# Re: Calculating array sizes in C - was: Re: Build regressions/improvements in v6.2-rc1

- From: Michael.Karcher \<hidden\>
- Date: 2023-01-19 22:26:33 
- List: linuxppc-dev
- Page: https://ratatoskr.run/linuxppc-dev/2023/01/2073335
- Thread: https://ratatoskr.run/amd-gfx/2022/12/2049012/t.md

```text
Isn't this supposed to be caught by this check:
>>>
>>>          a, __same_type(a, NULL)
>>>
>>> ?
>>
>> Yeah, but gcc thinks it is smarter than us...
>> Probably it drops the test, assuming UB cannot happen.
> Hmm, sounds like a GGC bug to me then. Not sure how to fix this then.


I don't see a clear bug at this point. We are talking about the C expression

  __same_type((void*)0, (void*)0)? 0 : sizeof((void*)0)/sizeof(*((void*0))

This expression is valid (assuming __same_type works, which is a GCC extension), and should return 0. As of now, I have no indication that this expression does not return 0. Also, it is true that this expression contains the suspicious pattern "sizeof(void*)/sizeof(void)", which is does not calculate the size of any array. GCC is free to emit as much warnings is it wants for any kind of expressions. From a C standard point of view, it's just a "quality of implementation" issue, and an implementation that emits useless warnings is of low quality, but not non-conforming.

In this case, we requested that gcc refuses to compile if it emits any kind of warning, which instructs gcc to reject programs that would be valid according to the C standard, but are deemed to be "likely incorrect".

I suggest to file a bug against gcc complaining about a "spurious warning", and using "-Werror -Wno-error-sizeof-pointer-div" until gcc is adapted to not emit the warning about the pointer division if the result is not used.


Regards,
  Michael Karcher

```