Re: Newbie: try, catch problem

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From: Mikael Gustavsson ("Mikael)
Date: 02/24/04 

Date: Tue, 24 Feb 2004 13:04:50 +0100

Hi Sam!

If you just add a MessageBox.Show(test.ToString()) after you have done the
parse you will see why no exception is thrown.

When I did what you did and displayed the result this is what I got

192.0.0.168

It seems is if the pattern is like this

IPAddress.Parse("192");
Result = 0.0.0.192

IPAddress.Parse("192.168");
Result = 192.0.0.168

IPAddress.Parse("192.168.2");
Result = 192.168.0.2

And finally
IPAddress.Parse("192.168.2.1");
Result = 192.168.2.1

Why it works this way I have no clue. It would seem logical that it should
have been
192.168.0.0 instead of 192.0.0.168 but logic and real life does not always
compute :)

Good luck anyway!

//Mikael

"Sam Henson" <anonymous@discussions.microsoft.com> wrote in message
news:2534CD88-E5B6-4260-822D-7A44C3728AA0@microsoft.com...
> Hi!
>
> the following code should throw an exception, but it doesn't! can anyone
tell me why?
>
> try
> {
> IPAddress test = IPAddress.Parse("192.168"); //invalid input
> }
> catch(Exception mye)
> {
> MessageBox.Show("error");
> return;
> }
>
> Any ideas?
>
> Sam Henson

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