net: atm: fix crash due to unvalidated vcc pointer in sigd_send()

net: atm: fix crash due to unvalidated vcc pointer in sigd_send()

Reproducer available at [1].

The ATM send path (sendmsg -> vcc_sendmsg -> sigd_send) reads the vcc
pointer from msg->vcc and uses it directly without any validation. This
pointer comes from userspace via sendmsg() and can be arbitrarily forged:

    int fd = socket(AF_ATMSVC, SOCK_DGRAM, 0);
    ioctl(fd, ATMSIGD_CTRL);  // become ATM signaling daemon
    struct msghdr msg = { .msg_iov = &iov, ... };
    *(unsigned long *)(buf + 4) = 0xdeadbeef;  // fake vcc pointer
    sendmsg(fd, &msg, 0);  // kernel dereferences 0xdeadbeef

In normal operation, the kernel sends the vcc pointer to the signaling
daemon via sigd_enq() when processing operations like connect(), bind(),
or listen(). The daemon is expected to return the same pointer when
responding. However, a malicious daemon can send arbitrary pointer values.

Fix this by introducing find_get_vcc() which validates the pointer by
searching through vcc_hash (similar to how sigd_close() iterates over
all VCCs), and acquires a reference via sock_hold() if found.

Since struct atm_vcc embeds struct sock as its first member, they share
the same lifetime. Therefore using sock_hold/sock_put is sufficient to
keep the vcc alive while it is being used.

Note that there may be a race with sigd_close() which could mark the vcc
with various flags (e.g., ATM_VF_RELEASED) after find_get_vcc() returns.
However, sock_hold() guarantees the memory remains valid, so this race
only affects the logical state, not memory safety.

[1]: https://gist.github.com/mrpre/1ba5949c45529c511152e2f4c755b0f3
Fixes: 1da177e4c3f4 ("Linux-2.6.12-rc2")
Reported-by: syzbot+1f22cb1769f249df9fa0@syzkaller.appspotmail.com
Closes: https://lore.kernel.org/all/69039850.a70a0220.5b2ed.005d.GAE@google.com/T/
Signed-off-by: Jiayuan Chen <jiayuan.chen@shopee.com>
Link: https://patch.msgid.link/20260205095501.131890-1-jiayuan.chen@linux.dev
Signed-off-by: Paolo Abeni <pabeni@redhat.com>