Try, Catch & Finalise blocks

[Vijayalakshmi Chipada]

Please let me know how this works out:
class Uest{
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("E1");
} finally {
++i;
}
}catch (Exception outer) {
i += 3;
}finally {
--i;
}
}
}
}
When I compile I am getting 4 & 5 as answers, Please help me out in this regard.

[Vijayalakshmi Chipada]

Please help me out in this regard
class Uest{
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("E1");
} finally {
++i;
}
}catch (Exception outer) {
i += 3;
}finally {
--i;
}
}
}
}
When compilation I am getting 4 & 5 as answers

[Yamuna Pattathil]

I have the same code with a lot of sop statements. You can easily understand the program flow.
The one with label "original" is the only one you had in your code.

class Utest{
public static void main (String[]args){
for (int i=0; i<10; ++i) {
try{
System.out.println("Outer try");
try{
System.out.println("Inner try");
if (i % 3 == 0) {
System.out.println("First if : "+i);
throw new Exception("EO");
}
System.out.println("original  "+i);
}catch (Exception inner) {
i *=2;
System.out.println("Inner catch :"+i);
if(i % 3 ==0) 
throw new Exception("E1");
} finally {
++i;
System.out.println("Inner finally :"+i);
}
}catch (Exception outer) {
i += 3;
System.out.println("Outer catch :"+i);
}finally {
--i;
System.out.println("Outer finally :"+i);
}
}
}
}

Also note that 0, 3, 6, 9 etc are values of i(<10) which causes if condition to be executed and hence Exception.
[This message has been edited by Yamuna Pattathil (edited December 04, 2000).]

[saumil shukla]

is there any way to trace this?

[Yamuna Pattathil]

I have given the output and explanations below

class Utest{
    public static void main (String[]args){
       for (int i=0; i<10; ++i) {
            try{
                System.out.println("Outer try");
                try{
                    System.out.println("Inner try");
                    if (i % 3 == 0) {
                        System.out.println("First if : "+i);// This is executed only if i is 0, 3, 6 or 9.
                        throw new Exception("EO");// An exception is thrown then.
                    }
                    System.out.println("original  "+i);//
                }catch (Exception inner) {
                    i *=2;
                    System.out.println("Inner catch :"+i);
                    if(i % 3 ==0) 
                        throw new Exception("E1");
                } finally {
                    ++i;
                    System.out.println("Inner finally :"+i);
                }
            }catch (Exception outer) {
                i += 3;
                System.out.println("Outer catch :"+i);
            }finally {
                --i;
                System.out.println("Outer finally :"+i);
            }
        }
    }
}
// The output is given below.
/*
Outer try// Enters the first try
Inner try// Enters the second try
First if : 0// i = 0, an exception(EO) is thrown hence the print "original" wont occur.
Inner catch :0// Program enters the inner catch block and i is still zero(0*2) and hence another exception occurs(EI)
Inner finally :1//Finally block is always executed which increments i to 1.
Outer catch :4// Now it enters the catch block (due to the second exception) where i is made 4
Outer finally :3// Finally, always executes making i = 3.
Outer try// Next loop starts Since i is 3 now its incremented to 4.
Inner try
original  4// Exceptions are not thrown this time since 4%3 <> 0 and hence this is printed.
Inner finally :5// As usual finally block is always executed, makes i = 5
Outer finally :4// Second finally
Outer try// Next loop
Inner try
original  5// Now i = 5 which means no exception, hence works the same way as when i = 4.
Inner finally :6// 
Outer finally :5//
Outer try// Next loop,  i = 6 now, note that 6%3 == 0
Inner try//
First if : 6//An exception
Inner catch :12// caught and made i = 6 * 2 = 12, remaining steps are similar to when i is 0.
Inner finally :13// 
Outer catch :16 //
Outer finally :15 //
Since i = 15 now loop condition fails and loop ends.
*/

Hope this helps.
[This message has been edited by Yamuna Pattathil (edited December 04, 2000).]