C: SIGALRM - alarm to display message every second

Question

So I'm trying to call an alarm to display a message "still working.." every second. I included signal.h.

Outside of my main I have my function: (I never declare/define s for int s)

void display_message(int s);   //Function for alarm set up
void display_message(int s) {
     printf("copyit: Still working...\n" );
     alarm(1);    //for every second
     signal(SIGALRM, display_message);
}

Then, in my main

while(1)
{
    signal(SIGALRM, display_message);
    alarm(1);     //Alarm signal every second.

That's in there as soon as the loop begins. But the program never outputs the 'still working...' message. What am I doing incorrectly? Thank you, ver much appreciated.

Answer 1

Signal handlers are not supposed to contain "business logic" or make library calls such as printf. See C11 §7.1.4/4 and its footnote:

Thus, a signal handler cannot, in general, call standard library functions.

All the signal handler should do is set a flag to be acted upon by non-interrupt code, and unblock a waiting system call. This program runs correctly and does not risk crashing, even if some I/O or other functionality were added:

#include <signal.h>
#include <stdio.h>
#include <stdbool.h>
#include <unistd.h>

volatile sig_atomic_t print_flag = false;

void handle_alarm( int sig ) {
    print_flag = true;
}

int main() {
    signal( SIGALRM, handle_alarm ); // Install handler first,
    alarm( 1 ); // before scheduling it to be called.
    for (;;) {
        sleep( 5 ); // Pretend to do something. Could also be read() or select().
        if ( print_flag ) {
            printf( "Hello\n" );
            print_flag = false;
            alarm( 1 ); // Reschedule.
        }
    }
}

Answer 2

Move the calls to signal and alarm to just before your loop. Calling alarm over and over at high speed keeps resetting the alarm to be in one second from that point, so you never reach the end of that second!

For example:

#include <stdio.h>
#include <signal.h>
#include <unistd.h>

void display_message(int s) {
     printf("copyit: Still working...\n" );
     alarm(1);    //for every second
     signal(SIGALRM, display_message);
}

int main(void) {
    signal(SIGALRM, display_message);
    alarm(1);
    int n = 0;
    while (1) {
        ++n;
    }
    return 0;
}

Answer 3

Do not call alarm() twice, just call it once in main() to initiate the callback, then once in display_message(). Try this code on Linux (Debian 7.8) :

#include  <stdio.h>
#include  <signal.h>

void display_message(int s);   //Function for alarm set up

void display_message(int s)
{
     printf("copyit: Still working...\n" );
     alarm(1);    //for every second
     signal(SIGALRM, display_message);
}

int main()
{
    signal(SIGALRM, display_message);
    alarm(1);     // Initial timeout setting

     while (1) 
     {   
          pause();
     }   
}

The result will be the following one :

copyit: Still working...
copyit: Still working...
copyit: Still working...
copyit: Still working...
copyit: Still working...
copyit: Still working...
copyit: Still working...
copyit: Still working...
copyit: Still working...
copyit: Still working...
copyit: Still working...
copyit: Still working...

Answer 4

The alarm() call is for a one off signal.

To repeat an alarm, you have to call alarm() again each time the signal occurs.

Another issue, also, is that you're likely to get EINTR errors. Many system functions get interrupted when you receive a signal. This makes for much more complicated programming since many of the OS functions are affected.

In any event, the correct way to wait for the next SIGALRM is to use the pause() function. Something the others have not mentioned (instead they have tight loops, ugly!)

That being said, what you are trying to do would be much easier with a simple sleep() call as in:

// print a message every second (simplified version)
for(;;)
{
    printf("My Message\n");
    sleep(1);
}

and such a loop could appear in a separate thread. Then you don't need a Unix signal to implement the feat.

Note: The sleep() function is actually implemented using the same timer as the alarm() and it is clearly mentioned that you should not mix both functions in the same code.

sleep(3) may be implemented using SIGALRM; mixing calls to alarm() and sleep(3) is a bad idea.

