Thread: signals interrupt and quit issue

I thought that this will run until I sent a signal from my keyboard!

The code is from another person and he said that he tested on a mac.

I run it in Linux, because I think this can't be done from my windows 7.

As you can see I have commented fflush(stdout), but uncommeting it has no effect.

The code

Code:

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

void handlerC(int sig)
{
        printf("Crtl-C pressed\n");
        printf("Received signal nr %i.\n",sig);
        psignal(sig,"handler");
}

void handlerQ(int sig)
{
        printf("Received signal nr %i.\n",sig);
        psignal(sig,"handler");
        printf("bb\n");
        exit(1);
}


void waitforever()
{
        signal(SIGINT,handlerC); // Ctrl-C
        signal(SIGQUIT,handlerQ); //ctrl-\
        while(1)
        {
                putchar('.');
                //fflush(stdout);
                sleep(2);
        }
}

int main(int argc, char* argv[])
{
        waitforever();
        return 0; // will never be called
}

Output

Code:

linux05:/home/users/std10093>./px
.linux05:/home/users/std10093>

So, only one dot is being printed and then the loop exits and we stop. Why? :/

Originally Posted by std10093

Output

Code:

linux05:/home/users/std10093>./px
.linux05:/home/users/std10093>

So, only one dot is being printed and then the loop exits and we stop. Why? :/

I think you inadvertently commented out your loop:

Code:

signal(SIGQUIT,handlerQ); //ctrl-\
        while(1)
        {

backslash followed by newline is interpreted as extending the single-line comment onto the next line. At least it is by GCC. If you compile with -Wall it will give you a warning about this.

I compiled without -Wall.

I deserved what happened to me! But from our mistakes, we learn!

Thank you both !