Gulp passing parameters between tasks?

Question

gulp.js

Greetings Toaster Members!
I will try to explain in more detail the essence of my problem.
In general, we have two asset location directories related to two applications:
$assets_dir = ${project_dir}/apps/public/assets
$assets_dir = ${project_dir}/apps/admin/assets
then the path to the css-style source is as follows:
${assets_dir }/stylesheets/src/style.scss
I need to put the assembled style bundle in a directory below src:
${assets_dir}/stylesheets/bundle.css

// gulpfile.js
const gulp = require('gulp');
const path = require('path');

const assets_dir = './apps/*/assets';

gulp.task('sass', function(cb) {
  // get watched full_path?
});

gulp.task('watch', function () {
  gulp.watch(assets_dir + '/stylesheets/src/**/*.scss', ['sass']);
});

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

How can I get the full path of the modified file in the 'sass' task?
Found, you can intercept the 'onchange' event and extract the path to the file, something like this:

ulp.watch(assets_dir + '/stylesheets/src/**/*.scss').on('change', function(file) { console.log(file.path); });

But then I need to somehow pass the resulting file to the 'sass' task for further processing.
It's just reluctance to describe tasks for each application.

Answer 1

In general, I revised, it turns out everything is simpler:

gulp.task('sass', function () {
  gulp.watch(assets_dir + '/stylesheets/src/**/*.scss', function(e) {
    console.log(e.path);
    // Building
  });
});

gulp.task('default', ['sass']);