dan exam doubt 14

[amit taneja]

hii
in one question

.............................

Question 10

class Level1Exception extends Exception {}
class Level2Exception extends Level1Exception {}
class Level3Exception extends Level2Exception {}
class Purple {
  public static void main(String args[]) {
    int a,b,c,d,f,g,x;
    a = b = c = d = f = g = 0;
    x = 1;
    try {
      throw new Level1Exception();
      try {
        switch (x) {
          case 1: throw new Level1Exception();
          case 2: throw new Level2Exception();
          case 3: throw new Level3Exception();
        }
        a++;
      }
      catch (Level2Exception e) {b++;}
      finally {c++;}
    }
    catch (Level1Exception e) { d++;}
    catch (Exception e) {f++;}
    finally {g++;}
    System.out.print(a+","+b+","+c+","+d+","+f+","+g);
  }
}

What is the result of attempting to compile and run the program?

a. Prints: 1,1,1,0,0,1
b. Prints: 0,1,1,0,0,1
c. Prints: 0,1,0,0,0,0
d. Prints: 0,1,0,0,0,1
e. Prints: 0,0,1,0,0,1
f. Compile-time error
g. Run-time error
h. None of the above


its answer is : f with explanation

A throw statement is the first statement in the outer try block. A throw statement appearing in a try block causes control to pass out of the block. Consequently, statements can not be reached if they appear in the block after the throw statement. The switch statement that appears after the throw statement is unreachable and results in a compile-time error.

I am not able to understand ...the why this option is choosen as well as its explanation

[ Code tags added and, I hope, formatting improved ]
[ February 24, 2005: Message edited by: Barry Gaunt ]

[Jay Pawar]

Amit,
Explanation is enough to understand the problem in the code. What part you didn't understand ?

[amit taneja]

i m not able to understand that why "c" is not the option as after "throw" in out try ...the flow of programe should directly transfer to Level1Exception e ..???


and i don't get any point in explanation what Dan is expaining in his answer ....sorry..

[amit taneja]

sorry in above i wrote why "c" is not the option ..i mean why "h" is not the option ?

[Jay Pawar]

Amit,
Lets simplify the problem. Compile the following code and see if you can understand what's happening.

class test {
public static void main(String args[]) {
try {
throw new Exception();
System.out.println("Never reached");
}
catch( Exception e) {
}
 
}
}

Here is the link in JLS about unreachable statements.
click here

Hope this helps you
[ February 24, 2005: Message edited by: Jay Pawar ]

[Carol Enderlin]

I can understand that you may have questions about the explanation, but the answer of (f) seems pretty straightforward if you try to compile and run it like the question suggests.

C:\_Work\java>javac Purple.java
Purple.java:12: unreachable statement
        try
        ^
1 error

[amit taneja]

thanx for reply...but can anybody please explain the explanation given by dan ...in more details...

see we are just sharing our thougths....so..it will be good for us to explore our world...

Thanx and Regards,
amit

[Carol Enderlin]

Which part are you not getting? Sure, someone could explain this line by line, but wouldn't you get more out of it by working through it?

Dan exam explanation about line 12 being unreachable:
1. A throw statement is the first statement in the outer try block.

2. A throw statement appearing in a try block causes control to pass out of the block.

3. Consequently, statements can not be reached if they appear in the block after the throw statement.

4. The switch statement that appears after the throw statement is unreachable and results in a compile-time error.

-- you can wait for someone else to explain it, but you said you want high score so I'm trying to push you.
[ February 25, 2005: Message edited by: Carol Enderlin ]

[amit taneja]

ok got it... it means the Exception1 e is catching up...and thats why compile time error is comming...

guys you know that was my earlier ..but i misundestood and make and query..
i forgot that ..catching an exception will give compile time error ?


is i m right guys.?

[Carol Enderlin]

I'm not sure what you're saying in the last post.

The compile time error is about an unreachable statement.

Because an exception is thrown you can never get to the try block with the switch statement in it.
[ February 27, 2005: Message edited by: Carol Enderlin ]

[amit taneja]

i think that doesn't matter....that ..inner try block is never executed !!
i think what matter is as soon as exception is thrown..wheather explicitly(as in this case ) or implicitly ( due to our programming error) ...
complier will show complier error

please correct me ...if i am wrong..

amit

[ankur rathi]

Amit , when you are throwing exception intentionally , it means you don't want to compiler to go deep inside ...
So other statements are unreachable ....
hope it may help .

but I have one doubt from JLS :

while(false) {
System.out.println("Hi");
}

if(false) {
System.out.println("Hi");
}

why first one gives compile time error & second one compiles fine ...


one more thing is , JLS is deep enough for just a small topic ( unreachable ) K&B book gives not enough information about like that ...
is it risk to follow K&B ??

