• CSAPP Lab2: Binary Bomb


    著名的CSAPP实验:二进制炸弹

    就是通过gdb和反汇编猜测程序意图,共有6关和一个隐藏关卡

    只有输入正确的字符串才能过关,否则会程序会bomb终止运行

    隐藏关卡需要输入特定字符串方会开启

    实验材料下载地址: http://csapp.cs.cmu.edu/2e/labs.html

    下面通关解法:

    反汇编:

    objdump -d bomb > bomb_assembly_32.S

    Phase 1:

        打开bomb_assembly_32.S,定位到<phase_1>函数,可以看到以下代码:

          

     1     8048b26:    8b 45 08                 mov    0x8(%ebp),%eax
     2 
     3     8048b29:    83 c4 f8                 add    $0xfffffff8,%esp
     4 
     5     8048b2c:    68 c0 97 04 08           push   $0x80497c0
     6 
     7     8048b31:    50                       push   %eax
     8 
     9     8048b32:    e8 f9 04 00 00           call   8049030 <strings_not_equal>
    10 
    11     8048b37:    83 c4 10                 add    $0x10,%esp
    12 
    13     8048b3a:    85 c0                    test   %eax,%eax
    14 
    15     8048b3c:    74 05                    je     8048b43 <phase_1+0x23>
    16 
    17     8048b3e:    e8 b9 09 00 00           call   80494fc <explode_bomb>

        可以看出,用户输入字串指针保存在0x8(%ebp), <phase_1>把此指针放入eax,

        然后把$0x80497c0压栈,再把eax也就是用户字串指针压栈,

        然后调用<strings_not_equal>

        待<strings_not_equal>返回后,测试返回值,

        若equal则进入下一phase,否则<explode_bomb>

        从<strings_not_equal>可知该函数用于比较两函数的值,因此需要两个字串作为输入,

        上面代码中,push %eax用于传递用户字串指针,

        则push $0x80497c0自然是传递比较字串的指针了.

        打开gdb,x/s 0x80497c0, 可以直接查看到该指针指向的子符串:

        Public speaking is very easy.

    Phase 2:

        打开bomb_assembly_32.S,定位到<phase_2>函数,留意以下几行:

     1     8048b50:    8b 55 08                 mov    0x8(%ebp),%edx
     2 
     3     8048b53:    83 c4 f8                 add    $0xfffffff8,%esp
     4 
     5     8048b56:    8d 45 e8                 lea    -0x18(%ebp),%eax
     6 
     7     8048b59:    50                       push   %eax
     8 
     9     8048b5a:    52                       push   %edx
    10 
    11     8048b5b:    e8 78 04 00 00           call   8048fd8 <read_six_numbers>

        mov 0x8(%ebp),%edx 将用户字串指针存入edx,

        lea -0x18(%ebp),%eax 把ebp-0x18这个地址存入eax,

        则最后三句

        push %eax

        push %edx

        call 8048fd8 <read_six_numbers>

        相当于read_six_numbers( 用户字串指针地址, ebp-0x18 )

        现在我们切换到<read_six_numbers>,看看这个函数是干什么的:

        先来看下面2行:

    1     8048fde:    8b 4d 08                 mov    0x8(%ebp),%ecx
    2 
    3     8048fe1:    8b 55 0c                 mov    0xc(%ebp),%edx

        把用户字串指针存入ecx, ebp-0x18存入edx

        往下看:

    1     8048fe4:    8d 42 14                 lea    0x14(%edx),%eax

        eax存入了 edx+0x14 这个值

        再往下:

     1     8048fe7:    50                       push   %eax
     2 
     3     8048fe8:    8d 42 10                 lea    0x10(%edx),%eax
     4 
     5     8048feb:    50                       push   %eax
     6 
     7     8048fec:    8d 42 0c                 lea    0xc(%edx),%eax
     8 
     9     8048fef:    50                       push   %eax
    10 
    11     8048ff0:    8d 42 08                 lea    0x8(%edx),%eax
    12 
    13     8048ff3:    50                       push   %eax
    14 
    15     8048ff4:    8d 42 04                 lea    0x4(%edx),%eax
    16 
    17     8048ff7:    50                       push   %eax
    18 
    19     8048ff8:    52                       push   %edx

