Is destructor called if SIGINT or SIGSTP issued?

Question

I have a class with a user-defined destructor. If the class was instantiated initially, and then SIGINT is issued (using CTRL+C in unix) while the program is running, will the destructor be called? What is the behaviour for SIGSTP (CTRL + Z in unix)?

Answer 1

No, by default, most signals cause an immediate, abnormal exit of your program.

However, you can easily change the default behavior for most signals.

This code shows how to make a signal exit your program normally, including calling all the usual destructors:

#include <iostream>
#include <signal.h>
#include <unistd.h>
#include <cstring>
#include <atomic>

std::atomic<bool> quit(false);    // signal flag

void got_signal(int)
{
    quit.store(true);
}

class Foo
{
public:
    ~Foo() { std::cout << "destructor\n"; }
};

int main(void)
{
    struct sigaction sa;
    memset( &sa, 0, sizeof(sa) );
    sa.sa_handler = got_signal;
    sigfillset(&sa.sa_mask);
    sigaction(SIGINT,&sa,NULL);

    Foo foo;    // needs destruction before exit
    while (true)
    {
        // do real work here...
        sleep(1);
        if( quit.load() ) break;    // exit normally after SIGINT
    }
    return 0;
}

If you run this program and press control-C, you should see the word "destructor" printed. Be aware that your signal handler functions (got_signal) should rarely do any work, other than setting a flag and returning quietly, unless you really know what you are doing.

Most signals are catchable as shown above, but not SIGKILL, you have no control over it because SIGKILL is a last-ditch method for killing a runaway process, and not SIGSTOP which allows a user to freeze a process cold. Note that you can catch SIGTSTP (control-Z) if desired, but you don't need to if your only interest in signals is destructor behavior, because eventually after a control-Z the process will be woken up, will continue running, and will exit normally with all the destructors in effect.