question on try catch block

[poornima viswa]

Read the following code
public class test{
public static void main(String[] args){
for(int i=0; i<10;++i){
try{
try{
if(i%3 ==0)throw new Exception("EO");
System.out.println(i);
}catch(Exception inner){
i*=2;
if(i%3==0)throw new Exception("EI");
}
finally{
++i;
}
}catch(Exception outer){
i+=3;
}finally{
--i;
}
}
}
}
can anyone explain how i get the output as 4 5
thankx in advance

[Bibhuti Dutta]

The reason is that for loop executes only for the values 0,4,5 and 6. When the loop executes for the 1st time it throws the exceptions and the value of i becomes 4 next time it enters into the for loop again.
Actually the body of the loop changes the value of i from 0 to 3, and the increment statement of the for loop makes it 4.
4 and 5 are printed and when i becomes 6 exceptions are thrown again and the value of i becomes greater than 10, there by terminating the loop.
Bibhuti Dutta

[poornima viswa]

Thank you i understood the flow now
I was just missing the increment that takes place in the for loop