How to fix error when running task in gulp?

Question

gulp.js

When running task build in gulp, an error occurs:
The following tasks did not complete: build, style
Did you forget to signal async completion?

gulp version:
gulp -v
CLI version: 2.2.0
Local version: 4.0.2

gulpfile.js content:

var gulp = require("gulp");
var less = require("gulp-less");
var plumber = require("gulp plumber");
var postcss = require("gulp-postcss");
var posthtml = require("gulp-posthtml");
var include = require("posthtml-include");
var autoprefixer = require("autoprefixer");
var minify = require("gulp-csso");
var imagemin = require("gulp-imagemin");
var svgstore = require("gulp-svgstore");
var rename = require("gulp-rename");
var server = require("browser-sync").create();
var run = require("run sequence");
var del = require("del");

gulp.task("style", function() {
gulp.src("source/less/style.less")
.pipe(plumber())
.pipe(less())
.pipe(postcss([
autoprefixer()
] ))

.pipe(gulp.dest("build/css"))
.pipe(minify())
.pipe(rename("style.min.css"))
.pipe(gulp.dest("build/css") )
.pipe(server.stream());
});

gulp.task("sprite", function () {
return gulp.src("source/img/icon-*.svg") //Write the names of the files to put in the sprite
.pipe(svgstore({
inlineSvg: true
}))
.pipe(rename("sprite.svg" ))
.pipe(gulp.dest("build/img"));
});

gulp.task("html", function () {
return gulp.src("source/*.html")
.pipe(posthtml([
include()
]))
.pipe(gulp.dest("build"));
});

gulp.task("images", function () {
return gulp.src("source/img/**/*.{png,jpg,svg}")
.pipe(imagemin([
imagemin.optipng({optimizationLevel: 3 }),
imagemin.jpegtran({progressive:
]))
.pipe(gulp.dest("source/img"));
});

gulp.task("webp", function () {
return gulp.src("source/img/**/*.{png,jpg}")
.pipe(webp({quality: 90}))
.pipe(gulp .dest("source/img"));
});

gulp.task("serve", function() {
server.init({
server: "build/"
});

gulp.watch("sourse/less/**/*.less", ["style"]);
gulp.watch("source/*.html", ["html"]);
});

gulp.task("copy", function () {
return gulp.src([
"source/fonts/**/*.{woff,woff2}",
"source/img/**",
"
.pipe(gulp.dest("build"));
});

gulp.task("clean", function () {
return del("build");
});

gulp.task("build", function (done) {
run(
"clean",
"copy",
"style",
"sprite",
"html",
done
);
});

Package.json content:

{
"name": "some",
"version": "1.0.0",
"description": "Trial project",
"main": "gulpfile.js",
"dependencies": {},
"devDependencies":
"gulp-csso": "^4.0.1",
"gulp-imagemin": "^7.1.0",
"gulp-less": "^4.0.1",
"gulp-plumber": "^1.2.1 ",
"gulp-postcss": "^8.0.0",
"gulp-posthtml": "^3.0.4",
"gulp-rename": "^2.0.0",
"gulp-series": "^1.0 .2",
"gulp-svgstore": "^7.0.1",
"gulp-webp": "^4.0.1",
"posthtml-include": "^1.3.3",
"run-sequence": " ^2.2.1"
},
"scripts": {
"editorconfig": "editorconfig-cli",
"stylelint": "stylelint 'source/less/**/*.less' --syntax less",
"test": "npm run editorconfig && npm run stylelint",
"build": "gulp build",
"start" : "npm run build && gulp serve"
},
"repository": {
"type": "git",
"url": "......"
},
"author": "Nina",
"license": "ISC",
"bugs": {
"url": " https://github.com/....... "
},
"homepage": " https://github.com/....... "
}

Answer 1

Add a return:

gulp.task("style", function() {
    return gulp.src("source/less/style.less")
    ...

Wrap the code in your question with a button < / >
Your error "Did you forget to signal async completion" is Googled as easily as possible.