sem_wait not unblocking with EINTR

问题

I'm new with semaphores and want to add multithreading to my program, but I cannot get around the following problem: sem_wait() should be able to receive a EINTR and unblock, as long as I didn't set the SA_RESTART flag. I am sending a SIGUSR1 to the worker thread that is blocking in sem_wait(), it does receive the signal and get interrupted, but it will then continue to block and so it will never give me a -1 return code together with errno = EINTR. However, if I do a sem_post from the main thread, it will unblock, give me an errno of EINTR but a RC of 0. I am totally puzzled with this behavior. Is it some weird NetBSD implementation or am I doing something wrong here? According to the man page, sem_wait is conform POSIX.1 (ISO/IEC 9945-1:1996). A simple code:

#include <stdio.h>
#include <stdlib.h>
#include <errno.h>
#include <signal.h>
#include <pthread.h>
#include <semaphore.h>

typedef struct workQueue_s
{
   int full;
   int empty;
   sem_t work;
   int sock_c[10];
} workQueue_t;

void signal_handler( int sig )
{
   switch( sig )
   {
      case SIGUSR1:
      printf( "Signal: I am pthread %p\n", pthread_self() );
      break;
   }
}

extern int errno;
workQueue_t queue;
pthread_t workerbees[8];

void *BeeWork( void *t )
{
   int RC;
   pthread_t tid;
   struct sigaction sa;
   sa.sa_handler = signal_handler;
   sigaction( SIGUSR1, &sa, NULL );

   printf( "Bee: I am pthread %p\n", pthread_self() );
   RC = sem_wait( &queue.work );
   printf( "Bee: got RC = %d and errno = %d\n", RC, errno );

   RC = sem_wait( &queue.work );
   printf( "Bee: got RC = %d and errno = %d\n", RC, errno );
   pthread_exit( ( void * ) t );
}

int main()
{
   int RC;
   long tid = 0;
   pthread_attr_t attr;
   pthread_attr_init( &attr );
   pthread_attr_setdetachstate( &attr, PTHREAD_CREATE_JOINABLE );

   queue.full = 0;
   queue.empty = 0;
   sem_init( &queue.work, 0, 0 );

   printf( "I am pthread %p\n", pthread_self() );
   pthread_create( &workerbees[tid], &attr, BeeWork, ( void * ) tid );
   pthread_attr_destroy( &attr );

   sleep( 2 );
   sem_post( &queue.work );
   sleep( 2 );
   pthread_kill( workerbees[tid], SIGUSR1 );
   sleep( 2 );

   // Remove this and sem_wait will stay blocked
   sem_post( &queue.work );
   sleep( 2 );
   return( 0 );
}

I know the printf is not aloud in the signal handler, but just for the heck of it, if I remove it I get the same results.

These are the results without sem_post:

I am pthread 0x7f7fffc00000
Bee: I am pthread 0x7f7ff6c00000
Bee: got RC = 0 and errno = 0
Signal: I am pthread 0x7f7ff6c00000

And with the sem_post:

I am pthread 0x7f7fffc00000
Bee: I am pthread 0x7f7ff6c00000
Bee: got RC = 0 and errno = 0
Signal: I am pthread 0x7f7ff6c00000
Bee: got RC = 0 and errno = 4

I know I don't really need to unblock and can simply do an exit from main, but I want to see it working anyway. The reason I'm using sem_wait is because I want to keep the worker threads alive and wake the one up waiting the longest from the main thread with sem_post, as soon as there is a new client connection from Postfix. I don't want to do pthread_create all the time, since I will receive calls multiple times per second and I don't want to lose speed and make Postfix unresponsive to new smtpd clients. It is a policydaemon for Postfix and the server is quite busy.

Am I missing something here? Is NetBSD just messed up with this?

回答1:

My post is about behaviour on Linux, but I think you may have similar behaviour, or at least I thought could be helpful. If not, let me know, I'll remove this useless 'noise'.

