Question

  struct sigaction psa;

I have enabled my signal handler in the main function as shown below:

    memset (&psa, 0, sizeof (psa));
    psa.sa_handler = pSigHandler;
    sigaction (SIGALRM, &psa, NULL);
    sigaction(SIGVTALRM, &psa, NULL);
    sigaction(SIGPROF, &psa, NULL);

My signal handler is like this:

static void pSigHandler(int signo){
    printf("Pareint signum: %d", signo);// debug
    switch (signo) {
        case SIGALRM:
            printf("P SIGALRM handler");//debug
            break;
        case SIGVTALRM:
            printf("P SIGVTALRM handler");//debug
            break;
        case SIGPROF:
            printf("P SIGPROF handler");//debug
            break;
        default: /*Should never get this case*/
            break;
    }
    return;
}

Now my question may be obvious to some people, why didn’t I see the printed debug lines when I run this? In fact, nothing was printed. Thank you very much for helping me to understand this. I’m running it on Linux, used Eclipse to program.

Answer 1

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

static void pSigHandler(int signo){
    switch (signo) {
            case SIGTSTP:
            printf("TSTP");
            fflush(stdout);
            break;
    }
}

int main(void)
{
    struct sigaction psa;
    psa.sa_handler = pSigHandler;
    sigaction(SIGTSTP, &psa, NULL);
    for(;;) {}
    return 0;
}

Because you need to fflush(stdout)

try with C-z

I’m not even sure if it’s safe to use stdio in a signal handler though.

Update: http://bytes.com/topic/c/answers/440109-signal-handler-sigsegv

According to that link, you should not do this.