        上面几行依次把 edx+0x14, edx+0x10, edx+0xc, edx+0x8, edx+4, edx 这6个地址值压栈

        注意edx是<phase_2>的stack frame的 ebp-0x18 这个地址值

        

    1     8048ff9:    68 1b 9b 04 08           push   $0x8049b1b
    2 
    3     8048ffe:    51                       push   %ecx
    4 
    5     8048fff:    e8 5c f8 ff ff           call   8048860 <sscanf@plt>

        前2行把 $0x8049b1b 和 ecx(用户字串指针) 压栈, 然后调用sscanf

        sscanf的原型是int sscanf(const char *str, const char *format, ...);

        按format的格式解释str,然后把得到的值放入后面省略号所代表的变量中

        因此, 按刚才压栈的顺序, str是用户输入字串, $0x8049b1b 是format的地址,

        edx, edx+4,...,edx+0x14是对应的变量.

        先用gdb查看format, x/s $0x8049b1b, "%d %d %d %d %d %d".

        可知,需要从用户字串中提取6个整数,存入(edx)--(edx+0x14)中.

        综上, <read_six_numbers> 作用就是从用户字串中提取6个数字, 存入<phase_2>stack frame中的(ebp-0x18)中

        回到<phase_2>接着看:

    1     8048b63:    83 7d e8 01              cmpl   $0x1,-0x18(%ebp)
    2 
    3     8048b67:    74 05                    je     8048b6e <phase_2+0x26>
    4 
    5     8048b69:    e8 8e 09 00 00           call   80494fc <explode_bomb>

        测试(ebp-0x18)是否等于1, 不等则bomb, 因此用户输入的第一个数字应为1.

    1     8048b6e:    bb 01 00 00 00           mov    $0x1,%ebx
    2 
    3     8048b73:    8d 75 e8                 lea    -0x18(%ebp),%esi

        令ebx=1, esi = ebp-18

     1     8048b76:    8d 43 01                 lea    0x1(%ebx),%eax
     2 
     3     8048b79:    0f af 44 9e fc           imul   -0x4(%esi,%ebx,4),%eax
     4 
     5     8048b7e:    39 04 9e                 cmp    %eax,(%esi,%ebx,4)
     6 
     7     8048b81:    74 05                    je     8048b88 <phase_2+0x40>
     8 
     9     8048b83:    e8 74 09 00 00           call   80494fc <explode_bomb>
    10 
    11     8048b88:    43                       inc    %ebx
    12 
    13     8048b89:    83 fb 05                 cmp    $0x5,%ebx
    14 
    15     8048b8c:    7e e8                    jle    8048b76 <phase_2+0x2e>

        注意, esi是存放6个数字中第1数字的地址,

        因此 -0x4(%esi,%ebx,4) 表示第ebx个数字,

        (%esi,ebx,4)表示第ebx+1个数字

        因此上面第3-6行代码检查 a[ebx]*(ebx+1) == a[ebx+1], 其中a[n]表示第n个数字

        若不等则bomb,否则ebx增1并循环

        因此<phase_2>需要输入一个数列, a[1]=1, a[n+1] = a[n]*(n+1), n<=6

        1, 2, 6, 24, 120, 720

    Phase 3:

        打开bomb_assembly_32.S,定位到<phase_3>函数,可以看到以下代码:

