Marking async task as failed and have gulp exit with non-zero exit code

Question

Consider this simple async gulp task (this is obviously not a real-life task but nicely illustrates a general problem of marking async tests as failed):

gulp.task('foo', function (done) {
    setTimeout(function () {
        //HERE: how to fail the gulp run and have it exit wit non-zero status?
    }, 1000);
});

I would like to know how to mark the foo task as failed so Gulp aborts the build and return non-zero exit code status to the shell. The things I've tried:

• done(1) - finishes the task but gulp exists with code 0
• done(new Error(fail)) - finishes the task but gulp exists with code 0
• throw new Error(fail) - does the trick but is seems "a bit brutal"

What would be an idiomatic way of marking such an async task as failed?

Answer 1

It turned out to be a bug in the gulp itself that was fixed in the version 3.8.4 via https://github.com/gulpjs/gulp/commit/d8d7542f58046f15d88795e3c48f952f54a02c3c

Starting from 3.8.4 one can write:

gulp.task('foo', function (done) {
    setTimeout(function () {
        done('error');
    }, 1000);
});

and have gulp properly exiting process with the return status 1.