Re: [PATCH 1/2] compiler.h: Introduce ptr_eq() to preserve address dependency

Alan Huang
2024-09-30T09:28:05Z

2024年9月30日 17:15，Alan Huang <mmpgouride@gmail.com> 写道：
>
> 2024年9月30日 16:57，Jonas Oberhauser <jonas.oberhauser@huaweicloud.com> 写道：
>>
>>
>>
>> Am 9/29/2024 um 12:26 AM schrieb Alan Huang:
>>> 2024年9月28日 23:55，Mathieu Desnoyers <mathieu.desnoyers@efficios.com> wrote：
>>>>
>>>> On 2024-09-28 17:49, Alan Stern wrote:
>>>>> On Sat, Sep 28, 2024 at 11:32:18AM -0400, Mathieu Desnoyers wrote:
>>>>>> On 2024-09-28 16:49, Alan Stern wrote:
>>>>>>> On Sat, Sep 28, 2024 at 09:51:27AM -0400, Mathieu Desnoyers wrote:
>>>>>>>> equality, which does not preserve address dependencies and allows the
>>>>>>>> following misordering speculations:
>>>>>>>>
>>>>>>>> - If @b is a constant, the compiler can issue the loads which depend
>>>>>>>> on @a before loading @a.
>>>>>>>> - If @b is a register populated by a prior load, weakly-ordered
>>>>>>>> CPUs can speculate loads which depend on @a before loading @a.
>>>>>>>
>>>>>>> It shouldn't matter whether @a and @b are constants, registers, or
>>>>>>> anything else. All that matters is that the compiler uses the wrong
>>>>>>> one, which allows weakly ordered CPUs to speculate loads you wouldn't
>>>>>>> expect it to, based on the source code alone.
>>>>>>
>>>>>> I only partially agree here.
>>>>>>
>>>>>> On weakly-ordered architectures, indeed we don't care whether the
>>>>>> issue is caused by the compiler reordering the code (constant)
>>>>>> or the CPU speculating the load (registers).
>>>>>>
>>>>>> However, on strongly-ordered architectures, AFAIU, only the constant
>>>>>> case is problematic (compiler reordering the dependent load), because
>>>>> I thought you were trying to prevent the compiler from using one pointer
>>>>> instead of the other, not trying to prevent it from reordering anything.
>>>>> Isn't this the point the documentation wants to get across when it says
>>>>> that comparing pointers can be dangerous?
>>>>
>>>> The motivation for introducing ptr_eq() is indeed because the
>>>> compiler barrier is not sufficient to prevent the compiler from
>>>> using one pointer instead of the other.
>>> barrier_data(&b) prevents that.
>>
>> I don't think one barrier_data can garantuee preventing this, because right after doing the comparison, the compiler still could do b=a.
>>
>> In that case you would be guaranteed to use the value in b, but that value is not the value loaded into b originally but rather the value loaded into a, and hence your address dependency goes to the wrong load still.
>
> After barrier_data(&b), *b will be loaded from memory, you mean even if *b is loaded from memory, the address dependency goes to the wrong load still?

Sorry, *b should b.

>
>>
>> However, doing
>>
>> barrier_data(&b);
>> if (a == b) {
>> barrier();
>> foo(*b);
>> }
>>
>> might maybe prevent it, because after the address of b is escaped, the compiler might no longer be allowed to just do b=a;, but I'm not sure if that is completely correct, since the compiler knows b==a and no other thread can be concurrently modifying a or b. Therefore, given that the compiler knows the hardware, it might know that assigning b=a would not cause any race-related issues even if another thread was reading b concurrently.
>>
>> Finally, it may be only a combination of barrier_data and making b volatile could be guaranteed to solve the issue, but the code will be very obscure compared to using ptr_eq.
>>
>> jonas