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What is gulp.start?

See original GitHub issue

I see that gulp.run was deprecated, but I also see that there is a gulp.start command that seems to do the same thing as gulp.run, but just is more aptly named and doesn’t come with the deprecation warning.

I couldn’t find anything in the docs about gulp.start. Am I correct in thinking that gulp.start is a replacement for gulp.run (because of the better name), or is it some unreliable internal API that some people have stumbled upon and it’s gonna disappear someday and people shouldn’t rely on it?

If it’s just missing from the docs, I’d be willing to do a PR as far as I understand how it works.

Issue Analytics

  • State:closed
  • Created 9 years ago
  • Reactions:11
  • Comments:55 (24 by maintainers)

github_iconTop GitHub Comments

4reactions
yocontracommented, Apr 23, 2014

gulp.run had a thing where if you passed it no arguments it would run the default task. This was a stupid wrapper around gulp.start, so gulp.run was deprecated in favor of

  1. Using task dependencies
  2. Using gulp.start if you absolutely need to

Both of these will be outmoded by the new task system in gulp 4 however, which uses composable functions instead of complex task dependency trees.

2reactions
yocontracommented, Jul 4, 2014

@avanderhoorn Using gulp.watch: gulp.watch('./build/**/*', ['pack'])

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