I want a taks to run 2 subtasks in parallel, and finish when they are both finished.

I was hoping that I could use the callback argument on gulp.start like this:

gulp.task('one', function(cb) {
    console.log('one')
    setTimeout(cb, 100)
})

gulp.task('two', function(cb) {
    console.log('two')
    setTimeout(cb, 200)
})

gulp.task('default', function(cb) {
    gulp.start('one', 'two', function() {
        console.log('gulp.start callback');
        cb()
    })
})

But the gulp.start callback is never called. Gulp quits immediately after task 'two' is done, and the default task never finishes. When I change the default task to

gulp.task('default', function(cb) {
    gulp.start('one', 'two', function() {
        console.log('gulp.start callback');
    })
    cb()
})

then the default task finishes right after starting both subtasks as expected. But then the gulp.start callback is executed after both subtasks are finished.

As another test I also tried

gulp.task('default', function() {
    gulp.start('one', 'two', function() {
        gulp.start('one', 'two')
    })
})

This causes the subtasks to repeatedly be run, in other words, the inner gulp.start reuses the callback of the outer gulp.start.