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CVE intelligence

CVE-2026-31576

Vulnerability intelligence

Published Apr 24, 2026Sources checked Sep 27, 2026
7.8HIGHCVSS out of 10
What this means

Review the available evidence

CVE-2026-31576 and is rated High severity with a CVSS score of 7.8. It is not in the current CISA KEV record we collected. That does not prove exploitation has not occurred.

CISA KEVNot listedBased on the latest collected catalog
EPSS0.2%Estimated 30-day exploitation probability
Ransomware useNot markedCISA KEV ransomware field
Threat actors0Source-linked actor relationships
Overview

What is CVE-2026-31576?

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.