Gulp watch not triggering minify on file change - gulp

Question

I have simple starter app, I created gulpfile.js file with content below,
let gulp = require('gulp');
let cleanCSS = require('gulp-clean-css');
// Task to minify css using package cleanCSs
gulp.task('minify-css', () => {
// Folder with files to minify
return gulp.src('src/assets/styles/*.css')
//The method pipe() allow you to chain multiple tasks together
//I execute the task to minify the files
.pipe(cleanCSS())
//I define the destination of the minified files with the method dest
.pipe(gulp.dest('src/assets/dist'));
});
//We create a 'default' task that will run when we run `gulp` in the project
gulp.task('default', function() {
// We use `gulp.watch` for Gulp to expect changes in the files to run again
gulp.watch('./src/assets/styles/*.css', function(evt) {
gulp.task('minify-css');
});
});
if I run gulp minify-css it works expected, but I need it to minify on file change
But all its do log a message in cmd windows like 'Starting ...'
I don't even know what does it mean...
package.json:
..
"gulp": "^4.0.2",
"gulp-clean-css": "^4.2.0"

Answer 1

I think you need to add return when running task minify-css, so that system knows when previous task was completed.
gulp.task('default', function() {
// We use `gulp.watch` for Gulp to expect changes in the files to run again
gulp.watch('./src/assets/styles/*.css', function(evt) {
return gulp.task('minify-css');
});
});