• 北京理工大学信息安全与对抗竞赛crackme01分析


    只是从crackme的情况来看,这次的题目没有成信院的难。

    首先,E-Debug Events找到按钮事件。

     1 004010A1  /.  55            push    ebp
     2 004010A2  |.  8BEC          mov     ebp, esp
     3 004010A4  |.  81EC 08000000 sub     esp, 8
     4 004010AA  |.  6A FF         push    -1
     5 004010AC  |.  6A 08         push    8
     6 004010AE  |.  68 02000116   push    16010002
     7 004010B3  |.  68 01000152   push    52010001
     8 004010B8  |.  E8 00010000   call    004011BD                         ;  获取用户输入的注册码
     9 004010BD  |.  83C4 10       add     esp, 10
    10 004010C0  |.  8945 FC       mov     dword ptr [ebp-4], eax
    11 004010C3  |.  68 4C9B4600   push    00469B4C                         ;  ASCII "Manson"
    12 004010C8  |.  FF75 FC       push    dword ptr [ebp-4]                ;  假码入栈
    13 004010CB  |.  E8 34FFFFFF   call    00401004                         ;  比较
    14 004010D0  |.  83C4 08       add     esp, 8
    15 004010D3  |.  83F8 00       cmp     eax, 0                           ;  结果
    16 004010D6  |.  B8 00000000   mov     eax, 0
    17 004010DB  |.  0F94C0        sete    al
    18 004010DE  |.  8945 F8       mov     dword ptr [ebp-8], eax           ;  比较结果放入某个位置
    19 004010E1  |.  8B5D FC       mov     ebx, dword ptr [ebp-4]
    20 004010E4  |.  85DB          test    ebx, ebx
    21 004010E6  |.  74 09         je      short 004010F1
    22 004010E8  |.  53            push    ebx
    23 004010E9  |.  E8 C9000000   call    004011B7
    24 004010EE  |.  83C4 04       add     esp, 4
    25 004010F1  |>  837D F8 00    cmp     dword ptr [ebp-8], 0
    26 004010F5  |.  0F84 35000000 je      00401130                         ;  关键跳
    27 004010FB  |.  6A 00         push    0
    28 004010FD  |.  6A 00         push    0
    29 004010FF  |.  6A 00         push    0
    30 00401101  |.  68 01030080   push    80000301
    31 00401106  |.  6A 00         push    0
    32 00401108  |.  68 00000000   push    0
    33 0040110D  |.  68 04000080   push    80000004
    34 00401112  |.  6A 00         push    0
    35 00401114  |.  68 539B4600   push    00469B53
    36 00401119  |.  68 03000000   push    3
    37 0040111E  |.  BB 00134000   mov     ebx, 00401300
    38 00401123  |.  E8 89000000   call    004011B1
    39 00401128  |.  83C4 28       add     esp, 28
    40 0040112B  |.  E9 30000000   jmp     00401160
    41 00401130  |>  6A 00         push    0
    42 00401132  |.  6A 00         push    0
    43 00401134  |.  6A 00         push    0
    44 00401136  |.  68 01030080   push    80000301
    45 0040113B  |.  6A 00         push    0
    46 0040113D  |.  68 10000000   push    10
    47 00401142  |.  68 04000080   push    80000004
    48 00401147  |.  6A 00         push    0
    49 00401149  |.  68 5C9B4600   push    00469B5C
    50 0040114E  |.  68 03000000   push    3
    51 00401153  |.  BB 00134000   mov     ebx, 00401300
    52 00401158  |.  E8 54000000   call    004011B1
    53 0040115D  |.  83C4 28       add     esp, 28
    54 00401160  |>  8BE5          mov     esp, ebp
    55 00401162  |.  5D            pop     ebp
    56 00401163  \.  C3            retn

    直接就是明码比较,所以很简单。

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  • 原文地址:https://www.cnblogs.com/tk091/p/2509396.html
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