[Carol Enderlin]

It's OK to throw an exception. This code compiles without error:

 
class Level1Exception extends Exception {}
class Level2Exception extends Level1Exception {}
class Level3Exception extends Level2Exception {}
class Purple3
{
    public static void main(String args[])
    {
        int a,b,c,d,f,g,x;
        a = b = c = d = f = g = 0;
        x = 1;
        try
        {
            throw new Level1Exception();
        }
        catch (Level1Exception e) { d++;}
        catch (Exception e) {f++;}
        finally {g++;}
        System.out.print(a+","+b+","+c+","+d+","+f+","+g);
    }
}
 

The compiler will complain if there is a statement that will never be reached:

  
class Level1Exception extends Exception {}
class Level2Exception extends Level1Exception {}
class Level3Exception extends Level2Exception {}
class Purple4
{
    public static void main(String args[])
    {
        int a,b,c,d,f,g,x;
        a = b = c = d = f = g = 0;
        x = 1;
        try
        {
            throw new Level1Exception();
            a = 6;
        }
        catch (Level1Exception e) { d++;}
        catch (Exception e) {f++;}
        finally {g++;}
        System.out.print(a+","+b+","+c+","+d+","+f+","+g);
    }
}
 
C:\_Work\java>javac Purple4.java
Purple4.java:14: unreachable statement
            a = 6;
            ^
1 error

[Carol Enderlin]

Rathi,

I guess you just needed to read a little more on unreachable statements in JLS (14.20 Unreachable Statements ), it says it's to allow conditional compilation.

The rationale for this differing treatment is to allow programmers to define "flag variables" such as:


static final boolean DEBUG = false;

and then write code such as:

if (DEBUG) { x=3; }

The idea is that it should be possible to change the value of DEBUG from false to true or from true to false and then compile the code correctly with no other changes to the program text.

[ankur rathi]

Carol , There is lot in JLS ..
But I am sure that there would be some logic behind this , that I am not able to get , please if you or any body got the idea then please help ...

Thanks .

[Carol Enderlin]

Rathi,

I pasted the relevant part of JLS. JLS says it is to allow conditional compilation.

[amit taneja]

thanx ....
i got the point..i posted originally


but i don't get any point which rathiji post ...

can any body explain in easy words and details..?

[Animesh Shrivastava]

Regarding the doubt by Rathi Ji,

In JLS for non reachable statements in a while loop it states as:

A while statement can complete normally iff at least one of the following is true:
The while statement is reachable and the condition expression is not a constant expression with value true.
There is a reachable break statement that exits the while statement.
The contained statement is reachable iff the while statement is reachable and the condition expression is not a constant expression whose value is false.

Just read the above bold statement carefully, it says that if the condition expression is not a constant expression whose value is false, the statement is reachable

  
while(false) { // Constant expression
    System.out.println("Hii"); // Not reachable
}

but

  
int z = 15; 
while(z < 2) { //Not a constant expression
   int y = 2; // Reachable
}

also

  
final int z = 15; // This becomes constant now
while(z < 2) { //constant expression
   int y = 2; // Not Reachable
}

Now look for the rule for if statement

The actual rules for the if statement are as follows:

ACTUAL: An if-then statement can complete normally iff it is reachable.
The then-statement is reachable iff the if-then statement is reachable.

Look out for the bold statement above, the then statement(The statement just after if statement) is reachable if and only if if-then is reachable. Here nothing is mentioned regarding constant expressions....

Thats why

  
if(false) {
   k = 4; // No compilation error
}

Hope this makes u clear

[meena latha]

hi Animesh ...is this true

  for(;false :wink: 
   {
      System.out.println("Welcome");//compile error unreachable statement
    }

as the condition expression is a constant expression and evaluates to false.

[Animesh Shrivastava]

I suggest u better look into JLS for more info.
Just look up for the rules for unreachable in a for loop.
The link is already given by Carol Enderlin in a previous message