end child gracefully when doing Ctrl-C in NodeJS

Question

I have a architecture with one parent that spawns a child as follow :

this.process = child.spawn(this.cmd, this.args);

this.process.on('exit', (code: number, signal: string) => {
    this.exitCode = code;
    console.log(`exit code: ${code}`);
});

there is no particular option because i want to keep a link with the child. So, when i press Ctr-C to kill the parent, it catches SIGINT to (1) end the child, (2) exit properly. But SIGINT is also propagated to the child, so the parent cannot end it gracefully. So, is there a way to do so ? maybe by preventing SIGINT to be propagated to the child ? or telling the child to ignore the signal ?

i know that something is possible by adding option detached: true and stdio: 'ignore' when spawning, but i don't want to do that because if the parent dies, i end up with zombies process. Whereas keeping the links ensure that the child is killed if the parent dies unexpectedly. Finally, i want to avoid catching SIGINT in the child as i want to keep it dumb.

EDIT: the parent already have a process.on('SIGINT', () => { ... } and the child is in python.

justin.m.chase justin.m.chase · Accepted Answer · 2020-05-27T14:01:11

You can catch exit codes like this:

process.on('SIGINT', () => {
  // handle it yourself
})

You can propagate it to children like so:

process.on('SIGINT', () => {
  this.child.kill('SIGINT')
})

The child can of course also choose to handle the signal, it is wise to not assume the child process will exit simply because you sent a single signal. You may need to set a timeout and send repeat or different signals to fully kill a process. You will also want to listen to the 'exit' message of the child process to know when its actually killed before continuing.

end child gracefully when doing Ctrl-C in NodeJS

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