(From Linux man alarm)

void alarm_handler(int)
{
    alarm(1);    // recurring alarm
}
    
int main(int argc, char *argv[])
{
    signal(SIGALRM, alarm_handler);
    alarm(1);

    for(;;)
    {
        printf("My Message\n");
        // ...do other work here if needed...
        pause();
    }

    // not reached (use Ctrl-C to exit)
    return 0;
}

You can create variations. For example, if you want the first message to happen after 1 second instead of immediately, move the pause() before the printf().

The "other work" comment supposes that your other work does not take more than 1 second.

It is possible to get the alarm signal on a specific thread if work is required in parallel, however, this can be complicated if any other timers are required (i.e. you can't easily share the alarm() timer with other functions.)

P.S. as mentioned by others, doing your printf() inside the signal handler is not a good idea at all.

There is another version where the alarm() is reset inside main() and the first message appears after one second and the loop runs for 60 seconds (1 minute):

void alarm_handler(int)
{
}
    
int main(int argc, char *argv[])
{
    signal(SIGALRM, alarm_handler);

    for(int seconds(0); seconds < 60; ++seconds)
    {
        alarm(1);
        // ...do other work here if needed...
        pause();
        printf("My Message\n");
    }

    // reached after 1 minute
    return 0;
}

Note that with this method, the time when the message will be printed is going to be skewed. The time to print your message is added to the clock before you restart the alarm... so it is always going to be a little over 1 second between each call. The other loop is better in that respect but it still is skewed. For a perfect (much better) timer, the poll() function is much better as you can specify when to wake up next. poll() can be used just and only with a timer. My Snap library uses that capability (look for the run() function, near the bottom of the file). In 2019. I moved that one .cpp file to the eventdispatcher library. The run() function is in the communicator.cpp file.

Answer 5

POSIX permits certain of its functions to be called from signal handling context, the async-signal safe functions, search for "async-sgnal safe" here. (These may be understood as "system calls" rather than library calls). Notably, this includes write(2).

So you could do

void
display_message (int s) {
     static char const working_message [] = "copyit: Still working...\n";
     write (1, working_message, sizeof working_message - sizeof "");
     alarm(1);    /* for every second */
}

By the way, precise periodic alarms are better implemented using setitimer(2), since these will not be subject to drift. Retriggering the alarm via software, as done here, will unavoidably accumulate error over time because of the time spent executing the software as well as scheduling latencies.

In POSIX sigaction(2) superceedes signal(2) for good reason: the original Unix signal handling model was simple. In particular, a signal handler was reset to its original "deposition" (e.g., terminate the process) once it was fired. You would have to re-associate SIGALRM with display_message() by calling signal() just before calling alarm() in display_message().

An even more important reason for using sigaction(2) is the SA_RESTART flag. Normally, system calls are interrupted when a signal handler is invoked. I.e., when then signal handler returns, the system call returns an error indication (often -1) and errno is set to EINTR, interrupted system call. (One reason for this is to be able to use SIGALRM to effect time outs, another is to have a higher instance, such as a user, to "unblock" the current process by sending it a signal, e.g., SIGINT by pressing control-C at the terminal).

In your case, you want signal handling to be transparent to the rest of the code, so you would set the SA_RESTART flag when invoking sigaction(2). This means the kernel should restart the interrupted system call automatically.

Answer 6

ooga is correct that you keep reloading the alarm so that it will never go off. This works. I just put a sleep in here so you don't keep stepping on yourself in the loop but you might want to substitute something more useful depending on where you are headed with this.

void display_message(int s)
{
     printf("copyit: Still working...\n" );
    // alarm(1);    //for every second
    // signal(SIGALRM, display_message);
}

int main(int argc, char *argv[])
{
    int ret;

    while(1)
    {
        signal(SIGALRM, display_message);
        alarm(1);

        if ((ret = sleep(3)) != 0)
        {
            printf("sleep was interrupted by SIGALRM\n");
        }
    }

    return (0);
}