Exploit sudo 1.8.0 < 1.8.3p1 - Format String

[Exploiter]

EDB-ID

18436

Проверка EDB

1. Пройдено

Автор

JOERNCHEN

Тип уязвимости

DOS

Платформа

LINUX

CVE

cve-2012-0809

Дата публикации

2012-01-31

Phenoelit Advisory <wir-haben-auch-mal-was-gefunden #0815 +--++>

[ Authors ]
        joernchen       <joernchen () phenoelit de>

        Phenoelit Group (http://www.phenoelit.de)

[ Affected Products ]
        sudo 1.8.0 - 1.8.3p1 (http://sudo.ws)

[ Vendor communication ]
        2012-01-24 Send vulnerability details to sudo maintainer
        2012-01-24 Maintainer is embarrased
        2012-01-27 Asking maintainer how the fixing goes
        2012-01-27 Maintainer responds with a patch and a release date
                   of 2012-01-30 for the patched sudo and advisory
        2012-01-30 Release of this advisory

[ Description ]

        Observe src/sudo.c:

void
sudo_debug(int level, const char *fmt, ...)
{
    va_list ap;
    char *fmt2;

    if (level > debug_level)
        return;

    /* Backet fmt with program name and a newline to make it a single 
    write */
    easprintf(&fmt2, "%s: %s\n", getprogname(), fmt);
    va_start(ap, fmt);
    vfprintf(stderr, fmt2, ap);
    va_end(ap);
    efree(fmt2);
}

        Here getprogname() is argv[0] and by this user controlled. So 
        argv[0] goes to fmt2 which then gets vfprintf()ed to stderr. The
        result is a Format String vulnerability.   

[ Example ]
        /tmp $ ln -s /usr/bin/sudo %n
        /tmp $ ./%n -D9
        *** %n in writable segment detected ***
        Aborted
        /tmp $

       A note regarding exploitability: The above example shows the result
       of FORTIFY_SOURCE which makes explotitation painful but not 
       impossible (see [0]). Without FORTIFY_SOURCE the exploit is straight
       forward:
         1. Use formatstring to overwrite the setuid() call with setgid()
         2. Trigger with formatstring -D9 
         3. Make use of SUDO_ASKPASS and have shellcode in askpass script
         4. As askpass will be called after the formatstring has 
            overwritten setuid() the askepass script will run with uid 0
         5. Enjoy the rootshell
 
[ Solution ]
        Update to version 1.8.3.p2 

[ References ]
        [0] http://www.phrack.org/issues.html?issue=67&id=9

[ end of file ]