Re: [PATCH v3 3/4] block/badblocks: fix badblocks loss when badblocks combine

在 2023/6/21 22:09, Ashok Raj 写道:

On Thu, Jun 22, 2023 at 01:20:51AM +0800, linan666@huaweicloud.com wrote:

quoted

From: Li Nan <redacted>

badblocks will loss if we set it as below:

   # echo 1 1 > bad_blocks
   # echo 3 1 > bad_blocks
   # echo 1 5 > bad_blocks
   # cat bad_blocks
     1 3

In badblocks_set(), if there is an intersection between p[lo] and p[hi],
we will combine them. The end of new badblocks is p[hi]'s end now. but
p[lo] may cross p[hi] and new end should be the larger of p[lo] and p[hi].

Reconsider rewriting the commit log. It seems you converted code to
sentence ;-).

I will rewrite log.

Also it might help to show after the patch how the above example would be
for cat bad_blocks

after patch:

# cat bad_blocks
   1 5

I will show it in next version. Thanks for your suggestion.

quoted

Fixes: 9e0e252a048b ("badblocks: Add core badblock management code")
Signed-off-by: Li Nan <redacted>
---
  block/badblocks.c | 10 ++++------
  1 file changed, 4 insertions(+), 6 deletions(-)

diff --git a/block/badblocks.c b/block/badblocks.c
index 7e6ebe2ac12c..2c2ef8284a3f 100644
--- a/block/badblocks.c
+++ b/block/badblocks.c

@@ -267,16 +267,14 @@ int badblocks_set(struct badblocks *bb, sector_t s, int sectors,
  	if (sectors == 0 && hi < bb->count) {
  		/* we might be able to combine lo and hi */
  		/* Note: 's' is at the end of 'lo' */
-		sector_t a = BB_OFFSET(p[hi]);
-		int lolen = BB_LEN(p[lo]);
-		int hilen = BB_LEN(p[hi]);
-		int newlen = lolen + hilen - (s - a);
+		sector_t a = BB_OFFSET(p[lo]);
+		int newlen = max(s, BB_OFFSET(p[hi]) + BB_LEN(p[hi])) - a;
  
-		if (s >= a && newlen < BB_MAX_LEN) {
+		if (s >= BB_OFFSET(p[hi]) && newlen < BB_MAX_LEN) {
  			/* yes, we can combine them */
  			int ack = BB_ACK(p[lo]) && BB_ACK(p[hi]);
  
-			p[lo] = BB_MAKE(BB_OFFSET(p[lo]), newlen, ack);
+			p[lo] = BB_MAKE(a, newlen, ack);
  			memmove(p + hi, p + hi + 1,
  				(bb->count - hi - 1) * 8);
  			bb->count--;

-- 
2.39.2

.

-- 
Thanks,
Nan