gulp error. Compiling sass to css. Where is the mistake?

Question

css

const gulp = require('gulp'),
    sass = require('gulp-sass'),
    cssmin = require('gulp-cssmin'),
    rename = require('gulp-rename'),
    plumber = require('gulp-plumber'),
    watch = require('gulp-watch');

gulp.task('sass', function () {
//указываем за какой папкой следить
  return gulp.src('./sass/**/*.sass')
    .pipe(plumber())
        //указываем куда сохранять CSS
    .pipe(gulp.dest('./css'));
});

gulp.task('cssmin', function () {
//указываем где лежит CSS
  gulp.src('./css/*.css')
    .pipe(plumber())
    .pipe(cssmin())
    .pipe(rename({suffix: '.min'}))
    .pipe(gulp.dest('dist'));
});

gulp.task('watch', function(){
  gulp.watch('./sass/**/*.sass', ['sass']);
  gulp.watch('./css/**/*.css', ['cssmin']);
});

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

Answer 1

I think the problem is in the watch function. depending on what version of gallp you have.
my watch function is written like this

gulp.task('watch', function() {
  gulp.watch('src/sass/*.scss', gulp.series('sass'));
});

Answer 2

Most likely it's a version issue, here, it seems, this problem was sorted out:
Gulp does not work, gives an error?
Try like this:

const gulp = require('gulp'),
    sass = require('gulp-sass'),
    cssmin = require('gulp-cssmin'),
    rename = require('gulp-rename'),
    plumber = require('gulp-plumber'),
    watch = require('gulp-watch');

gulp.task('sass', function(done) {
    gulp.src("./sass/**/*.sass")
        .pipe(plumber())
        .pipe(sass())
        .pipe(gulp.dest("./css"))
        .pipe(browserSync.stream());


    done();
});

gulp.task('cssmin', function () {
//указываем где лежит CSS
  gulp.src('./css/*.css')
    .pipe(plumber())
    .pipe(cssmin())
    .pipe(rename({suffix: '.min'}))
    .pipe(gulp.dest('dist'));
});

gulp.task('watch', function(){
  gulp.watch('./sass/**/*.sass', gulp.series('sass'));
  gulp.watch('./css/**/*.css', gulp.series('cssmin'));
});

gulp.task('default', gulp.series('sass', 'cssmin', 'watch'));