Why isn't the overflow flag set?

[Simen Kjærås]

On Fri, 11 Nov 2011 19:19:03 +0100, Simon <s.d.hammett at gmail.com> wrote:

> On 10/11/2011 18:46, Simen Kjærås wrote:
>> On Thu, 10 Nov 2011 18:56:37 +0100, Steven Schveighoffer
>> <schveiguy at yahoo.com> wrote:
>>
>>> On Thu, 10 Nov 2011 12:24:30 -0500, Simen Kjærås
>>> <simen.kjaras at gmail.com> wrote:
>>>
>>>> I'm creating a Checked!T struct for integral values. The version
>>>> attached
>>>> works
>>>> for the most part, but some parts confuse me, namely that the  
>>>> following
>>>> operations do not set the overflow flag:
>>>>
>>>> uint.max + 1
>>>> ulong.max + 1
>>>>
>>>> int.min - 1
>>>> uint.min - 1
>>>> long.min - 1
>>>> ulong.min - 1
>>>>
>>>> uint.max << 1
>>>> long.max << 1
>>>> ulong.max << 1
>>>>
>>>> int.min << 1
>>>> long.min << 1
>>>
>>> These are likely folded into a single constant. I think it would be
>>> inappropriate for these to set the flag (it would be something
>>> unnatural anyways).
>>
>> Sorry, my example was simplified. These are all on the form
>>
>> uint a = uint.max;
>> a = a + 1;
>>
>> It works for all integral types not mentioned here.
>>
>>
>>>> int a = int.max; a *= 2
>>>> long a = long.max; a *= 2
>>>
>>> These I would think could set the overflow flag, but not necessarily
>>> when optimized.
>>
>> No optimizations are on, and -debug is passed. I find it unlikely the
>> compiler is doing much magic here. If it does, I am more than willing
>> to call it a bug.
>
> Don't know if it's changed but the last time I looked at the compiler,  
> it still does constant folding even in debug mode.

That may be, but these are not folded, at least not for the types not
mentioned above.