Re: [PATCH v3 4/4] fast-export: make sure refs are updated properly

[Jonathan Nieder <jrnieder@gmail.com>]

Felipe Contreras wrote:

> --- a/builtin/fast-export.c
> +++ b/builtin/fast-export.c
> @@ -523,11 +523,16 @@ static void get_tags_and_duplicates(struct object_array *pending,
>  				typename(e->item->type));
>  			continue;
>  		}
> -		if (commit->util) {
> -			/* more than one name for the same object */
> +
> +		/*
> +		 * This ref will not be updated through a commit, lets make
> +		 * sure it gets properly upddated eventually.
> +		 */
> +		if (commit->util || commit->object.flags & SHOWN) {
>  			if (!(commit->object.flags & UNINTERESTING))
>  				string_list_append(extra_refs, full_name)->util = commit;
> -		} else
> +		}
> +		if (!commit->util)
>  			commit->util = full_name;

Here's an explanation of why the above makes sense to me.

get_tags_and_duplicates() gets called after the marks import and
before the revision walk.  It walks through the revs from the
commandline and for each one:

 - peels it to a refname, and then to a commit
 - stores the refname so fast-export knows what arg to pass to
   the "commit" command during the revision walk
 - if it already had a refname stored, instead adds the
   (refname, commit) pair to the extra_refs list, so fast-export
   knows to add a "reset" command later.

If the commit already has the SHOWN flag set because it was pointed to
by a mark, it is not going to come up in the revision walk, so it will
not be mentioned in the output stream unless it is added to
extra_refs.  That's what this patch does.

Incidentally, the change from "else" to "if (!commit->util)" is
unnecessary because if a commit is already SHOWN then it will not be
encountered in the revision walk so commit->util does not need to be
set.

If the commit does not have the SHOWN or UNINTERESTING flag set but it
is going to get the UNINTERESTING flag set during the walk because of
a negative commit listed on the command line, this patch won't help.

Jonathan