javascript - Ignoring task dependencies in gulp.js -

i have following task

gulp.task('sass-build', ['clean-build'], function () {     // stuff }); 

which requires clean-build finished before starting. however, use task gulp.watch without clean-build dependency.

i'm aware solve helper task suggested in skipping tasks listed dependencies in gulp, require using gulp.run (which deprecated) or gulp.start (which isn't documented because it's considered bad practise).

what better way of doing this? or way use gulp.run or gulp.start?

the approach showed in linked question not bad. instead of gulp.run can use simple functions same:

var sassfunction = function() {     return gulp.src('blahbla')         .pipe(bla()) }  gulp.task('sass-build', ['clean-build'], function () {     return sassfunction() });  gulp.task('sass-dev', function () {     return sassfunction(); });   gulp.task('watch', function() {     gulp.watch('your/sass/files/**/*.scss', ['sass-dev']) }); 

your other approach not bad , incorporated gulp 4.

gulp.task('default', gulp.series('clean-build', gulp.parallel('sass-build'))); 

something out for, since changes dependency trees completely.