Why does gulp (v4) throw "Did you forget to signal async completion?"?

Question

JavaScript

Hello!
gulpfile.js:

'use strict';

const gulp = require('gulp');
const sass = require('gulp-sass');
const clean = require('gulp-clean');

var arrBlocksNames = [
    'sass.common.blocks'
];

function cleanSassBlocks() {
    arrBlocksNames.forEach(function(blocksName) {
        return gulp.src(blocksName + '/**/*.css' )
            .pipe(clean({force: true}));
    });
}

gulp.task('default', gulp.series(cleanSassBlocks));

The console throws an error/warning:

[13:26:11] Starting 'default'...
[13:26:11] Starting 'cleanSassBlocks'...
[13:26:11] The following tasks did not complete: default, cleanSassBlocks
[13:26:11] Did you forget to signal async completion?

At the same time, everything is deleted as it should. What is the reason, tell me please?

Answer 1

Thanks for the answers, but I figured it out myself and solved the problem like this:

arrBlocksNames.forEach(function(blocksName) {
    gulp.task(blocksName, function() {
        return gulp.src(blocksName + '/**/*.css')
            .pipe(clean({force: true}));
    });
});

Answer 2

You need to add done to the function argument.
And at the end of the function, write done();
That's all

Answer 3

return arrBlocksNames.map(...

Answer 4

You need to start the continuation of the process:

function cleanSassBlocks(f) { f();
    arrBlocksNames.forEach(function(blocksName) {
        return gulp.src(blocksName + '/**/*.css' )
            .pipe(clean({force: true}));
    }); 
...

(as suggested in https://github.com/sindresorhus/del/issues/45#issu... )