7.8
CVE-2022-48649
- EPSS 0.02%
- Published 28.04.2024 13:15:07
- Last modified 10.01.2025 17:49:53
- Source 416baaa9-dc9f-4396-8d5f-8c081f
- CVE-Watchlists
- Open
In the Linux kernel, the following vulnerability has been resolved:
mm/slab_common: fix possible double free of kmem_cache
When doing slub_debug test, kfence's 'test_memcache_typesafe_by_rcu'
kunit test case cause a use-after-free error:
BUG: KASAN: use-after-free in kobject_del+0x14/0x30
Read of size 8 at addr ffff888007679090 by task kunit_try_catch/261
CPU: 1 PID: 261 Comm: kunit_try_catch Tainted: G B N 6.0.0-rc5-next-20220916 #17
Hardware name: QEMU Standard PC (i440FX + PIIX, 1996), BIOS 1.15.0-1 04/01/2014
Call Trace:
<TASK>
dump_stack_lvl+0x34/0x48
print_address_description.constprop.0+0x87/0x2a5
print_report+0x103/0x1ed
kasan_report+0xb7/0x140
kobject_del+0x14/0x30
kmem_cache_destroy+0x130/0x170
test_exit+0x1a/0x30
kunit_try_run_case+0xad/0xc0
kunit_generic_run_threadfn_adapter+0x26/0x50
kthread+0x17b/0x1b0
</TASK>
The cause is inside kmem_cache_destroy():
kmem_cache_destroy
acquire lock/mutex
shutdown_cache
schedule_work(kmem_cache_release) (if RCU flag set)
release lock/mutex
kmem_cache_release (if RCU flag not set)
In some certain timing, the scheduled work could be run before
the next RCU flag checking, which can then get a wrong value
and lead to double kmem_cache_release().
Fix it by caching the RCU flag inside protected area, just like 'refcnt'Data is provided by the National Vulnerability Database (NVD)
Linux ≫ Linux Kernel Version >= 5.19.8 < 5.19.12
| Type | Source | Score | Percentile |
|---|---|---|---|
| EPSS | FIRST.org | 0.02% | 0.032 |
| Source | Base Score | Exploit Score | Impact Score | Vector string |
|---|---|---|---|---|
| nvd@nist.gov | 7.8 | 1.8 | 5.9 |
CVSS:3.1/AV:L/AC:L/PR:L/UI:N/S:U/C:H/I:H/A:H
|
CWE-415 Double Free
The product calls free() twice on the same memory address, potentially leading to modification of unexpected memory locations.
CWE-416 Use After Free
The product reuses or references memory after it has been freed. At some point afterward, the memory may be allocated again and saved in another pointer, while the original pointer references a location somewhere within the new allocation. Any operations using the original pointer are no longer valid because the memory "belongs" to the code that operates on the new pointer.