Overriden method throwing Exception.

[m ali]

Hi Ranchers,

 
class Ping extends Utils {
 public static void main(String [] args) throws Exception {
 Utils u = new Ping();
 System.out.print(u.getInt(args[0]));
 }
 int getInt(String arg) {
 return Integer.parseInt(arg);
 }
 }
 
 class Utils {
 int getInt(String x) throws Exception { return 7; }
 }

If we run it with cmd arguemnt 0 it will print 0. We are sure that in the above code the overridden getInt() method(Subclass version) is calling. Then why we declare main method throws exception,when subclass getInt() method not throwing any exception. Please anyone explain?

Thanks in advance.

[Jesse Custer]

Decision about wich overridden method will be called is made at runtime. The compiler doesn't know that the variable of type Utils actually contains a Ping. So we are forced to handle the exception or declare it like done here.

[Burkhard Hassel]

Hi ranchers,

yes, when you say
Ping u1 = new Ping() the code compiles without the "throws".

What's a bit confusing about the code, is that you are refering on the arguments of main. It would be easier to understand when you just say

...
public static void main(String[] args) { // no throws
Ping  u1 = new Ping();// OK
Utils u2 = new Ping();// OK
System.out.print(u1.getInt("5"));// OK
System.out.print(u2.getInt("5"));   // no compile
}
...

Yours,
Bu.

[Burkhard Hassel]

Hi ranchers,

and now for something completely different.

Try to post what is here in code tags without code tags to the java ranch:

Ping u = new Ping();

Ping you,
Bu.

[marc weber]

I cannot divulge that secret.
[ April 17, 2007: Message edited by: marc weber ]