Описание
af_unix: Don't leave consecutive consumed OOB skbs.
FAQ
Is Azure Linux the only Microsoft product that includes this open-source library and is therefore potentially affected by this vulnerability?
One of the main benefits to our customers who choose to use the Azure Linux distro is the commitment to keep it up to date with the most recent and most secure versions of the open source libraries with which the distro is composed. Microsoft is committed to transparency in this work which is why we began publishing CSAF/VEX in October 2025. See this blog post for more information. If impact to additional products is identified, we will update the CVE to reflect this.
EPSS
7.8 High
CVSS3
Связанные уязвимости
In the Linux kernel, the following vulnerability has been resolved: af_unix: Don't leave consecutive consumed OOB skbs. Jann Horn reported a use-after-free in unix_stream_read_generic(). The following sequences reproduce the issue: $ python3 from socket import * s1, s2 = socketpair(AF_UNIX, SOCK_STREAM) s1.send(b'x', MSG_OOB) s2.recv(1, MSG_OOB) # leave a consumed OOB skb s1.send(b'y', MSG_OOB) s2.recv(1, MSG_OOB) # leave a consumed OOB skb s1.send(b'z', MSG_OOB) s2.recv(1) # recv 'z' illegally s2.recv(1, MSG_OOB) # access 'z' skb (use-after-free) Even though a user reads OOB data, the skb holding the data stays on the recv queue to mark the OOB boundary and break the next recv(). After the last send() in the scenario above, the sk2's recv queue has 2 leading consumed OOB skbs and 1 real OOB skb. Then, the following happens during the next recv() without MSG_OOB 1. unix_stream_read_generic() peeks the first consumed OOB skb 2. manage_oob() returns the next c...
In the Linux kernel, the following vulnerability has been resolved: af_unix: Don't leave consecutive consumed OOB skbs. Jann Horn reported a use-after-free in unix_stream_read_generic(). The following sequences reproduce the issue: $ python3 from socket import * s1, s2 = socketpair(AF_UNIX, SOCK_STREAM) s1.send(b'x', MSG_OOB) s2.recv(1, MSG_OOB) # leave a consumed OOB skb s1.send(b'y', MSG_OOB) s2.recv(1, MSG_OOB) # leave a consumed OOB skb s1.send(b'z', MSG_OOB) s2.recv(1) # recv 'z' illegally s2.recv(1, MSG_OOB) # access 'z' skb (use-after-free) Even though a user reads OOB data, the skb holding the data stays on the recv queue to mark the OOB boundary and break the next recv(). After the last send() in the scenario above, the sk2's recv queue has 2 leading consumed OOB skbs and 1 real OOB skb. Then, the following happens during the next recv() without MSG_OOB 1. unix_stream_read_generic() peeks the first consumed OOB skb 2. manage_oob() returns the next c...
In the Linux kernel, the following vulnerability has been resolved: af_unix: Don't leave consecutive consumed OOB skbs. Jann Horn reported a use-after-free in unix_stream_read_generic(). The following sequences reproduce the issue: $ python3 from socket import * s1, s2 = socketpair(AF_UNIX, SOCK_STREAM) s1.send(b'x', MSG_OOB) s2.recv(1, MSG_OOB) # leave a consumed OOB skb s1.send(b'y', MSG_OOB) s2.recv(1, MSG_OOB) # leave a consumed OOB skb s1.send(b'z', MSG_OOB) s2.recv(1) # recv 'z' illegally s2.recv(1, MSG_OOB) # access 'z' skb (use-after-free) Even though a user reads OOB data, the skb holding the data stays on the recv queue to mark the OOB boundary and break the next recv(). After the last send() in the scenario above, the sk2's recv queue has 2 leading consumed OOB skbs and 1 real OOB skb. Then, the following happens during the next recv() without MSG_OOB 1. unix_stream_read_generic() peeks the first consumed OOB skb 2. ma
In the Linux kernel, the following vulnerability has been resolved: a ...
In the Linux kernel, the following vulnerability has been resolved: af_unix: Don't leave consecutive consumed OOB skbs. Jann Horn reported a use-after-free in unix_stream_read_generic(). The following sequences reproduce the issue: $ python3 from socket import * s1, s2 = socketpair(AF_UNIX, SOCK_STREAM) s1.send(b'x', MSG_OOB) s2.recv(1, MSG_OOB) # leave a consumed OOB skb s1.send(b'y', MSG_OOB) s2.recv(1, MSG_OOB) # leave a consumed OOB skb s1.send(b'z', MSG_OOB) s2.recv(1) # recv 'z' illegally s2.recv(1, MSG_OOB) # access 'z' skb (use-after-free) Even though a user reads OOB data, the skb holding the data stays on the recv queue to mark the OOB boundary and break the next recv(). After the last send() in the scenario above, the sk2's recv queue has 2 leading consumed OOB skbs and 1 real OOB skb. Then, the following happens during the next recv() without MSG_OOB 1. unix_stream_read_generic() peeks the first consumed OOB skb 2....
EPSS
7.8 High
CVSS3