Aksal/nuttx
3
1
Fork 1
mirror of https://github.com/apache/nuttx.git synced 2026-08-01 20:28:58 +00:00
Code Issues Projects Releases Packages Wiki Activity Actions 2

libs/libc/obstack/lib_obstack_free.c: fix object check within chunk

Browse source 
The original condition incorrectly used &h->chunk instead of
h->chunk in the calculation whether the object is in the chunk. This
could lead to the wrong behavior as the first branch gave incorrect
result and thus sometimes the entire obstack was freed even though
object was not NULL.

The commit also simplifies the logic, we can use pointer arithmetic
here and just do h->chunk + 1 as it gives the same result as
(FAR char *)h->chunk + sizeof(struct _obstack_chunk). This saves
unnecessary cast and sizeof.

The second branch should be less than or equal, not just less than.
This ensures the object is correctly located in the chunk even after
previous obstack_finish was called.

Signed-off-by: Michal Lenc <michallenc@seznam.cz>
Co-authored-by: Karel Kočí <kkoci@elektroline.cz>
This commit is contained in:
Michal Lenc 2025-12-12 16:29:15 +01:00 • committed by Xiang Xiao
parent 73232509e8
commit 1e1185c63e
1 changed files with 7 additions and 3 deletions
Download patch file Download diff file Expand all files Collapse all files

10
libs/libc/obstack/lib_obstack_free.c
View file

@ -53,9 +53,13 @@ void obstack_free(FAR struct obstack *h, FAR void *object)
while (h->chunk)
{
if (object >=
(FAR void *)((FAR char *)&h->chunk + sizeof(struct _obstack_chunk))
&& object < (FAR void *)h->chunk->limit)
/* Object has to be after chunk + chunk' header. We can use pointer
* arithmetic here as h->chunk + 1 is the same as
* (FAR char *)h->chunk + sizeof(struct _obstack_chunk)
*/
if (object >= (FAR void *)(h->chunk + 1)
&& object <= (FAR void *)h->chunk->limit)
{
/* The object is in this chunk so just move object base.
* Note: this keeps the last chunk allocated. This is desirable