Re:constructors throwing exceptions

[sunil kumre]

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.

A)It can be achived 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.
B)It can be achived by avoiding explicit calls to the base class constructor.
C)It cannot be done in the Java Laungage with the above definition of the base class.

why in this case the answer is C).
thanks in advance.

[bill bozeman]

This is a thought, I haven't checked it, but this is what I think:
A is wrong because a call to super() needs to be the first statement in the subclasses constructor. If you have to catch the exception, then you are putting a try statement first.
B is wrong because if you don't make an explicit call, then an implicit call will be made. Since the implicit or explicit call throw an exception, it must be caught.
Bill

[sunil kumre]

thanks Bill...for your wonderful explanation.