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SubjectRe: [RFC PATCH for 4.21 01/16] rseq/selftests: Add reference counter to coexist with glibc
----- On Oct 11, 2018, at 12:20 PM, Szabolcs Nagy wrote:

> On 11/10/18 16:13, Mathieu Desnoyers wrote:
>> ----- On Oct 11, 2018, at 6:37 AM, Szabolcs Nagy wrote:
>>> On 10/10/18 20:19, Mathieu Desnoyers wrote:
>>>> In order to integrate rseq into user-space applications, add a reference
>>>> counter field after the struct rseq TLS ABI so many rseq users can be
>>>> linked into the same application (e.g. librseq and glibc). The
>>>> reference count ensures that rseq syscall registration/unregistration
>>>> happens only for the most early/late user for each thread, thus ensuring
>>>> that rseq is registered across the lifetime of all rseq users for a
>>>> given thread.
>>> ...
>>>> +__attribute__((visibility("hidden"))) __thread
>>>> +volatile struct libc_rseq __lib_rseq_abi = {
>>> ...
>>>> +extern __attribute__((weak, alias("__lib_rseq_abi"))) __thread
>>>> +volatile struct rseq __rseq_abi;
>>> ...
>>>> @@ -70,7 +86,7 @@ int rseq_register_current_thread(void)
>>>> sigset_t oldset;
>>>> signal_off_save(&oldset);
>>>> - if (refcount++)
>>>> + if (__lib_rseq_abi.refcount++)
>>>> goto end;
>>>> rc = sys_rseq(&__rseq_abi, sizeof(struct rseq), 0, RSEQ_SIG);
>>> why do you use a local refcounter instead of the __rseq_abi one?
>> There is no refcount in struct rseq (the ABI between kernel and user-space).
>> The registration refcount was part of an earlier version of the rseq system
>> call,
>> but we decided against keeping it in the kernel.
>> So I'm adding one _after_ struct rseq, purely to allow interaction between
>> various user-space components (program/libraries).
> then all those components must use the same
> rseq_register_current_thread
> rseq_unregister_current_thread
> functions and not call the syscall on their own.

Not quite. Each user (programs and shared objects) must handle the refcount in a
similar way if they wish to invoke the syscall by themselves. They can
alternately use the librseq APIs if they do not wish to have a local implementation
of the reference counting and syscall registration/unregistration.

> in which case the refcount could be a static __thread variable.

Yes, but I want to limit the number of symbols we need to export
from glibc by appending the refcount field at the end of struct rseq.

> but it's in a magic struct that's called "abi" which is confusing,
> the counter is not abi, it's in a hidden object.

No, it is really an ABI between user-space apps/libs. It's not meant to be
hidden. glibc implements its own register/unregister functions (it does not
link against librseq). librseq exposes register/unregister functions as public
APIs. Those also use the refcount. I also plan to have existing libraries, e.g.
liblttng-ust and possibly liburcu flavors, implement the
registration/unregistration and refcount handling on their own, so we don't
have to add a requirement on additional linking on librseq for pre-existing

So that refcount is not an ABI between kernel and user-space, but it's a
user-space ABI nevertheless (between program and shared objects).

>>> what prevents calling rseq_register_current_thread more than 4G times?
>> Nothing. It would indeed be cleaner to error out if we detect that refcount is
>> at
>> INT_MAX. Is that what you have in mind ?
> yes

Allright, will fix.

>>> why cant the kernel see that the same address is registered again and succeed?
>> It can, and it does. However, refcounting at user-level is needed to ensure
>> the registration "lifetime" for rseq covers its entire use. If we have two
>> libraries
>> using rseq, we end up with the following scenario:
>> Thread 1
>> libA registers rseq
>> libB registers rseq
>> libB unregisters rseq
>> libA uses rseq -> bug! it's been unregistered by libB.
>> libA unregisters rseq -> unexpected, it's already been unregistered.
>> same applies if libA unregisters rseq before libB (and libB try to use rseq
>> after libA has unregistered).
>> The refcount in user-space fixes this.
> i see.

Thanks for the feedback!


>> Thoughts ?
>> Thanks,
>> Mathieu

Mathieu Desnoyers
EfficiOS Inc.

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