Why does this compile?

[tni@nc.rr.com]

int x = 0; int y = 0; int z = 0;

try
{
z = x / y;
}
catch (Exception) // THIS IS NOT A (DECLARATION)!! !!
{

}

Thomas

[Michael Bray]

Re: Why does this compile?

Why shouldn't it compile? If you want to catch exceptions, you don't have
to tell it a specific exception variable to store it in...

-mdb

tni@nc.rr.com wrote in news:1137706760 .748579.244950
@g14g2000cwa.go oglegroups.com:
[color=blue]
> int x = 0; int y = 0; int z = 0;
>
> try
> {
> z = x / y;
> }
> catch (Exception) // THIS IS NOT A (DECLARATION)!! !!
> {
>
> }
>
> Thomas
>
>[/color]

[Jon Skeet [C# MVP]]

Re: Why does this compile?

<tni@nc.rr.co m> wrote:[color=blue]
> int x = 0; int y = 0; int z = 0;
>
> try
> {
> z = x / y;
> }
> catch (Exception) // THIS IS NOT A (DECLARATION)!! !!
> {
>
> }[/color]

C# isn't like Java - you don't have to declare a variable to catch the
exception "into".

From the ECMA spec:

<quote>
When a catch clause specifies a class-type, the type must be
System.Exceptio n or a type that derives from System.Exceptio n.

When a catch clause specifies both a class-type and an identifier, an
exception variable of the given name and type is declared. The
exception variable corresponds to a local variable with a scope that
extends over the catch block. During execution of the catch block, the
exception variable represents the exception currently being handled.
For purposes of definite assignment checking, the exception variable is
considered definitely assigned in its entire scope.

Unless a catch clause includes an exception variable name, it is
impossible to access the exception object in the catch block.
</quote>

--
Jon Skeet - <skeet@pobox.co m>
http://www.pobox.com/~skeet Blog: http://www.msmvps.com/jon.skeet
If replying to the group, please do not mail me too

[Joerg Jooss]

Re: Why does this compile?

Hello tni@nc.rr.com,
[color=blue]
> int x = 0; int y = 0; int z = 0;
>
> try
> {
> z = x / y;
> }
> catch (Exception) // THIS IS NOT A (DECLARATION)!! !!
> {
> }[/color]

If you don't use the exception instance within the catch block, you don't
need to declare an identifier (C# lang spec §8.10). Of course you cannot
access the original exception in this case ;-)

Cheers,
--
Joerg Jooss
news-reply@joergjoos s.de

[rossum]

Re: Why does this compile?

On 19 Jan 2006 13:39:20 -0800, tni@nc.rr.com wrote:
[color=blue]
> int x = 0; int y = 0; int z = 0;
>
> try
> {
> z = x / y;
> }
> catch (Exception) // THIS IS NOT A (DECLARATION)!! !!
> {
>
> }
>
>Thomas[/color]
Because there aren't any errors in it.

rossum

--

The ultimate truth is that there is no ultimate truth

[Nick Hounsome]

Re: Why does this compile?


<tni@nc.rr.co m> wrote in message
news:1137706760 .748579.244950@ g14g2000cwa.goo glegroups.com.. .[color=blue]
> int x = 0; int y = 0; int z = 0;
>
> try
> {
> z = x / y;
> }
> catch (Exception) // THIS IS NOT A (DECLARATION)!! !!
> {
>
> }[/color]

For 2 related reasons: Code like the following is common

catch(Exception Type1)
{
// the only interesting thing about the exception is its type
DoStuff();
}
catch(Exception Type2)
{
// cleanup stuff here but don't use exeception
throw; // rethrows exception
}


and

catch(Exception ex)
{
}

will give a compiler warning about unused variable ex.

Of course it seems quite reasonable to me that the compiler should not warn
in this particular case but that's how it is.