Simon,
>why is the comparison to 0 done twice?
Well, you should think of the second 0 as the constant for false. So an
expression x != 0 gets deduced like this:
x != 0
!(x == 0)
(x == 0) == false
(x == 0) == 0
x == 0 == 0
You could also reason the other way around: what if there was only one
comparison, like you are implying to be what you were expecting?
Wouldn't you then just simply check for equality?
HTH,
Stefan
>-----Original Message-----
>From: [EMAIL PROTECTED]
>[mailto:[EMAIL PROTECTED] On Behalf
>Of Simon Hewitt
>Sent: Thursday, February 27, 2003 11:02 AM
>To: [EMAIL PROTECTED]
>Subject: Compilation question....
>
>
>Given this source code
>
>C# source:
> public bool IsIndexed {
> get { return (SearchFlags & (int)
>AttributeSearchFlags.Index) != 0; }
> }
>
>
>why is the comparison to 0 done twice?
>
>
>
>Anakrino decompilation:
>public bool get_IsIndexed() {
> return this.SearchFlags & 1 == 0 == 0;
>}
>
>
>ILDASM decompilation:
>{
> // Code size 15 (0xf)
> .maxstack 2
> IL_0000: ldarg.0
> IL_0001: call instance int32
>SimmoTech.ActiveDirectory.ADAttributeSchema::get_SearchFlags()
> IL_0006: ldc.i4.1
> IL_0007: and
> IL_0008: ldc.i4.0
> IL_0009: ceq
> IL_000b: ldc.i4.0
> IL_000c: ceq
> IL_000e: ret
>} // end of method ADAttributeSchema::get_IsIndexed
>
