7.5
CVE-2025-12105
- EPSS 0.06%
- Veröffentlicht 23.10.2025 09:14:14
- Zuletzt bearbeitet 17.12.2025 15:15:48
- Quelle secalert@redhat.com
- CVE-Watchlists
- Unerledigt
A flaw was found in the asynchronous message queue handling of the libsoup library, widely used by GNOME and WebKit-based applications to manage HTTP/2 communications. When network operations are aborted at specific timing intervals, an internal message queue item may be freed twice due to missing state synchronization. This leads to a use-after-free memory access, potentially crashing the affected application. Attackers could exploit this behavior remotely by triggering specific HTTP/2 read and cancel sequences, resulting in a denial-of-service condition.
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HerstellerGNOME
≫
Produkt
libsoup
Default Statusunaffected
Version <=
3.6.5
Version
0
Status
affected
HerstellerRed Hat
≫
Produkt
Red Hat Enterprise Linux 10
Default Statusaffected
Version <
*
Version
0:3.6.5-3.el10_1.7
Status
unaffected
HerstellerRed Hat
≫
Produkt
Red Hat Enterprise Linux 10.0 Extended Update Support
Default Statusaffected
Version <
*
Version
0:3.6.5-3.el10_0.10
Status
unaffected
HerstellerRed Hat
≫
Produkt
Red Hat Enterprise Linux 6
Default Statusunknown
HerstellerRed Hat
≫
Produkt
Red Hat Enterprise Linux 7
Default Statusunaffected
HerstellerRed Hat
≫
Produkt
Red Hat Enterprise Linux 8
Default Statusunaffected
HerstellerRed Hat
≫
Produkt
Red Hat Enterprise Linux 9
Default Statusunaffected
| Typ | Quelle | Score | Percentile |
|---|---|---|---|
| EPSS | FIRST.org | 0.06% | 0.195 |
| Quelle | Base Score | Exploit Score | Impact Score | Vector String |
|---|---|---|---|---|
| secalert@redhat.com | 7.5 | 3.9 | 3.6 |
CVSS:3.1/AV:N/AC:L/PR:N/UI:N/S:U/C:N/I:N/A:H
|
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 Any operations using the original pointer are no longer valid because the memory "belongs" to the code that operates on the new pointer.