Описание
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
- unix_stream_read_generic() peeks the first consumed OOB skb
- ma
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Уязвимые конфигурации
Одно из
EPSS
7.8 High
CVSS3
Дефекты
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: 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