tail -f, without using polling

Question

I am looking for the most performant way to to tail a file using Node.js.

There are some libaries that do this, but they seem to use polling as they ask for an interval to check for file changes.

I am wondering if I can just spawn a process that does tail -f and read from the stdout out that process. Does anyone know if the tail -f command on a linux machines uses polling or does it somehow hook into lower level observables? How does tail -f work?

In Node.js, I can do this:

const cp = require('child_process');

module.exports = function(file){

    const n = cp.spawn('tail',['-f',file]);

    n.on('error', function(err){
        console.error(err.stack || err);
    });

    return n.stdout;


};

but I am wondering if there as a more efficient way to do this

Answer 1

On Linux, the GNU coreutils version of tail uses inotify to block waiting for changed to the file. With FreeBSD, tail uses kevent to do the same. This will be better than repeatedly checking for changes and sleeping. Depending on how often you need to do this and the frequency with which changes occur, it may or may not be worth the trouble to spawn an external process from node.js just to call tail.

If you are not using Linux or FreeBSD (or possibly MacOS?), the tail utility is probably no better than what you can do directly in JavaScript by repeatedly checking and sleeping.

Alternatively, if you are using Linux or FreeBSD, you may be able to use inotify or kevent directly from node.js with some sort of extension/plugin/module without spawning a process. I don't know if that exists.