How do I write a Gulp task that simply calls a function

Question

I'm striving to create a Gulp task that does nothing except calling a custom function. No, I have, no source files, and no, I have no destination files. I just want to call a custom function in a standalone task, so I can have other tasks depending on it.

For the love of me, I've checked Google and SO and I couldn't find an example. The closest I've come with is this

var through = require('through2');

gulp.task(
    'my-custom-task',
    function ()
    {
        return through.obj(
            function write(chunk, enc, callback) {
                // here is where the custom function is called
                myCustomFunction('foo', 'bar');
                callback(null, chunk);
            }
        );
    }
);

Now this does call myCustomFunction, but when I run the task with gulp my-custom-task, I can see the task starting but not finishing.

[10:55:37] Starting 'clean:tmp'...
[10:55:37] Finished 'clean:tmp' after 46 ms
[10:55:37] Starting 'my-custom-task'...

How should I write my task correctly?

Answer 1

If you just want a task that runs some function, then just do that:

gulp.task('my-custom-task', function () {
    myCustomFunction('foo', 'bar');
});

If your function does something asynchronously, it should call a callback at the end, so gulp is able to know when it’s done:

gulp.task('my-custom-task', function (callback) {
    myCustomFunction('foo', 'bar', callback);
});

---

As for why your solution does not work, using through is a way to work with streams. You can use it to create handlers which you can .pipe() into gulp streams. Since you have nothing actually using gulp streams, there is no need for you to create a stream handler and/or use through here.