     1     ;; edx stores pointer of user input
     2 
     3     8048b9f:    8b 55 08                 mov    0x8(%ebp),%edx
     4 
     5     8048ba2:    83 c4 f4                 add    $0xfffffff4,%esp
     6 
     7     ;; push ebp-4 onto stack
     8 
     9     8048ba5:    8d 45 fc                 lea    -0x4(%ebp),%eax
    10 
    11     8048ba8:    50                       push   %eax
    12 
    13     ;; push ebp-5 onto stack
    14 
    15     8048ba9:    8d 45 fb                 lea    -0x5(%ebp),%eax
    16 
    17     8048bac:    50                       push   %eax
    18 
    19     ;; push ebp-12 onto stack
    20 
    21     8048bad:    8d 45 f4                 lea    -0xc(%ebp),%eax
    22 
    23     8048bb0:    50                       push   %eax
    24 
    25     ;; push $0x80497de onto stack
    26 
    27     ;; gdb x/s 0x80497de: "%d %c %d"
    28 
    29     8048bb1:    68 de 97 04 08           push   $0x80497de
    30 
    31     ;; push pointer of user input onto stack
    32 
    33     8048bb6:    52                       push   %edx
    34 
    35     8048bb7:    e8 a4 fc ff ff           call   8048860 <sscanf@plt>

        具体代码请看注释,一开始主要是sscanf(用户字串指针, "%d %c %d", ebp-12, ebp-5, ebp-4)

        继续看下去:

     1     ;; (ebp-12) stores the first int, compare to 7
     2 
     3     ;; cmpl takes (ebp-12) as unsigned int
     4 
     5     8048bc9:    83 7d f4 07              cmpl   $0x7,-0xc(%ebp)
     6 
     7     ;; (unsigned)(ebp-12) > 7, jump to 0x8048c88, which will bomb
     8 
     9     8048bcd:    0f 87 b5 00 00 00        ja     8048c88 <phase_3+0xf0>
    10 
    11     ;; jump to *( 0x80497e8 + 4*(the first int) )
    12 
    13     8048bd3:    8b 45 f4                 mov    -0xc(%ebp),%eax
    14 
    15     8048bd6:    ff 24 85 e8 97 04 08     jmp    *0x80497e8(,%eax,4)

        关键在于最后的跳转,根据输入的第一个整数确定跳转地址,

        地址存储在(0x80497e8 + 4*(the first int)).

        容易联想到(0x80497e8)存储着一个跳转表,用gdb查看之,x/10wx 0x80497e8:

        0x80497e8:    0x08048be0    0x08048c00    0x08048c16    0x08048c28
    
        0x80497f8:    0x08048c40    0x08048c52    0x08048c64    0x08048c76
    
        0x8049808:    0x67006425    0x746e6169

        可以看到表中有很多个地址,先来看第一个地址指向的语句(对应的输入整数为0):

     1     ;; bl = 0x71
     2 
     3     8048be0:    b3 71                    mov    $0x71,%bl
     4 
     5     ;; if 0x309==777==the last int,
     6 
     7     ;; jump to 0x8048c8f, which will compare the char
     8 
     9     8048be2:    81 7d fc 09 03 00 00     cmpl   $0x309,-0x4(%ebp)
    10 
    11     8048be9:    0f 84 a0 00 00 00        je     8048c8f <phase_3+0xf7>
    12 
    13     8048bef:    e8 08 09 00 00           call   80494fc <explode_bomb>

        可以看出,先把0x71存入bl,

        然后若输入的最后一个整数==777的话,则跳转到0x8048c8f

     1     ;; after compare the last int, jump here
     2 
     3     ;; bl = 0x71 = 'q', compare to the char
     4 
     5     ;; if ==, jump to 0x8048c99, and leave this function
     6 
     7     8048c8f:    3a 5d fb                 cmp    -0x5(%ebp),%bl
     8 
     9     8048c92:    74 05                    je     8048c99 <phase_3+0x101>
    10 
    11     8048c94:    e8 63 08 00 00           call   80494fc <explode_bomb>

        比较输入的字符是否等于'q',若等于则defuse成功

        因此,输入应为: "0 q 777"

        当然此题应该有不止一个答案,选择跳转表中不同的地址会导致不同的输入.

