Re: git push doesn't use local branch name as default

From: Felipe Contreras <felipe.contreras@gmail.com>
Date: May 28, 2021, 17:52 UTC
Message-ID: <60b12dc1f19b1_5190820879@natae.notmuch>
In-Reply-To: <f9f38d42-2b93-0d3b-798b-4c6f44eb111d@gmail.com>

Mathias Kunter wrote:

Show 10 quoted lines

> > you only get the "current" behavior when fetch_remote is non-NULL.
> 
> Well, then my suggestion actually is to also use the `current` behavior 
> when fetch_remote is NULL - i.e. change
> 
> > return (fetch_remote && fetch_remote != remote);
> 
> to
> 
> > return (!fetch_remote || fetch_remote != remote);

It's not quite that easy. You need to see the context of that code:

  static int is_workflow_triangular(struct remote *remote)
  {
    struct remote *fetch_remote = remote_get(NULL);
    return (fetch_remote && fetch_remote != remote);
  }

That would affect many pathways.

Show 7 quoted lines

> I'd argue that if `simple` pushing is used, then the expected behavior 
> of the command
> 
> > git push <remote_name>
> 
> on a branch without upstream would actually be to use the `current` 
> behavior instead of bailing out with an error.

I agree, but this mis-mash of modes makes the logic very hard to see.

I'll send patches to cleanup the logic, it makes no sense to have a frankenstein of two modes and that is in fact the default mode. The logic of the default mode should be crystal clear.

Cheers.

-- 
Felipe Contreras