CVEvulnerability

CVE-2026-31576

7.8 HIGH Published 2026-04-24 Modified 2026-06-17 Official page ↗
Status
Modified
Published
2026-04-24
Last modified
2026-06-17
Assigner
416baaa9-dc9f-4396-8d5f-8c081fb06d67
CVSS 3.1 ([email protected])
7.8 HIGH — CVSS:3.1/AV:L/AC:L/PR:L/UI:N/S:U/C:H/I:H/A:H
Weaknesses
CWE-416

Description

In the Linux kernel, the following vulnerability has been resolved: media: hackrf: fix to not free memory after the device is registered in hackrf_probe() In hackrf driver, the following race condition occurs: ``` CPU0 CPU1 hackrf_probe() kzalloc(); // alloc hackrf_dev .... v4l2_device_register(); .... fd = sys_open("/path/to/dev"); // open hackrf fd .... v4l2_device_unregister(); .... kfree(); // free hackrf_dev .... sys_ioctl(fd, ...); v4l2_ioctl(); video_is_registered() // UAF!! .... sys_close(fd); v4l2_release() // UAF!! hackrf_video_release() kfree(); // DFB!! ``` When a V4L2 or video device is unregistered, the device node is removed so new open() calls are blocked. However, file descriptors that are already open-and any in-flight I/O-do not terminate immediately; they remain valid until the last reference is dropped and the driver's release() is invoked. Therefore, freeing device memory on the error path after hackrf_probe() has registered dev it will lead to a race to use-after-free vuln, since those already-open handles haven't been released yet. And since release() free memory too, race to use-after-free and double-free vuln occur. To prevent this, if device is registered from probe(), it should be modified to free memory only through release() rather than calling kfree() directly.

Affected configurations

• linux linux kernel (< 6.6.136) • linux linux kernel (>= 6.7 < 6.12.83) • linux linux kernel (>= 6.13 < 6.18.24) • linux linux kernel (>= 6.19 < 6.19.14) • linux linux kernel (>= 7.0 < 7.0.1)

Related

Weakness · CWE (1)

References