    Phase 4:

        打开bomb_assembly_32.S,定位到<phase_4>函数,可以看到以下代码:

     1     ;; edx = pointer of input string
     2 
     3     8048ce6:    8b 55 08                 mov    0x8(%ebp),%edx
     4 
     5     8048ce9:    83 c4 fc                 add    $0xfffffffc,%esp
     6 
     7     ;; eax = ebp-4
     8 
     9     8048cec:    8d 45 fc                 lea    -0x4(%ebp),%eax
    10 
    11     ;; push ebp-4
    12 
    13     8048cef:    50                       push   %eax
    14 
    15     ;; push $0x8049808
    16 
    17     ;; x/s 0x804980: "%d"
    18 
    19     8048cf0:    68 08 98 04 08           push   $0x8049808
    20 
    21     ;; push pointer of input string
    22 
    23     8048cf5:    52                       push   %edx
    24 
    25     8048cf6:    e8 65 fb ff ff           call   8048860 <sscanf@plt>

        就是读入一个整数,存入ebp-4

       

     1  ;; func4( input_number )
     2 
     3     8048d11:    8b 45 fc                 mov    -0x4(%ebp),%eax
     4 
     5     8048d14:    50                       push   %eax
     6 
     7     8048d15:    e8 86 ff ff ff           call   8048ca0 <func4>
     8 
     9 
    10 
    11     8048d1a:    83 c4 10                 add    $0x10,%esp
    12 
    13     ;; eax should contain the return value of <func4>
    14 
    15     ;; if eax == 0x37 == 55, defused
    16 
    17     8048d1d:    83 f8 37                 cmp    $0x37,%eax
    18 
    19     8048d20:    74 05                    je     8048d27 <phase_4+0x47>
    20 
    21     8048d22:    e8 d5 07 00 00           call   80494fc <explode_bomb>

        然后比较 func4( input_number )==55, 若等于则成功defuse.

        接下来看看<func4>:

        

     1     ;; ebx = input_number
     2 
     3     8048ca8:    8b 5d 08                 mov    0x8(%ebp),%ebx
     4 
     5     ;; if input_number<=1, <func4> return 1
     6 
     7     8048cab:    83 fb 01                 cmp    $0x1,%ebx
     8 
     9     8048cae:    7e 20                    jle    8048cd0 <func4+0x30>
    10 
    11 
    12 
    13     8048cb0:    83 c4 f4                 add    $0xfffffff4,%esp
    14 
    15     ;; esi == func4( input_number-1 )
    16 
    17     8048cb3:    8d 43 ff                 lea    -0x1(%ebx),%eax
    18 
    19     8048cb6:    50                       push   %eax
    20 
    21     8048cb7:    e8 e4 ff ff ff           call   8048ca0 <func4>
    22 
    23     8048cbc:    89 c6                    mov    %eax,%esi
    24 
    25 
    26 
    27     8048cbe:    83 c4 f4                 add    $0xfffffff4,%esp
    28 
    29 
    30 
    31     ;; esi += func4( input_number-2 )
    32 
    33     8048cc1:    8d 43 fe                 lea    -0x2(%ebx),%eax
    34 
    35     8048cc4:    50                       push   %eax
    36 
    37     8048cc5:    e8 d6 ff ff ff           call   8048ca0 <func4>
    38 
    39     8048cca:    01 f0                    add    %esi,%eax

        很明显是Fibonacci数列,  func4(n) = func4(n-1) + func4(n-2)

        注意f(0)=f(1)=1, 通过简单计算知f(9)=55

        因此输入应为55

    Phase 5:

        打开bomb_assembly_32.S,定位到<phase_5>函数,可以看到以下代码:

     1     ;; ebx = pointer of input
     2 
     3     ;; push ebx onto stack
     4 
     5     ;; call string_length
     6 
     7     8048d34:    8b 5d 08                 mov    0x8(%ebp),%ebx
     8 
     9     8048d37:    83 c4 f4                 add    $0xfffffff4,%esp
    10 
    11     8048d3a:    53                       push   %ebx
    12 
    13     8048d3b:    e8 d8 02 00 00           call   8049018 <string_length>
    14 
    15 
    16 
    17     8048d40:    83 c4 10                 add    $0x10,%esp
    18 
    19     ;; eax stores the return value of string_length
    20 
    21     ;; if eax == 6, jump to 0x8048d4d 
    22 
    23     8048d43:    83 f8 06                 cmp    $0x6,%eax
    24 
    25     8048d46:    74 05                    je     8048d4d <phase_5+0x21>
    26 
    27     8048d48:    e8 af 07 00 00           call   80494fc <explode_bomb>

        从上面代码可知,输入需要6个字符.

