Abhilash Quiz

[tvs sundaram]

Question 72.
Read the following piece of code carefully.
import java.io.IOException;

public class Question72
{
public Question72() throws IOException
{
throw new IOException();
}
}
Assume that the defination of Question72E begins with the line
public class Question72E extends Question72
It is required that none of the constructors of Question72E should throw any checked exception.

1. It can be achieved by placing the call to the superclass with a super keyword , which is placed in a try block with a catch block to handle the IOException thrown by the super class.
2. It can be achived by avoiding explicit calls to the base class constructor.
3. It cannot be done in the Java Laungage with the above definition of the base class.
--------------------------------------------------------------------------------
The Given answer is 3.
First of all, I haven't understood the question very well.
Ranchers, pls help me in first understanding the Question & then let's discuss the results.
tvs sundaram

[Ashik Uzzaman]

Hi TVS,
A nice code snippet to discuss abt. First we have to remember that the constructor Question72() is throwing a checked Exception. So when u r going to throw a checked exception u have to catch it somewhere or ur program will give a compile time error. So when u r declaring a subclass Question72E which has no constructor, it will implicitely call the super class constructor which throws such an exception that is not caught in the calling constructor(Subclass's ctrc). So there will be a compile time error complaining uncaught exception. But it will work well if u modify ur subclass in this way.
class Question72E extends Question72{
Question72E() throws IOException{}
}
Carry on this thread...
------------------
azaman

[tvs sundaram]

But can we define a class throwing an Exception?
I don't think so..
tvs sundaram

[Thomas Paul]

1. doesn't work because the call to super() must be the first thing in the constructor. Option 1 requires a "try" to be the first thing.

<font size="+0">
import java.io.IOException; 

public class Question72 
{ 
    public Question72() throws IOException 
    { 
        throw new IOException(); 
    } 
}
class Question72E extends Question72 {
    public Question72E() 
    { 
        try {
            super();
        } catch (IOException e) {}
    } 
}
Gives this error:
Question72.java:16: call to super must be first statement in constructor
            super();
                 ^
1 error

2. even if you don't explicitly call the default constructor, it will be called anyway. So option 2 doesn't work either.
Which leaves 3 as the correct answer.
[This message has been edited by Thomas Paul (edited August 20, 2001).]

[Ashik Uzzaman]

Hi TVS,
No classes can't throw exception. But my slightly modified code shows u can declare to throw any exception in method header and need not catch them until u r not using these methods implicitely/explicitely. See the following code.

import java.io.IOException;
class Question72 
{ 
public Question72() throws IOException 
{ 
throw new IOException(); 
} 
}
class Question72E extends Question72{
Question72E() throws IOException{}
}
<font color="red">
class Demo
{
public static void main(String []a)
{
try{
new Question72();
new Question72E();
}
catch(Exception e)
{
System.out.println("Problem in main()");
}
}
}</font>

If u remove the Demo class (red colored) in the code it will just compile fine. But if u want to run the program u need a main() where u have to catch the exceptions thrown by the constructors. It's true for any other methods as main() when using this constructors. Hope it clears....
Anyway, i just checked that it's my 300th post within one and half months. How r u preapared tvs?
------------------
azaman
[This message has been edited by Ashik uzzaman (edited August 21, 2001).]

[tvs sundaram]

Hi guys, Thanks for your time & effort.
tvs sundaram