1");
}finally{
++i;
}
}catch (Exception outer) {
i+=3;
}finally{
--i;
}
}
}
}

[Valentin Crettaz]

jellyman,
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[Selvi Ranjan]

Hi,
The flow of the program is as such
1. for loop i = 0
2. i%3 is 0 Goes to Exception handler inner
3. Exception handler inner
i *= 2 is 0
i%3 is 0 goes to Exception handler outer
finally for inner try
++i is 1
4. Exception handler outer
i+=3 is 4
finally for outer try
--i is 3
5. for loop i = 4 (i will be incremented by one and check i<10 true
6. i%3 is not equal to 0 so print 4
7. Inner try finally ++i is 5
8. Outer try finally --i is 4
9. for loop i = 5 (i will be incremented by one and check i<10 true)
10. i%3 is not equal to 0 so print 5
11. Inner try finally ++i is 6
12. Outer try finally --i is 5
13. for loop i = 6 (i will be incremented by one and check i<10 true)
14. i%3 is 0 now the program flow is as such from 1-4
15. for loop i =16 so the loop ends
Hope this will help u understand the program flow and explains why the program prints 4 and 5