Exception

[Parameswaran Thangavel]

 

import java.io.IOException; 
 
class Q
{ 
    Q() throws IOException 
 
    { 
        throw new IOException(); 
    } 
}
 

 

class Param extends Q
 { 
int a;
 
Param() throws IOException
{
 
}
      public static void main(String args[]) 
      { 
try{
Param p=new Param();}
catch(Exception e)
{System.out.println("caught");}
 
System.out.println(p.a); //1
 
      } 
 }
 
  

compiler error at line 1

please explain me


P.S:i donno why peoples are not replying to my post please do reply.... though it is having silly questions...

[vidya sagar]

Hi paramesh

I think scope of object p in the try block is over when block ends
if u move the println statement next to object creation statement it works

  
public static void main(String args[])      
{ try{
Param p=new Param();}catch(Exception e){System.out.println("caught");} 

[vidya sagar]

(sorry friends i click the enter button mistakenly
Hi paramesh

I think scope of object p in the try block is over when block ends
if u move the println statement next to object creation statement it works

  
public static void main(String args[])      
{ try{
          Param p=new Param();
         [B]System.out.println(p.a)[/B]   
         }catch(Exception e){System.out.println("caught");} 

[Anonymous]

Hi

The compilation error is caused by a problem in the variable scope "p".

p is defined inside the try{} and is not accesible after the "catch".

If you want to use this variable outside the catch you must declare it outside the try:


Param p = null;
try{
p = new Param();
} catch {

}

System.out.println(p.a);

[vidya sagar]

P.S:i donno why peoples are not replying to my post please do reply.... though it is having silly questions...

Take it in a correct sense

your are asking basic doubt......i hope UR not understood basic concepts correctly otherwise UR not thinking for sometime before asking