---- Reported by michael@synthesyssolutions.com 2006-09-11 17:33:22 MST ----

Description of Problem:

The following code compiles and executes using MS.NET, but not Mono:

using System;

public class NullableInt
{
         public static void Main()
         {
                 object x = null;

                 int? y = x as int?;  /* Causes CS0077 */

                 Console.WriteLine("y: '{0}'", y);
                 Console.WriteLine("y.HasValue: '{0}'", y.HasValue);
         }
}

Steps to reproduce the problem:
1. Copy and paste code above into a source code file
2. Attempt to compile source file with gmcs
3. Receive compiler error CS0077

Actual Results:

test.cs(9,14): error CS0077: The as operator must be used with a 
reference type (`int?' is a value type)
Compilation failed: 1 error(s), 0 warnings

Expected Results:

Successful compilation.  The program outputs the following using MS.NET:

y: ''
y.HasValue: 'False'

How often does this happen? 

Every time.

Additional Information:

Using svn TRUNK



---- Additional Comments From marek.safar@seznam.cz 2006-09-23 17:38:08 MST ----

It looks like csc issue or C# standard was not updated correctly.

Based on the sections 11.4 and 14.9.11 it should not be allowed.




---- Additional Comments From marek.safar@seznam.cz 2006-10-03 10:46:15 MST ----

*** https://bugzilla.novell.com/show_bug.cgi?id=MONO79566 has been marked as a duplicate of this bug. ***



---- Additional Comments From marek.safar@seznam.cz 2007-03-24 05:39:49 MST ----

This bit is specified in Microsoft version of the standard.

24.3.6 The as operator
The as operator (§7.9.10) is extended to support nullable types. In an
operation of the form e as T, e must be an expression and T must be a
reference type, a type parameter known to be a reference type, or a
nullable type. Furthermore, at least one of the following must be
true, or otherwise a compile-time error occurs:
•	An identity (§6.1.1), implicit reference (§6.1.4), boxing (§6.1.5,
§24.2.3), explicit reference (§6.2.3), or unboxing (§6.2.4, §24.2.3)
conversion exists from the type of e to T.
•	The type of e or T is an open type.
•	e is the null literal.
The operation e as T produces the same result as
e is T ? (T)(e) : (T)null
except that e is only evaluated once. The compiler can be expected to
optimize e as T to perform at most one dynamic type check as opposed
to the two dynamic type checks implied by the expansion above.




---- Additional Comments From martin@ximian.com 2007-04-11 14:18:23 MST ----

Fixed in SVN.


Unknown bug field "cf_op_sys_details" encountered while moving bug
   <cf_op_sys_details>Debian Etch</cf_op_sys_details>