     1     ;; edx = 0
     2 
     3     8048d4d:    31 d2                    xor    %edx,%edx
     4 
     5     ;; ecx = ebp-8
     6 
     7     8048d4f:    8d 4d f8                 lea    -0x8(%ebp),%ecx
     8 
     9     ;; esi = 0x804b220
    10 
    11     8048d52:    be 20 b2 04 08           mov    $0x804b220,%esi
    12 
    13     ;; edx is a counter from 0 to 5
    14 
    15     ;; al = (edx + ebx), then al reads a char each time
    16 
    17     8048d57:    8a 04 1a                 mov    (%edx,%ebx,1),%al
    18 
    19     ;; extract the low 4 bit of al
    20 
    21     8048d5a:    24 0f                    and    $0xf,%al
    22 
    23     ;; sign-extend al to eax
    24 
    25     8048d5c:    0f be c0                 movsbl %al,%eax
    26 
    27     ;; al = ( eax + 0x804b220 )
    28 
    29     ;; x/16c 0x804b220:
    30 
    31     ;; 0x804b220:    105 'i'    115 's'    114 'r'    118 'v'    101 'e'    97 'a'    119 'w'    104 'h'
    32 
    33     ;; 0x804b228:    111 'o'    98 'b'    112 'p'    110 'n'    117 'u'    116 't'    102 'f'    103 'g'
    34 
    35     8048d5f:    8a 04 30                 mov    (%eax,%esi,1),%al
    36 
    37     ;; edx + ecx = al,
    38 
    39     ;; notice that, ecx = ebp-8
    40 
    41     ;; and edx is a counter from 0 to 5
    42 
    43     8048d62:    88 04 0a                 mov    %al,(%edx,%ecx,1)
    44 
    45     8048d65:    42                       inc    %edx
    46 
    47     ;; loop
    48 
    49     8048d66:    83 fa 05                 cmp    $0x5,%edx
    50 
    51     8048d69:    7e ec                    jle    8048d57 <phase_5+0x2b>
    52 
    53 
    54 
    55     ;; ebp-2 = 0, a terminal of string started from ebp-8
    56 
    57     8048d6b:    c6 45 fe 00              movb   $0x0,-0x2(%ebp)
    58 
    59     8048d6f:    83 c4 f8                 add    $0xfffffff8,%esp

        

        上面代码的作用是循环读取6个输入字符中的每一字符input[k],

        提取input[k]的低四位,把这四位构成的整数index当作索引,

        查找0x804b220开始16个字节中存储的字符.

        用gdb查看, x/16c 0x804b220:

    1     0x804b220:    105 'i'    115 's'    114 'r'    118 'v'    101 'e'    97 'a'    119 'w'    104 'h'
    2 
    3     0x804b228:    111 'o'    98 'b'    112 'p'    110 'n'    117 'u'    116 't'    102 'f'    103 'g'

        获取0x804b220[ input[k] & 0xf ]后,将之copy至 (ebp-8)[k]

        继续看:

     1     ;; x/s 0x804980b: "giants"
     2 
     3     ;; push "giants"
     4 
     5     8048d72:    68 0b 98 04 08           push   $0x804980b
     6 
     7     ;; push ebp-8
     8 
     9     8048d77:    8d 45 f8                 lea    -0x8(%ebp),%eax
    10 
    11     8048d7a:    50                       push   %eax
    12 
    13     ;; compare "giants" and the string started from ebp-8
    14 
    15     8048d7b:    e8 b0 02 00 00           call   8049030 <strings_not_equal>
    16 
    17     8048d80:    83 c4 10                 add    $0x10,%esp
    18 
    19     8048d83:    85 c0                    test   %eax,%eax
    20 
    21     ;; if two strings equal to each other, defused
    22 
    23     8048d85:    74 05                    je     8048d8c <phase_5+0x60>
    24 
    25     8048d87:    e8 70 07 00 00           call   80494fc <explode_bomb>

        上面代码便是将ebp-18开始的字串和"giants"比较,若相等,则defused.

