Re: [PATCH v3 10/10] selftests/io_uring: add bpf io_uring selftests

[Pavel Begunkov <asml.silence@gmail.com>]

On 11/20/25 01:41, Ming Lei wrote:
> On Wed, Nov 19, 2025 at 07:00:41PM +0000, Pavel Begunkov wrote:
>> On 11/14/25 13:08, Ming Lei wrote:
>>> On Thu, Nov 13, 2025 at 11:59:47AM +0000, Pavel Begunkov wrote:
>> ...
>>>> +	bpf_printk("queue nop request, data %lu\n", (unsigned long)reqs_to_run);
>>>> +	sqe = &sqes[sq_hdr->tail & (SQ_ENTRIES - 1)];
>>>> +	sqe->user_data = reqs_to_run;
>>>> +	sq_hdr->tail++;
>>>
>>> Looks this way turns io_uring_enter() into pthread-unsafe, does it need to
>>> be documented?
>>
>> Assuming you mean parallel io_uring_enter() calls modifying the SQ,
>> it's not different from how it currently is. If you're sharing an
>> io_uring, threads need to sync the use of SQ/CQ.
> 
> Please see the example:
> 
> thread_fn(struct io_uring *ring)
> {
> 	while (true) {
> 		pthread_mutex_lock(sqe_mutex);
> 		sqe = io_uring_get_sqe(ring);
> 		io_uring_prep_op(sqe);
> 		pthread_mutex_unlock(sqe_mutex);
> 
> 		io_uring_enter(ring);
> 
> 		pthread_mutex_lock(cqe_mutex);
> 		io_uring_wait_cqe(ring, &cqe);
> 		io_uring_cqe_seen(ring, cqe);
> 		pthread_mutex_unlock(cqe_mutex);
> 	}
> }
> 
> `thread_fn` is supposed to work concurrently from >1 pthreads:
> 
> 1) io_uring_enter() is claimed as pthread safe
> 
> 2) because of userspace lock protection, there is single code path for
> producing sqe for SQ at same time, and single code path for consuming sqe
> from io_uring_enter().
> 
> With bpf controlled io_uring patches, sqe can be produced from io_uring_enter(),
> and cqe can be consumed in io_uring_enter() too, there will be race between
> bpf prog(producing sqe, or consuming cqe) and userspace lock-protected
> code block.

BPF is attached by the same process/user that creates io_uring. The
guarantees are same as before, the user code (which includes BPF)
should protect from concurrent mutations.

In this example, just extend the first critical section to
io_uring_enter(). Concurrent io_uring_enter() will be serialised
by a mutex anyway. But let me note, that sharing rings is not
a great pattern in either case.

-- 
Pavel Begunkov