gulp.run is deprecated. How do I compose tasks?

Here is a composed task I don't know how to replace it with task dependencies.

...
gulp.task('watch', function () {
 var server = function(){
  gulp.run('jasmine');
  gulp.run('embed');
 };
 var client = function(){
  gulp.run('scripts');
  gulp.run('styles');
  gulp.run('copy');
  gulp.run('lint');
 };
 gulp.watch('app/*.js', server);
 gulp.watch('spec/nodejs/*.js', server);
 gulp.watch('app/backend/*.js', server);
 gulp.watch('src/admin/*.js', client);
 gulp.watch('src/admin/*.css', client);
 gulp.watch('src/geojson-index.json', function(){
  gulp.run('copygeojson');
 });
});

The corresponding changelog https://github.com/gulpjs/gulp/blob/master/CHANGELOG.md#35 [deprecate gulp.run]

Or you can do like this:

gulp.start('task1', 'task2');

gulp.task('watch', function () {
  var server = ['jasmine', 'embed'];
  var client = ['scripts', 'styles', 'copy', 'lint'];
  gulp.watch('app/*.js', server);
  gulp.watch('spec/nodejs/*.js', server);
  gulp.watch('app/backend/*.js', server);
  gulp.watch('src/admin/*.js', client);
  gulp.watch('src/admin/*.css', client);
  gulp.watch('src/geojson-index.json', ['copygeojson']);
});

You no longer need to pass a function (though you still can) to run tasks. You can give watch an array of task names and it will do this for you.

source: https://github.com/gulpjs/gulp/issues/755

gulp.start() was never meant to be a public api nor used. And as stated above in comments, the task management is being replaced in the next release....so gulp.start() will be breaking.

The true intention of the gulp design is to make regular Javascript functions, and only make the task for calling them.

Example:

function getJsFiles() {
    var sourcePaths = [
        './app/scripts/**/*.js',
        '!./app/scripts/**/*.spec.js',
        '!./app/scripts/app.js'
    ];

    var sources = gulp.src(sourcePaths, { read: false }).pipe(angularFilesort());

    return gulp.src('./app/index.html')
        .pipe(injector(sources, { ignorePath: 'app', addRootSlash: false }))
        .pipe(gulp.dest('./app'));
}

gulp.task('js', function () {
    jsStream = getJsFiles();
});

Forgive me for resurrecting an old question. The accepted answer does not address the issue of running tasks before setting the watches. The next answer uses gulp.start which is going away. The third answer points out that regular functions should be used but the example seems strange. I did some searching but did not find a simple example.

Here is my solution. The idea is to define regular js functions then register them as tasks. The functions can then be called directly if needed or from within a watch.

var 
  gulp     = require('gulp'),
  concat   = require('gulp-concat'),
  markdown = require('gulp-showdown')
;
var scriptFiles   = [ 'ang/app.js' ];
var markdownFiles = [ 'content/articles/*.md'];

var watchTask = function() 
{
  buildTask();

  gulp.watch(scriptFiles,  ['scripts' ]);
  gulp.watch(markdownFiles,['markdown']);
};
gulp.task('watch',watchTask);

var buildTask = function()
{
  scriptsTask();
  markdownTask();
};
gulp.task('build',buildTask);

var markdownTask = function() 
{
  gulp.src(markdownFiles)
    .pipe(markdown())
    .pipe(gulp.dest('web/articles'));
};
gulp.task('markdown',markdownTask);

var scriptsTask = function() 
{
  gulp.src(scriptFiles)
    .pipe(concat('app.js'))
    .pipe(gulp.dest('web/js'));

  gulp.src(
    [
      'bower_components/angular/angular.min.js',
      'bower_components/angular-route/angular-route.min.js'
    ])
    .pipe(concat('vendor.js'))
    .pipe(gulp.dest('web/js'));

  gulp.src(
    [
      'bower_components/angular/angular.min.js.map',
      'bower_components/angular-route/angular-route.min.js.map'
    ])
    .pipe(gulp.dest('web/js'));
};
gulp.task('scripts', scriptsTask);

I am new to gulp. Please let me know if I have overlooked something obvious.

gulp 4

gulp.parallel('taskName1', 'taskName2')()
gulp.series('taskName1', 'taskName2')()

I Like gulp4 !

gulp.run is deprecated. How do I compose tasks?

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