Correct transition of Gulp from version 3 to version 4?

Question

JavaScript

Automatically updated from version 3 to version 4 and, accordingly, tasks stopped running. Can you help rewrite the following three tasks to the new syntax for an example?

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

gulp.task('watch', function () {
    gulp.watch('src/img/*', ['compress']);
});

gulp.task('css', function () {
    return gulp.src('src/scss/*.scss')
        .pipe(plumber())
        .pipe(sourcemaps.init())
        .pipe(sass())
        .on('error', sass.logError)
        .pipe(cleanCSS())
        .pipe(autoprefixer({
            browsers: ['last 2 versions'],
            cascade: false
        }))
        .pipe(sourcemaps.write())
        .pipe(gulp.dest('dist/css'))
        .pipe(browserSync.stream());
});

There, as I understand it, the series function is now used, does the first task in my example also run through series?

Answer 1

const gulp = require('gulp');

function compress() {
    // compress task
}

function watch() {
    gulp.watch('src/img/*', compress);
}

function css() {
    return gulp
        .src('src/scss/*.scss')
        .pipe(plumber())
        .pipe(sourcemaps.init())
        .pipe(sass())
        .on('error', sass.logError)
        .pipe(cleanCSS())
        .pipe(
            autoprefixer({
                browsers: ['last 2 versions'],
                cascade: false
            })
        )
        .pipe(sourcemaps.write())
        .pipe(gulp.dest('dist/css'))
        .pipe(browserSync.stream());
}

module.exports.default = gulp.series(watch); // ну если один таск можно и просто `watch`.
module.exports.watch = watch;
module.exports.css = css;