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Subject: Re: Correct way to express to the compiler "this does not get
 clobbered"?
To: Andrea Corallo <andrea.corallo@arm.com>
Cc: gcc-help@gcc.gnu.org, nd@arm.com
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 <15f8e09a-9d21-f8f6-8611-2845bd8bc74e@foss.arm.com> <gkrtut3gjjz.fsf@arm.com>
From: Richard Earnshaw <Richard.Earnshaw@foss.arm.com>
Message-ID: <1632b7ed-dd89-a53e-8367-a8e29d34bdf4@foss.arm.com>
Date: Thu, 3 Dec 2020 13:06:45 +0000
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On 03/12/2020 12:28, Andrea Corallo wrote:
> Richard Earnshaw <Richard.Earnshaw@foss.arm.com> writes:
> 
>> On 03/12/2020 11:47, Andrea Corallo via Gcc-help wrote:
>>> Hi all,
>>>
>>> I've a piece of code that reduced looks like this:
>>>
>>> #+begin_src C
>>> typedef struct {
>>>   void (*fun_ptr)(void);
>>> } x_t;
>>>
>>> x_t *x;
>>>
>>> void
>>> f (void)
>>> {
>>>   const x_t const *y = x;
>>>   for (int i = 0; i < 1000; ++i)
>>>     y->fun_ptr ();
>>> }
>>> #+end_src
>>>
>>> What is the correct way (if any) to express to the compiler that the
>>> value of y->fun_ptr does not get clobbered by the function call itself
>>> so the corresponding load to obtain its value can be moved out of the
>>> loop?
>>>
>>> My understanding is that the const qualifier is more for diagnostic
>>> reasons and is not sufficient for GCC to make this assumption.  OTOH I
>>> cannot give 'fun_ptr' the attribute pure as it's not.
>>>
>>> Thanks
>>>
>>>   Andrea
>>>
>>
>> Why not just put the function pointer in a local variable?  Then the
>> compiler will know that the value can't change.
> 
> Hi Richard,
> 
> yeah that would do the job.
> 
> This is generated code and changing the code generator that way might
> not be completely trivial (e.g. where is the best position to perform
> the assignments to each local variable?).  Therefore before going for
> this solution I'd like to be convinced there's no way to express this
> directly to GCC.
> 
>   Andrea
> 

Perhaps you could declare y with the restrict qualifier?  But that would
need to apply to all of *y, not just fun_ptr.

R.