Should nothing-to-rebase still move active bookmarks?

[Kevin Bullock]

On Mar 13, 2013, at 12:40 PM, Siddharth Agarwal wrote:

> Consider the following repo, where A and B are bookmarks:
> 
>  o--o--o(B)
>   \
>    -o (A)
> 
> At this point, doing
> 
> $ hg update A
> $ hg rebase -d B
> 
> will move A's commit onto B, or in other words A ahead of B.
> 
>  o--o--o--o (A)
>       (B)
> 
> However, when A is an ancestor of B:
> 
>  --o--o--o (B)
>   (A)
> 
> the above commands will do nothing.
> 
> Compare to git, where even if A is an ancestor of B, if A is active rebasing it onto B will move A and B to the same commit.
> 
> One of our engineers hit this while trying to do "make A depend on B". Depending on whether A is an ancestor of B or not, the engineer has to either rebase or do something like `hg bookmark -f -r B A`.
> 
> I think git's behaviour makes a lot of sense here. Is it possible to reconcile that with hg's principles?

How differently would you feel about this situation if moving the bookmark didn't require --force?

This is largely the same consideration going on in a different thread, but regarding rebase instead of merge.

pacem in terris / мир / शान्ति / ‎‫سَلاَم‬ / 平和
Kevin R. Bullock