[Mesa-dev] [PATCH v2] glsls: Modify exec_list to avoid strict-aliasing violations

[Davin McCall]

On 26/06/15 17:08, Eirik Byrkjeflot Anonsen wrote:
> Davin McCall <davmac at davmac.org> writes:
>
>> On 26/06/15 14:31, Eirik Byrkjeflot Anonsen wrote:
>>> Erik Faye-Lund <kusmabite at gmail.com> writes:
>>>
>>>> On Fri, Jun 26, 2015 at 1:23 PM, Davin McCall <davmac at davmac.org> wrote:
>>>>> On 26/06/15 12:03, Davin McCall wrote:
>>>>>> ... The stored value of 'n' is not accessed by any other type than the
>>>>>> type of n itself. This value is then cast to a different pointer type. You
>>>>>> are mistaken if you think that the cast accesses the stored value of n. The
>>>>>> other "stored value" access that it occurs in that expression is to the
>>>>>> object pointed at by the result of the cast. [...]:
>>>>> I'm sorry, I think that was phrased somewhat abrasively, which I did not
>>>>> intend. Let me try this part again. If we by break up the expression in
>>>>> order of evaluation:
>>>>>
>>>>> From:
>>>>>      return ((const struct exec_node **)n)[0]
>>>>>
>>>>> In order of evaluation:
>>>>>
>>>>> n
>>>>> - which accesses the stored value of n, i.e. a value of type 'struct exec
>>>>> node *', via n, which is obviously of that type.
>>>>>
>>>>> (const struct exec_node **)n
>>>>>    - which casts that value, after it has been retrieved, to another type. If
>>>>> this were an aliasing violation, then casting any pointer variable to
>>>>> another type would be an aliasing violation; this is clearly not the case.
>>>>>
>>>>> ((const struct exec_node **)n)[0]
>>>>> - which de-references the result of the above cast, thereby accessing a
>>>>> stored value of type 'exec node *' using a glvalue of type 'exec node *'.
>>>> I think breaking this up is a mistake, because the strict-aliasing
>>>> rules is explicitly about the *combination* of these two things.
>>>>
>>>> You *are* accessing the underlying memory of 'n' through a different
>>>> type, and this is what strict aliasing is all about. But it takes two
>>>> steps, a single step isn't enough to do so.
>>>>
>>>> Those other spec-quotes doesn't undo the strict-aliasing definitions;
>>>> knowing how things are laid out in memory doesn't mean the compiler
>>>> cannot assume two differently typed variables doesn't overlap.
>>> So basically, you're saying that e.g.:
>>>
>>>      p->next = a;
>>>      q = exec_node_get_next_const(p);
>>>
>>> is equivalent to:
>>>
>>>      exec_node * p1 = p;
>>>      exec_node ** p2 = (exec_node**)p;
>>>      p1->next = a;
>>>      q = p2[0];
>>> [...]
>>>    Thus the two
>>> assignments can be "safely" reordered. Sounds plausible to me.
>> The assignments are to 'p1->next' and 'p2[0]', which *are* the same type
>> (struct exec_node *) and therefore the assignments *cannot* be
>> reordered. It is exactly this that I rely on in my patch to resolve the
>> aliasing issue with the current code.
> But, on the other hand, if *p1 and *p2 are known to be non-overlapping,
> it would mean that p1->next and p2[0] are non-overlapping. Thus the two
> statements can be reordered. Though I'd actually have to carefully read
> the spec to check whether a compiler would be allowed to do that.

In the example you've give *p1 and *p2 can still overlap, because *p2 is 
of type 'struct exec_node *' and *p1 is of type 'struct exec_node' which 
contains members of type 'struct exec_node *'.

But, yes, if it could be shown that they didn't overlap, the two 
statements could be re-ordered. It's just that in this case, that cannot 
be shown.

> [...]
>
> However, I'm assuming the intent of the rule is that two valid pointers
> of different types can never point to the same address. In other words,
> when the compiler sees two pointers of different types, it can assume
> they point to different memory locations.

There is one exception however (which I referred to above) - if one 
points at an aggregate type which contains a member whose type is the 
same as what the other points to.

>
> Thus, if the compiler thinks:
>
>     p->next = a  becomes  "write a to p+0"
>     b = q[0]     becomes  "read b from q+0"
>
> With p and q being known different values (due to having different
> types), it is clear that the read statement can not be affected by the
> write statement.
>
> Again, I'm not sure that's what the spec actually says, but I suspect
> strongly that this was the intention of the strict aliasing rules. And
> that precisely these optimizations is what compilers do when they use
> the strict aliasing rules.

Right, if you can eliminate the "membership" exception. :)

Davin