        注意到 (ebp-18)[k] = 0x804b220[ input[k] & 0xf ]

    1     0x804b220:    105 'i'    115 's'    114 'r'    118 'v'    101 'e'    97 'a'    119 'w'    104 'h'
    2 
    3     0x804b228:    111 'o'    98 'b'    112 'p'    110 'n'    117 'u'    116 't'    102 'f'    103 'g'

        因此,

    1     input[0]&0xf = 0xf, input[1]&0xf = 0x0,
    2 
    3     input[2]&0xf = 0x5, input[3]&0xf = 0xb,
    4 
    5     input[4]&0xf = 0xd, input[5]&0xf = 0x1,

        只要输入的各个字符的低四位符合上面就好,我个人选取了"opekma"

        

    Phase 6:

        写得太复杂了,各种内外循环,各种跳转,看得头晕,日后有闲再看.

        现在先把能看懂的部份写出来:

     1     ;; edx = pointer of input
     2 
     3     8048da1:    8b 55 08                 mov    0x8(%ebp),%edx
     4 
     5     ;; (ebp-0x34) = $0x804b26c
     6 
     7     8048da4:    c7 45 cc 6c b2 04 08     movl   $0x804b26c,-0x34(%ebp)
     8 
     9     8048dab:    83 c4 f8                 add    $0xfffffff8,%esp
    10 
    11     ;; read six numbers from input,
    12 
    13     ;; and storse in the area started from ebp-18
    14 
    15     8048dae:    8d 45 e8                 lea    -0x18(%ebp),%eax
    16 
    17     8048db1:    50                       push   %eax
    18 
    19     8048db2:    52                       push   %edx
    20 
    21     8048db3:    e8 20 02 00 00           call   8048fd8 <read_six_numbers>

        上面代码就是从输入读入6个整数,存入ebp-0x18,

        初步怀疑0x804b26c地址存放着一个链表.

     1     ;; edi = 0
     2 
     3     8048db8:    31 ff                    xor    %edi,%edi
     4 
     5     8048dba:    83 c4 10                 add    $0x10,%esp
     6 
     7     8048dbd:    8d 76 00                 lea    0x0(%esi),%esi
     8 
     9     ;; eax = (ebp-0x18 + 4*edi) = six-number[edi]
    10 
    11     ;; ebp-0x18 = the beginning address of the six numbers
    12 
    13     ;; edi is a counter from 0 to 5
    14 
    15     8048dc0:    8d 45 e8                 lea    -0x18(%ebp),%eax
    16 
    17     8048dc3:    8b 04 b8                 mov    (%eax,%edi,4),%eax
    18 
    19     ;; eax = six-number[edi]-1
    20 
    21     8048dc6:    48                       dec    %eax
    22 
    23     ;; if eax <= 5 , continue
    24 
    25     8048dc7:    83 f8 05                 cmp    $0x5,%eax
    26 
    27     8048dca:    76 05                    jbe    8048dd1 <phase_6+0x39>
    28 
    29     8048dcc:    e8 2b 07 00 00           call   80494fc <explode_bomb>
    30 
    31 
    32 
    33     ;; if edi+1 > 5, finish edi loop
    34 
    35     8048dd1:    8d 5f 01                 lea    0x1(%edi),%ebx
    36 
    37     8048dd4:    83 fb 05                 cmp    $0x5,%ebx
    38 
    39     8048dd7:    7f 23                    jg     8048dfc <phase_6+0x64>
    40 
    41 
    42 
    43     ;; (ebp-0x38) = edi*4
    44 
    45     8048dd9:    8d 04 bd 00 00 00 00     lea    0x0(,%edi,4),%eax
    46 
    47     8048de0:    89 45 c8                 mov    %eax,-0x38(%ebp)
    48 
    49 
    50 
    51     ;; esi = ebp-18 = the beginning address of the six numbers
    52 
    53     8048de3:    8d 75 e8                 lea    -0x18(%ebp),%esi
    54 
    55     ;; edx = (ebp-0x38) = edi*4
    56 
    57     ;; inner loops,
    58 
    59     ;; ebx is the counter from edi+1 to 5
    60 
    61     8048de6:    8b 55 c8                 mov    -0x38(%ebp),%edx
    62 
    63     ;; eax = edx + esi = six-number[edi]
    64 
    65     8048de9:    8b 04 32                 mov    (%edx,%esi,1),%eax
    66 
    67     ;; compare six-number[edi] and six-number[edi+ebx]
    68 
    69     8048dec:    3b 04 9e                 cmp    (%esi,%ebx,4),%eax
    70 
    71     ;; if six-number[edi] != six-number[edi+1], continue
    72 
    73     8048def:    75 05                    jne    8048df6 <phase_6+0x5e>
    74 
    75     8048df1:    e8 06 07 00 00           call   80494fc <explode_bomb>
    76 
    77     ;; ebx++
    78 
    79     ;; if ebx<=5, jump to 0x8048de6, ebx loops
    80 
    81     ;; else , finish ebx loop
    82 
    83     8048df6:    43                       inc    %ebx
    84 
    85     8048df7:    83 fb 05                 cmp    $0x5,%ebx
    86 
    87     8048dfa:    7e ea                    jle    8048de6 <phase_6+0x4e>

