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1009

answers:

3
#include  <stdio.h>
#include  <signal.h>


void  ALARMhandler(int sig)
{
  signal(SIGALRM, SIG_IGN);          /* ignore this signal       */
  printf("Hello");
  signal(SIGALRM, ALARMhandler);     /* reinstall the handler    */
}

int main(int argc, char *argv[])
{
  alarm(2);                     /* set alarm clock          */
  while (1)
    ;
  printf("All done");
}

I expect the program to print "hello" after 2 seconds, but instead the output is "zsh: alarm ./a.out"

Any idea what is going on?

+2  A: 

You are not installing the signal handler first.
You need to tell the system that you want to handle the signal before actually receiving it, so you need to call signal() from main before the signal comes.

int main(int argc, char *argv[])
{
  signal(SIGALRM, ALARMhandler);     /* install the handler    */
  alarm(2);                     /* set alarm clock          */
  while (1);
}
Arkaitz Jimenez
+3  A: 

You're forgetting to set the alarm handler initially. Change the start of main() like:

int main(int argc, char *argv[])
{
   signal(SIGALRM, ALARMhandler);
   ...

Also, the signal handler will probably print nothing. That's because the C library caches output until it sees an end of line. So:

void  ALARMhandler(int sig)
{
  signal(SIGALRM, SIG_IGN);          /* ignore this signal       */
  printf("Hello\n");
  signal(SIGALRM, ALARMhandler);     /* reinstall the handler    */
}

For a real-world program, printing from a signal handler is not very safe. A signal handler should do as little as it can, preferably only setting a flag here or there. And the flag should be declared volatile.

Andomar
Real-world example: I once worked on a system that used an Access database as a back end, and under certain circumstances a `printf()` call in a signal handler would write to the .mdb file instead of stdout, hosing the database beyond repair.
John Bode
+3  A: 

You're not setting the handler in your main function.

Before you do alarm(2), put the signal(SIGALRM, ALARMhandler); in your main.

It should work then.

Note that your "All Done" will never be printed, because you'll stay in the while(1) loop after the signal processor has run. If you want the loop to be broken, you'll need to have a flag that the signal handler changes.

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

/* number of times the handle will run: */
volatile int breakflag = 3;

void handle(int sig) {
    printf("Hello\n");
    --breakflag;
    alarm(1);
}

int main() {
    signal(SIGALRM, handle);
    alarm(1);
    while(breakflag) { sleep(1) }
    printf("done\n");
    return 0;
}
McPherrinM
+1 For the nice program. I think you'd have to call `alarm(2)` in the signal handler: normally one `alarm()` call only generates a single `SIGALRM`
Andomar
Of course, you're right. I obviously didn't test this code, and have't used signals in this manner in a while (Notably, they're useless in a threaded environment to use this way).
McPherrinM