Exception Handling

[Faqeer Khudaka]

Hi all,

Had a question about the following code:

  class  TooColdException extends RuntimeException{
 
}
public class Lsmul{
    public static void main(String argv[]){
Lsmul lsm = new Lsmul();
lsm.go();
System.out.println(lsm.aboat());
   }
    public void go(){
nfl();
    }
    public int  aboat(){
try{
    nfl();
    return 1;
}catch(TooColdException tce){
    System.out.println(tce.getMessage());
    return 2;
}finally{
return 3;
}
 
    }
 
    public void nfl() throws TooColdException{
System.out.println("seal");
 
    }
}
 

Prints " seal, seal, 3 " when compiled and run. I was wondering how is this code getting compiled cause try block in aboat method is returning 1 and then in the finally block, there is return value of 3. Isnt this wrong cause a method can return only one value and even if it can return 2 values, why 1 is not printed??

Regards

[Craig Tyler]

The reason why 3 is printed instead of 1 is that the finally block is run even though 1 was supposed to be returned in the try block. Since the finally block does a return statement as well, this return statement takes precedence over the return statement in the try block.

This would also be the case if the catch block was entered and the return 2 statement was reached. The finally block would run and its return statement would overrule the one in the catch block.

Remember: The finally block always runs unless System.exit() is called or something happens like turning off the computer.

[gaurav singhal]

I understood what is told but plz tell me why

class TooColdException extends RuntimeException{ }
public class Lsmul{

public static void main(String argv[]){
Lsmul lsm = new Lsmul();
lsm.go();
System.out.println(lsm.aboat());
}

public void go(){
nfl();
}

public int aboat(){
int a;
try{
nfl();
a=5;
return a;
}
catch(TooColdException tce){
System.out.println(tce.getMessage());
return 2;
}
finally
{
a=6;

}
}
public void nfl() throws TooColdException{
System.out.println("seal");
}
}


The output of this program is seal, seal and 5 if the finally is executed the value returned should be 6.

[Christophe Verré]

Try to output the value of 'a' in the finally block, and you'll understand that this variable is different from the one you declared in aboat();
The finally block actually acts a bit like a subroutine, with its own local variables, which will be called to each exiting points.

[gaurav singhal]

Ya i have tried printing a value but i want to know that the scope of a is local to the method aboat so it should be reflected while returning.

[gaurav singhal]

Plz explain me what do mean by finally block act as a subroutine with its own local variable because the variable it is changing is defined in function level.

[Christophe Verré]

http://www.javaworld.com/javaworld/jw-02-1997/jw-02-hood.html

[Abhinav Gogna]

  class TooColdException extends RuntimeException{ }
public class Lsmul{ 
 
public static void main(String argv[]){ 
Lsmul lsm = new Lsmul(); 
lsm.go(); 
System.out.println(lsm.aboat()); 
} 
 
public void go(){ 
nfl(); 
} 
 
public int aboat(){ 
int a;
try{ 
nfl();
a=5; 
return a; 
}
catch(TooColdException tce){ 
System.out.println(tce.getMessage()); 
return 2; 
}
finally
{ 
a=6;
[B]return a;[/B]
 
} 
} 
public void nfl() throws TooColdException{ 
System.out.println("seal"); 
}
} 
  

See above. Value is 5 because its not object reference you are returning but primitive. a is already returned before the finally block is run. If you put return a in finally block then it returns 6.