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

Mathias Kunter wrote:

quoted

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

quoted

return (fetch_remote && fetch_remote != remote);

to

quoted

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.

I'd argue that if `simple` pushing is used, then the expected behavior 
of the command

quoted

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