Re: [BUG] -Wuninitialized: initialize variable with itself

[Alejandro Colomar <alx.manpages@gmail.com>]

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Hi David,

On 11/14/22 10:41, David Brown wrote:
> On 13/11/2022 19:43, Alejandro Colomar via Gcc wrote:
>> Hi Andrew!
>>
>> On 11/13/22 19:41, Andrew Pinski wrote:
>>> On Sun, Nov 13, 2022 at 10:40 AM Andrew Pinski <pinskia@gmail.com> wrote:
>>>>
>>>> On Sun, Nov 13, 2022 at 10:36 AM Alejandro Colomar via Gcc
>>>> <gcc@gcc.gnu.org> wrote:
>>>>>
>>>>> Hi,
>>>>>
>>>>> While discussing some idea for a new feature, I tested the following example
>>>>> program:
>>>>>
>>>>>
>>>>>       int main(void)
>>>>>       {
>>>>>           int i = i;
>>>>>           return i;
>>>>>       }
>>>>
>>>> This is NOT a bug but a documented way of having the warning not being there.
>>>> See 
>>>> https://gcc.gnu.org/onlinedocs/gcc-12.2.0/gcc/Warning-Options.html#index-Winit-self
>>>> https://gcc.gnu.org/onlinedocs/gcc-12.2.0/gcc/Warning-Options.html#index-Wuninitialized
>>>> "If you want to warn about code that uses the uninitialized value of
>>>> the variable in its own initializer, use the -Winit-self option."
>>>
>>> I should note the main reason why I Know about this is because I fixed
>>> this feature years ago (at least for C front-end)
>>> and added the option to disable the feature.
>>
>> I'm curious: what are the reasons why one would want to disable such a warning?
>> Why is it not in -Wall or -Wextra?
>>
>> Thanks,
>>
>> Alex
>>
> 
> Warnings are not perfect - there is always the risk of false positives and false 
> negatives.  And different people will have different ideas about what code is 
> perfectly reasonable, and what code is risky and should trigger a warning.  Thus 
> gcc has warning flag groups (-Wall, -Wextra) that try to match common consensus, 
> and individual flags for personal fine-tuning.
> 
> Sometimes it is useful to have a simple way to override a warning in code, 
> without going through "#pragma GCC diagnostic" lines (which are powerful, but 
> not pretty).
> 
> So if you have :
> 
>      int i;
>      if (a == 1) i = 1;
>      if (b == 1) i = 2;
>      if (c == 1) i = 3;
>      return i;
> 
> the compiler will warn that "i" may not be initialised.  But if you /know/ that 
> one of the three conditions will match (or you don't care what "i" is if it does 
> not match), then you know your code is fine and don't want the warning.  Writing 
> "int i = i;" is a way of telling the compiler "I know what I am doing, even 
> though this code looks dodgy, because I know more than you do".

Ahh, that makes sense.  Since the default warnings warn about 'int i=i+1;' it 
makes sense to me.  Writing 'int i=i;' is just too stupid that can be considered 
a reasonable way to tell the compiler we know better.

Thanks!

Cheers,

Alex

> 
> It's just like writing "while ((*p++ = *q++));", or using a cast to void to turn 
> off an "unused parameter" warning.

-- 
<http://www.alejandro-colomar.es/>

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