CODEWITHSUNDEEP
linuxsignals
Bob5421
Bob5421

Reputation: 9173

SIGFPE handler loop call

Look at this Unix C program:

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

void handler(int signum)
{
    printf("Handler signum=%d\n",signum);
}

int main(int argc, char *argv)
{
    printf("Start\n");
    signal(SIGFPE, handler);
    int i=10/0;
    printf("Next\n");
    return 0;
}

As you can see, i am connecting SIGFPE to an handler. Then, i make a DIV0 erreur. The handler is fired, that is great. But, this handler is called in loop ! Why ?

Thanks

Upvotes: 2

Views: 216

Answers (1)

Jim Janney
Jim Janney

Reputation: 511

If you simply return from your handler, execution resumes at the point where the signal was thrown, which results in another divide by zero error, which results in the handler being called again, and so on. You need to arrange for execution to continue at some other point in the code. The traditional approach is to use setjmp/longjmp, something like this

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

jmp_buf buf;

void handler(int signum)
{
  longjmp(buf, signum);
}

int main(int argc, char *argv)
{
  int rc = setjmp(buf);
  if (rc == 0) {
    printf("Start\n");
    signal(SIGFPE, handler);
    int i=10/0;
  }

  printf("Handler signum=%d\n", rc);
  printf("Next\n");
  return 0;
}

Note: this approach is very old school, and probably someone can suggest a better way to handle it. Also, you are probably better off calling sigaction rather than signal, as the semantics of signal are not consistent across different versions of Unix.

Upvotes: 3

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