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Re: gcc 4.3.0, -Wconversion: assignment-by operators for shorter types
- From: "Manuel López-Ibáñez" <lopezibanez at gmail dot com>
- To: "Andrew Pinski" <pinskia at gmail dot com>
- Cc: "Andriy Gapon" <avg at icyb dot net dot ua>, gcc at gcc dot gnu dot org
- Date: Wed, 30 Jul 2008 10:50:36 +0200
- Subject: Re: gcc 4.3.0, -Wconversion: assignment-by operators for shorter types
- References: <483AF557.8050603@icyb.net.ua> <6c33472e0805270350p62d8f334v8e881e361fee4bb0@mail.gmail.com> <483C531A.5060507@icyb.net.ua> <de8d50360805271148u10525605ib84afa5c7957b2f7@mail.gmail.com> <483C5948.1060209@icyb.net.ua> <de8d50360805271200h3ced4e88n7cec1258cb0a7157@mail.gmail.com>
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.