Re: [PATCH v5 4/9] fetch: print left-hand side when fetching HEAD:foo

> On Wed, May 10, 2023 at 02:34:15PM +0200, Patrick Steinhardt wrote:
> 
> > @@ -946,7 +948,7 @@ static int update_local_ref(struct ref *ref,
> >  	if (oideq(&ref->old_oid, &ref->new_oid)) {
> >  		if (verbosity > 0)
> >  			display_ref_update(display_state, '=', _("[up to date]"), NULL,
> > -					   remote, ref->name, summary_width);
> > +					   remote_ref->name, ref->name, summary_width);
> >  		return 0;
> >  	}
> 
> Here (and in other hunks) we now dereference remote_ref unconditionally.
> But in existing parts of the code, we guard against remote_ref being
> NULL. E.g., later on:
> 
>           if (!current || !updated) {
>                   const char *msg;
>                   const char *what;
>                   int r;
>                   /*
>                    * Nicely describe the new ref we're fetching.
>                    * Base this on the remote's ref name, as it's
>                    * more likely to follow a standard layout.
>                    */
>                   const char *name = remote_ref ? remote_ref->name : "";
> 		  [...]
> 
> I'm not sure if the old code was being overly defensive, or if the new
> code is ripe for a segfault. But it's probably worth looking into (it
> was noticed by Coverity).
> 
> Looking at the caller, it is always store_update_refs() which passes its
> own "rm", a pointer iterating over ref_map. And it should always be
> non-NULL, since that's the loop condition.
> 
> So I think your code is fine, but you might want to double-check my
> logic. (And it may be worth cleaning up the existing redundant check to
> prevent confusion).