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

From: Mathias Kunter <mathiaskunter@gmail.com>
Date: May 28, 2021, 07:44 UTC
Message-ID: <f9f38d42-2b93-0d3b-798b-4c6f44eb111d@gmail.com>
In-Reply-To: <CABPp-BEeqqGwN3=MbWCYnsyryBm3WoaX5GZyXTWy18UiYUT4zg@mail.gmail.com>

> 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);

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.

Am 28.05.21 um 09:00 schrieb Elijah Newren:

Show 29 quoted lines

> On Thu, May 27, 2021 at 11:39 PM Mathias Kunter <mathiaskunter@gmail.com> wrote:
>>
>> Felipe,
>>
>> thanks for your reply.
>>
>>> Sounds like you want to change the default to `push.default=current`.
>>
>> Yes, but shouldn't `simple` pushing also work? The documentation says
>> about `push.default=simple`:
>>
>>> When pushing to a remote that is different from the remote you normally
>>> pull from, work as `current`.
> 
> Perhaps this wording should be clarified to read
> 
> When you have a remote that you normally pull from but you are pushing
> to a different remote then that one, then work as 'current'.
> 
>> If there is no upstream, then there also is no "remote I normally pull
>> from", and thus, according to the doc, `simple` should actually work
>> like `current` in this case. Am I wrong here?
> 
> The relevant code is
> 
>      return (fetch_remote && fetch_remote != remote);
> 
> so you only get the "current" behavior when fetch_remote is non-NULL.
>