[LU-2184] avoid breaking access to IGIF files when testing fid_is_norm Created: 15/Oct/12 Updated: 18/Jun/14 Resolved: 09/May/14 |
|
| Status: | Resolved |
| Project: | Lustre |
| Component/s: | None |
| Affects Version/s: | Lustre 2.4.0 |
| Fix Version/s: | Lustre 2.4.0 |
| Type: | Bug | Priority: | Minor |
| Reporter: | Richard Henwood (Inactive) | Assignee: | Richard Henwood (Inactive) |
| Resolution: | Fixed | Votes: | 0 |
| Labels: | None | ||
| Issue Links: |
|
||||||||
| Severity: | 3 | ||||||||
| Rank (Obsolete): | 5225 | ||||||||
| Description |
|
Andreas' comment on http://review.whamcloud.com/#patch,sidebyside,4255,3,lustre/mdt/mdt_handler.c suggests that fid_is_igif should be called when fid_is_norm is called to avoid breaking access to IGIF files. |
| Comments |
| Comment by Richard Henwood (Inactive) [ 09/May/14 ] |
|
Some time in the past, the call in question was replaced by a call to fid_is_namespace_visible which addresses the concerns in this ticket. fid_is_namespace_visible is defined in lustre_fid.h as: static inline int fid_is_namespace_visible(const struct lu_fid *fid) { const __u64 seq = fid_seq(fid); /* Here, we cannot distinguish whether the normal FID is for OST * object or not. It is caller's duty to check more if needed. */ return (!fid_is_last_id(fid) && (fid_seq_is_norm(seq) || fid_seq_is_igif(seq))) || fid_is_root(fid) || fid_seq_is_dot_lustre(seq); } |
| Comment by Andreas Dilger [ 18/Jun/14 ] |
|
Fix was in patch http://review.whamcloud.com/5764 for 2.4.0 |