gcc 4.3.0, -Wconversion: assignment-by operators for shorter types

[Manuel López-Ibáñez]

2008/5/27 Andrew Pinski <pinskia@gmail.com>:
> On Tue, May 27, 2008 at 11:56 AM, Andriy Gapon <avg@icyb.net.ua> wrote:
>> Thank you for the explanation! I didn't realize the difference.
>>
>> OTOH, do you think that those arithmetic warnings are practical (as opposed
>> to being correct)?
>
> I think so as the short int case has a defined overflow of the signed
> short type that is SHRT_MAX + 1 is defined while INT_MAX + 1 is not.

Andrew, I don't understand what you mean by "SHRT_MAX + 1 is defined
while INT_MAX + 1 is not".

Is there actually any difference in the following cases given   short
short_x = SHRT_MAX ?

       short_x = short_x + 1;
       short_x += 1;
       short_x++;
       ++short_x;

What about unsigned short ushort_x = USHRT_MAX ?

       ushort_x = ushort_x + 1;
       ushort_x += 1;
       ushort_x++;
       ++ushort_x;

Cheers,

Manuel.