The_Arhitect Posted January 31, 2012 Report Posted January 31, 2012 sudo 1.8.0 - 1.8.3p1 Format String VulnerabilityPhenoelit 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:voidsudo_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 ]Sursa: sudo 1.8.0 - 1.8.3p1 Format String Vulnerability Quote