ext2/ext3:useless code for return value

When rsv is the right hand side of goal, we should return NULL, because now rsv's prev is NULL, or we return rsv.

Signed-off-by Peter Pan(潘卫平) <[email protected]>

---
 fs/ext2/balloc.c |    6 +-----
 fs/ext3/balloc.c |    6 +-----
 2 files changed, 2 insertions(+), 10 deletions(-)

fs/ext2/balloc.c +0 -0

index 3cf038c..023990f 100644

Quoted 15 lines

When rsv is the right hand side of goal, we should return NULL, because now rsv's prev is NULL, or we return rsv.

Signed-off-by Peter Pan(潘卫平)<[email protected]>

---
  fs/ext2/balloc.c |    6 +-----
  fs/ext3/balloc.c |    6 +-----
  2 files changed, 2 insertions(+), 10 deletions(-)

fs/ext2/balloc.c +0 -0

index 3cf038c..023990f 100644

fs/ext2/balloc.c search_reserve_window(struct rb_root *root, · line 327

Quoted 10 lines

fs/ext2/balloc.c +1 -5

@@ -323,11 +323,7 @@ search_reserve_window(struct rb_root *root,
 ext2_fsblk_t goal)
  	 * side of the interval containing the goal.  If it's the RHS,
  	 * we need to back up one.
  	 */
-	if (rsv->rsv_start>  goal) {
-		n = rb_prev(&rsv->rsv_node);
-		rsv = rb_entry(n, struct ext2_reserve_window_node, rsv_node);
-	}
-	return rsv;
+    return (rsv->rsv_start<  goal) ? rsv : NULL;

Hmm, I'm not sure I understand your reasoning. Suppose we have an RB-tree
with two intervals 0-10, 20-30. Interval 0-10 is in the root. Now we search for goal 15. In the root we go to right because 10<15, in the next node we go to left because 15< 20. Then the loop terminates. Now your code would return NULL but previous code would return rb_prev of interval 20-30 which is 0-10. And that is what we want as far as I understand what we expect from the function...

Honza