Watching './**/*.js' causes excessive CPU usage

Question

I need to watch all folders including the current one for changes. Therefore I used

gulp.task('start', () => {
    if (node) node.kill()
    node = spawn('node', ['server.js'], {
        stdio: 'inherit'
    })
    node.on('close', function(code) {
        if (code === 8) {
            gulp.log('Error detected, waiting for changes...');
        }
    });
});

gulp.watch(['./**/*.js'], ['start']);

gulp.task('default', ['start', 'watch']);

process.on('exit', function() {
    if (node) node.kill()
});

but it causes 100% CPU usage. If I use it only on a single folder (which contains 99% of all files which need to be watched) like 'api/**/*.js` there's hardly any CPU usage.

What am I doing wrong?

Answer 1

You are watching all .js files in all directories and subdirectories of your project. Which includes node_modules. Depending on how many dependencies you have in your project and how many dependencies those dependencies have etc. you might be watching thousands of .js files.

You should limit your watch glob to only those directories where source files are located:

gulp.watch(['api/**/*.js', 'app/**/*.js', ...], ['start']);

Alternatively you can try to exclude your node_modules folder and any other folders that don't contain source files (like your build destination folder):

gulp.watch(['api/**/*.js', '!node_modules/**/*.js', ...], ['start']);