I tried to reproduce your setup and I was quite astonished of seeing what you describe happening. Looking deeper helped me figure out that there was actually something more subtil; if you have a look to strace, you'll see somthing like:

[pid  6984] futex(0x6020e8, FUTEX_WAIT_PRIVATE, 0, NULL <unfinished ...>
[pid  6983] rt_sigprocmask(SIG_BLOCK, [CHLD], [], 8) = 0
[pid  6983] rt_sigaction(SIGCHLD, NULL, {SIG_DFL, [], 0}, 8) = 0
[pid  6983] rt_sigprocmask(SIG_SETMASK, [], NULL, 8) = 0
[pid  6983] nanosleep({2, 0}, 0x7fffe5794a70) = 0
[pid  6983] tgkill(6983, 6984, SIGUSR1 <unfinished ...>
[pid  6984] <... futex resumed> )       = ? ERESTARTSYS (To be restarted if SA_RESTART is set)
[pid  6983] <... tgkill resumed> )      = 0
[pid  6984] --- SIGUSR1 {si_signo=SIGUSR1, si_code=SI_TKILL, si_pid=6983, si_uid=500} ---
[pid  6983] rt_sigprocmask(SIG_BLOCK, [CHLD],  <unfinished ...>
[pid  6984] rt_sigreturn( <unfinished ...>
[pid  6983] <... rt_sigprocmask resumed> [], 8) = 0
[pid  6984] <... rt_sigreturn resumed> ) = -1 EINTR (Interrupted system call)

see the lines with ERESTARTSYS and the EINTR: the sistem call being interrupted is actually rt_sigreturn resumed, not futex (the system call underlying the sem_wait) as you expected. I must say I was quite puzzled but reading the man gave some interesting clues (man 7 signal):

   If  a blocked call to one of the following interfaces is interrupted by
   a signal handler, then the call will be automatically  restarted  after
   the  signal  handler returns if the SA_RESTART flag was used; otherwise
   the call will fail with the error EINTR:
[...]

       * futex(2)  FUTEX_WAIT  (since  Linux  2.6.22;  beforehand,  always
         failed with EINTR).

So I guess you have a kernel that has a similar behaviour (see netBSD doc?) and you can observe that the system call automatically restart without any chance for you to see it.

That said, I completely removed the sem_post() from your program and just sent signal to 'break' the sem_wait() ans looking at strace I saw (filtering on the bee thread):

[pid  8309] futex(0x7fffc0470990, FUTEX_WAIT_PRIVATE, 0, NULL <unfinished ...>
[pid  8309] <... futex resumed> )       = ? ERESTARTSYS (To be restarted if SA_RESTART is set)
[pid  8309] --- SIGUSR1 {si_signo=SIGUSR1, si_code=SI_TKILL, si_pid=8308, si_uid=500} ---
[pid  8309] rt_sigreturn()              = -1 EINTR (Interrupted system call)
[pid  8309] madvise(0x7fd5f6019000, 8368128, MADV_DONTNEED) = 0
[pid  8309] _exit(0)

I must say I don't master the details, but the kernel seems to find out where I'm trying to stand and make the whole thing have the correct behaviour:

Bee: got RC = -1 and errno = Interrupted system call

回答2:

Thanks for your answer OznOg, if I remove the last sem_post and make the last sleep a little longer, I get this with ktrace:

PSIG  SIGUSR1 caught handler=0x40035c mask=(): code=SI_LWP sent by pid=10631, uid=0)
CALL  write(1,0x7f7ff7e04000,0x24)
GIO   fd 1 wrote 36 bytes "Signal: I am pthread 0x7f7ff7800000\n"
RET   write 36/0x24
CALL  setcontext(0x7f7ff7bff970)
RET   setcontext JUSTRETURN
CALL  ___lwp_park50(0,0,0x7f7ff7e01100,0x7f7ff7e01100)
RET   __nanosleep50 0
CALL  exit(0)
RET   ___lwp_park50 -1 errno 4 Interrupted system call

Seems like sem_wait will only return by either an exit or a sem_post....

来源：https://stackoverflow.com/questions/33853901/sem-wait-not-unblocking-with-eintr

标签

Linux

semaphore

NetBSD