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

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();
});

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.