Runaway SIGALRM signal handler


 
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# 1  
Old 04-09-2008
Runaway SIGALRM signal handler

I have written a program to demonstrate a problem I have encountered when using BSD style asynchronous input using the O_ASYNC flag in conjunction with a real time interval timer sending regular SIGALRM signals to the program. The SIGIO handler obeys all safe practices, using only an atomic update to a volatile sig_atomic_t variable.

The program is below. The idea is that when the user enters a 'q', the program is supposed to stop. It does not. If you uncomment the printf statement in teh main program's loop you will see that the loop is exited, but the return statement is never reached because the SIGALRMS seem to keep the program alive. Be prepared to kill it with a Ctrl-C.

If I use the AIO style asynchronous I/O there is no problem. It is related to the O_ASYNC flag. (In fact if you simply call fcntl() and tell the kernel to send signals to the process, without setting the flag, the same problem will occur! )

Can anyone explain what is wrong?

Code:
#include    <stdio.h>
#include    <signal.h>
#include    <sys/time.h>
#include    <fcntl.h>


void    on_alarm(int);    /* handler for alarm    */
void    on_input(int);    /* handler for keybd    */
int     set_timer( int which, long initial, long repeat );

volatile sig_atomic_t   finished  = 0;

int main( int argc, char * argv[])
{
    int  fd_flags;
    int k = 0;
    struct sigaction newhandler;    
    sigset_t         blocked;       

    newhandler.sa_handler = on_input;  /* handler function    */
    newhandler.sa_flags   = 0;   

    sigemptyset(&blocked);             
    newhandler.sa_mask = blocked;      
    if ( sigaction(SIGIO, &newhandler, NULL) == -1 )
        perror("sigaction");

    newhandler.sa_handler = on_alarm;  /* handler function    */
    if ( sigaction(SIGALRM, &newhandler, NULL) == -1 )
        perror("sigaction");


    fcntl(0, F_SETOWN, getpid());
    fd_flags = fcntl(0, F_GETFL);
    fcntl(0, F_SETFL, (fd_flags|O_ASYNC));

    set_timer(ITIMER_REAL, 500, 500); 


    while( !finished  )    {         
       pause();
       printf("still in loop and finished = %d\n", finished);
    }
    return 0;
}
void on_input(int signum)
{
    int  c = getchar();          
    if ( c == 'q' )
        finished = 1;
}

void on_alarm(int signum)
{
    static int k = 0;
    printf("call number %d\n", k++);
}


int set_timer( int which, long initial, long repeat )
{
    struct itimerval itimer;
    long secs;

    // initialize initial delay
    secs = initial / 1000 ; 
    itimer.it_value.tv_sec     = secs;    
    itimer.it_value.tv_usec    = (initial - secs*1000 ) * 1000 ;   

    // initialize repeat inveral
    secs = repeat / 1000 ; 
    itimer.it_interval.tv_sec  = secs;     
    itimer.it_interval.tv_usec = (repeat - secs*1000 ) * 1000 ;
    
    return setitimer(which, &itimer, NULL);
}


Last edited by Perderabo; 04-10-2008 at 11:30 PM.. Reason: Adding code tags... it's just 13 little keystrokes to readability.
# 2  
Old 04-09-2008
You are using 500 when you do integer division of 500 /1000 you get zero. That's what I see right off.
# 3  
Old 04-09-2008
Not that simple

The lines:

secs = initial / 1000 ;
itimer.it_value.tv_sec = secs;
itimer.it_value.tv_usec = (initial - secs*1000 ) * 1000 ;

set the seconds to 0 and the microseconds to 500,000. That is intentional integer division. It si the standard way to convert milliseconds to (secs,microsecs).

Stewart
# 4  
Old 04-10-2008
Works for me on both linux 2.6.x and FreeBSD 6.0-RELEASE.
# 5  
Old 04-10-2008
When you run it, the 'q' causes the program to quit? I have run this program on Linux 2.6..x (from 9 to 24) and on Solaris 9, and it does not terminate on 'q'. And I have run it on at least 6 different machines. You used the exact code?

Stewartw
# 6  
Old 04-10-2008
Yep. I did the intuitive interactive thing. I fear your problem is pretty standard for stdio buffered programs looking for a newline.
Code:
@skbsd::~/src>./progname
call number 0
still in loop and finished = 0
call number 1
still in loop and finished = 0
call number 2
still in loop and finished = 0
call number 3
still in loop and finished = 0
call number 4
still in loop and finished = 0
call number 5
still in loop and finished = 0
qcall number 6
qcall number 7
qcall number 8
qcall number 9
qcall number 10
<\r here>
still in loop and finished = 1

# 7  
Old 04-11-2008
You are doing something extremely non-standard and you aren't bothering to test the return codes. Solaris does not support O_ASYNC. Here is the Solaris 9 fcntl man page. I don't see any O_ASYNC flag mentioned. You actually got this to compile on Solaris??? I got to try that tomorrow.... I would expect it to run on FreeBSD. I'm not sure about Linux.... I'm don't have time right now to look up the man pages. Anyway this BSD async stuff is pretty much BSD only and not part of Posix.

Please test those return codes. If not all the time, at least break down and test them when you don't understand the behavior of the system calls.
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