Re: New warning in Xcode 7
Re: New warning in Xcode 7
- Subject: Re: New warning in Xcode 7
- From: Rick Mann <email@hidden>
- Date: Tue, 13 Oct 2015 16:01:13 -0700
> On Oct 13, 2015, at 15:50 , Greg Parker <email@hidden> wrote:
>
>
>> On Oct 13, 2015, at 3:24 PM, Graham Cox <email@hidden> wrote:
>>
>> In a piece of code I haven’t worked on in a while, I’m suddenly getting a new warning which I’ve never seen before:
>>
>> “Multiple unsequenced modifications to ‘ix’”
>>
>> Code is:
>>
>> ix = ++ix % guess.count;
>>
>> where ix is a NSUInteger.
>>
>> Is this telling me that the order of the preincrement and the mod operation is undefined? Surely a preincrement is defined to happen first, that’s why it’s called a PREincrement? Or does the warning refer to something else? I’ve used this form of expression for years without any issues, why is it suddenly one?
>
> The mod operator uses the value after the increment. That part is well-defined.
>
> The compiler is complaining about the two stores to `ix`, one from the preincrement and one from operator =. Storing to the same location twice in the same statement is undefined in many cases: the language does not specify which one wins. Look up "sequence point" for more details (although the current language standards no longer use the "sequence point" terminology).
>
> I thought that this particular form was allowed, but I'm not enough of a language lawyer to know for sure. You might get better answers from a compiler mailing list.
>
> You can avoid confusion (human and compiler) by writing the statement without the preincrement's side effect:
> ix = (ix+1) % guess.count;
Actually, yeah, this is a much better alternative.
The document I linked in my other answer (http://en.cppreference.com/w/c/language/eval_order), under "Rules," specifies where the sequence points are. As of C11, rule 10 applies: the value of the right-hand-side is sequenced before the assignment, but not the side effects of the right-hand side. So, it's possible that the assignment to ix could happen first, and then the increment. So, it would compute ix+1, use that value in evaluating the right hand side, assign that result to ix, then store the old ix+1 into ix.
--
Rick Mann
email@hidden
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