How do I update to Gulp 4?

Originally posted on Liquid Light

Gulp is a fantastic build tool for compiling your SCSS, Less and SVG sprites that we use at Liquid Light.

For a while now, the gulpJS team have been working on Version 4. This version features some nice additions to the build tool but does also feature some substantial breaking changes. Upgrading is simple, but does require a few tweaks.

Upgrading gulp to v4

Note: The $ shows the command is to be run on the command line and shouldn’t be typed

$ npm install -g gulp
$ gulp -v
CLI version 2.0.1

If you are in a gulp project, a local version will be listed below the CLI version

Install gulp 4 locally

Once globally installed, gulp v4 will then need to be installed on a per-project basis.

$ npm install gulp --save-dev

If in your package.json file gulp is listed under dependencies, then replace the --save-dev with just --save.

You should now have gulp v4 installed and ready to go. This can be verified by running the version command above

Running gulp -v once again should now give you

$ gulp -v
CLI version 2.0.1
Local version 4.0.0

Updating your gulpfile

Now you have v4 successfully installed, you’ll need to do a few updates to your gulpfile.js. The best thing to do is run gulp and follow the errors that you get.

When you get an error, you will be faced with several lines of what appears to be jargon.

The first thing you need to look for is the error message. This will be something like

Error: Error message here

Make a note of that. The second line to look for is the one that includes the path to your gulpfile with some numbers afterwards. This is what mine looks like:

at Object.<anonymous> /Sites/gulp-v4/gulpfile.js:418:6

This tells you the error was generated in your gulpfile at line 418, character 6.

I’ve tested below some errors I came across during the update and how to fix them.

AssertionError: Task function must be specified

This error for me was thrown because of:

gulp.task('default', ['del'], function() {
// default task code here

Where the del task is being run before the default task.

To resolve this, you need to specify that the del and following function are to be run in a series.

To resolve this, change the code to:

gulp.task('default', gulp.series('del', function() {
// default task code here

Note: because you are opening a parenthesis for gulp.series, don’t forget to add an extra closing one after the function.

Make sure you update the rest of your gulpfile to follow suit.

The gulp has updated this syntax for running tasks in series to add the functionality of running tasks in parallel with gulp.parallel. More can be read about it in the gulp docs.

Did you forget to signal async completion?

The following tasks did not complete: default, del
Did you forget to signal async completion?

This error occurred on an anonymous function (the one occurring after del in the example above).

Gulp v4 requires a stream, promise, event emitter, child process or observable to be returned from a function or task. This was resolved in the simplest case of passing a parameter into the function and firing it after the task is completed.


gulp.task('default', gulp.series(function(done) {
// task code here

(Note the done in-between the parenthesis when the function opens and then it firing at the end).

I specifically got this error when trying to run the del npm package.

My del package was set to call del.sync which returns an array, Gulp requires one of the stream, promise, event emitter, child process or observable to be returned, which del does by default.

return del(dirs);

Gulp watcher with change event and paths

With gulp v3, the watcher would return the path of the modified file within the function it called.

For example, the output for the below would be an object with the location of the file modified and event (e.g. changed).

gulp.task('watch', function(){'path/to/css/*.css').on('change', function(event) {
// code to execute

With gulp 4, this doesn't seem to be the case. Instead, the watcher can fire some standard functions, but if you need filename based operations, these need to be moved to the changed function.

gulp.task('watch', function(){'path/to/css/*.css')
.on('change', function(path, stats) {
// code to execute on change
.on('unlink', , function(path, stats) {
// code to execute on delete

If you did need to fire a generic function when a file is changed while being watched, the gulp tutorials have an example of this.

Just a small reminder, if you are upgrading to Gulp 4 and have a shared project, the other developers will need to upgrade also.

If you have encountered any other errors, then feel free to drop a comment below or tweet us - we'd be more than happy to help and include the solution above for others!

Update 13/08/2018: Two years later and Gulp 4 still hasn't been released!

View this post on Gitlab

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Mike Street

Written by Mike Street

Mike is a front-end developer from Brighton, UK. He spends his time writing, cycling and coding. You can find Mike on Twitter.