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Message-ID: <e2a129d122ee48b2b061fa0d3f89474a@huawei.com>
Date: Thu, 20 Jan 2022 02:19:51 +0000
From: "zhaohang (F)" <zhaohang14@...wei.com>
To: "musl@...ts.openwall.com" <musl@...ts.openwall.com>
CC: "zhangwentao (M)" <zhangwentao234@...wei.com>
Subject: 答复: Re:Re:Re: [pthread] pthread_barrier_wait  invalid case

Maybe the following patch can solve this lacking issue

diff --git a/src/time/timer_create.c b/src/time/timer_create.c
index 0a29f05c2..dcd24fdcc 100644
--- a/src/time/timer_create.c
+++ b/src/time/timer_create.c
@@ -103,6 +103,10 @@ static void *start(void *arg)
        union sigval val = args->sev->sigev_value;

        __child_sync(&args->b);
+
+       if (self->timer_id < 0)
+               return 0;
+
        for (;;) {
                siginfo_t si;
                while (sigwaitinfo(SIGTIMER_SET, &si) < 0);

发件人: zuotina [mailto:zuotingyang@....com]
发送时间: 2022年1月19日 22:56
收件人: musl@...ts.openwall.com
主题: [musl] Re:[musl] Re:Re: [musl] [pthread] pthread_barrier_wait invalid case


Hi Team,
Simple feedback on this issue
First,  replace pthread_barrier_wait in timer_create with a custom sync function (implemented by __wait, __wake),
then the problem of panic is solved
But I still think the best way is fixing pthread_barrier_wait.

In addition, it is also the problem of the timer_create function. Continue to ask for advice.
```c
timer_create:
case SIGEV_THREAD:
r = pthread_create(&td, &attr, start, &args);
    ...
if (syscall(SYS_timer_create, clk, &ksev, &timerid) < 0)
timerid = -1;
```
If this syscall fails, the 'start' thread will reside permanently,
so the above only sets timerid = -1, which should not be perfect ?
```c
start:
for (;;) {
while (sigwaitinfo(SIGTIMER_SET, &si) < 0);
}

```





At 2021-12-17 22:28:14, "zuotina" <zuotingyang@....com<mailto:zuotingyang@....com>> wrote:

At 2021-12-17 02:16:07, "Rich Felker" <dalias@...c.org<mailto:dalias@...c.org>> wrote:

>On Thu, Dec 16, 2021 at 11:25:35PM +0800, zuotina wrote:

>> Hi everrone

>>

>>

>> I encountered a panic problem when using timer_create recently.

>> Although the probability is small, it still happened.

>> Finaly I found there is a problem in the code of phtread_barrier_wait,

>> and review code found that there may be problems in the following place,

>> 81  a_store(&b->_b_lock, 0);

>> 82  if (b->_b_waiters) __wake(&b->_b_lock, 1, 1);

>> If scheduling occurs between lines 81 and 82, it will be not good.

>> So I did an experiment and modified the source code of pthread_barrier_wait to verify my guess

>> ```c

>> 81  a_store(&b->_b_lock, 0);

>>                  /* If it is scheduled out here, when another thread executes pthread_barrier_wait again,

>>                     it can go through the entire function happily, that is, it will not be blocked */

>>       syscall(yiled); // new add for test

>>                // When the dispatch comes back, this b has been released

>> 82  if (b->_b_waiters) __wake(&b->_b_lock, 1, 1);

>> ```

>

>The intent here is that it's not possible that b has been released,

>because all waiters have to synchronize on b->_b_inst. It's possible

>there's a bug here. I'll look. What arch are you running on?

 running on aarch64.

 Looking forward to fix, thank you

>> Here is an example of timer_create (src/time/timer_create.c)

>> There are two threads A and B call pthread_barrier_wait.

>> The call is as follows

>> A thread: (timer_create // parent thread)

>> {

>>        .....

>>       // new add for test---begin

>>        while(b->_b_inst == NULL) {

>>                 syscall(yield);

>>        }

>>      // new add for test---end

>>      pthread_barrier_wait();

>> }

>> B thread: (start // child thread)

>> {

>>        .....

>>       //  Ensure that this function is advanced to the if (!inst) {} branch of barrier_wait

>>       pthread_barrier_wait();

>> }

>>

>>

>> In short, the reason for panic is that pthread_barrier_wait is not blocked as expected;

>> I hope you help to confirm whether there is a problem with the implementation

>> of pthread_barrier_wait or am I wrong?

>>

>>

>> Looking forward to your reply. Thank you.

>

>Thanks for the report.

>

>Rich







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