不能捕获多个中断

Question

I want my program to run in a loop until it receives an alarm signal, in the meantime I want to run some code every time it receives an interrupt signal.

The following almost works:

bool volatile waiting = true;
bool volatile interrupted = false;

void catch_interrupt(int x)
{
    interrupted = true;
}

void catch_alarm(int x)
{
    waiting = false;
}

void alive()
{
    signal(SIGINT, catch_interrupt);
    signal(SIGALRM, catch_alarm);
    alarm(10);

    while (waiting)
    {
        if (interrupted)
        {
            printf("interrupted\n");
            interrupted = false;
        }
    }

    printf("done\n");
}

The problem is that it only works for the first interrupt signal. The second interrupt kills the program (without printing "done") regardless.

So the output I see is

^Cinterrupted
^C

when I want to see

^Cinterrupted
^Cinterrupted
done

Answer 1

Do not use signal() to install custom signal handlers, as its behavior in that case varies across implementations. In particular, on some systems, if signal() is used to set a custom signal handler for a given signal, then that signal's disposition is reset when the signal is received. That's what you seem to be observing, but you cannot portably rely on it.

Instead, install signal handlers via sigaction() . Among other things, it has a mechanism for specifying whether the handler should be reset upon receipt of the signal.

Answer 2

Signal handlers set with signal() are disarmed when the signal handler is called in response to a signal. You have to rearm the signals each time:

void catch_interrupt(int x)
{
    interrupted = true; 
    signal(SIGINT, catch_interrupt);

}

void catch_alarm(int x)
{
    waiting = false;
    signal(SIGALRM, catch_alarm);
}

Yes, that means there is a small window of vulnerability when the first signal has been handled but the handler has not yet been reinstated when a second signal will cause the program to react as if no signal handler is installed (because no signal handler is installed).