        内外两层循环,外层用edi计数,确保输入的6个整数不大于6,

        内层用ebx计数,保证所有数字两两不相等.

        再往后的代码异常混乱,各种链表离历,没空看....

        先从网上获得答案:4 2 6 3 1 5

    Secret Phase:

        首先要找到<secret_phase>的入口,经搜索发现入口是在<phase_defused>里面.

        先来看看<phase_defused>:

    1     ;; every time call read_line, ( 0x804b480 )++
    2 
    3     ;; only with 6 correct answer given ,will the secret phase appear
    4 
    5     8049533:    83 3d 80 b4 04 08 06     cmpl   $0x6,0x804b480
    6 
    7     804953a:    75 63                    jne    804959f <phase_defused+0x73>

        (0x804b480)是一个计数器,每当调用一次<read_line>每自增1,因此只有6关全通才能打开隐藏关卡.

     1     ;; push ebp-0x50
     2 
     3     804953c:    8d 5d b0                 lea    -0x50(%ebp),%ebx
     4 
     5     804953f:    53                       push   %ebx
     6 
     7     ;; push ebp-0x54
     8 
     9     8049540:    8d 45 ac                 lea    -0x54(%ebp),%eax
    10 
    11     8049543:    50                       push   %eax
    12 
    13     ;; (gdb) x/s 0x8049d03
    14 
    15     ;; 0x8049d03:    "%d %s"
    16 
    17     8049544:    68 03 9d 04 08           push   $0x8049d03
    18 
    19     ;; push the string stores in 0x804b770
    20 
    21     ;; the address of input of phase 4
    22 
    23     8049549:    68 70 b7 04 08           push   $0x804b770
    24 
    25     804954e:    e8 0d f3 ff ff           call   8048860 <sscanf@plt>
    26 
    27 
    28 
    29     ....
    30 
    31     
    32 
    33     ;; (gdb) x/s 0x8049d09
    34 
    35     ;; 0x8049d09:    "austinpowers"
    36 
    37     804955e:    68 09 9d 04 08           push   $0x8049d09
    38 
    39     ;; push the %s
    40 
    41     8049563:    53                       push   %ebx
    42 
    43     8049564:    e8 c7 fa ff ff           call   8049030 <strings_not_equal>

        省略号上方的代码调用sscanf( (char *)0x804b770, "%d %s", (int *)(ebp-0x54), (char *)ebp-0x50 )

        即从0x804b770读入一个整数和字串.

        再看省略号下方的代码,比较读入的字串和"austinpowers", 若相等,则打开<secret_phase>

        好了,现在问题是,如何把一个整数和"austinpowers"写入地址0x804b770?

        回想前几关,写入字串都是通过read_line,所以猜想可能是在某一关的输入中多输入些内容以写入地址0x804b770.

        用gdb查看前几关输入字串的指针,发现第4关的输入刚好是在地址0x804b770,而Phase 4只需输入一个数字,因此只需

        在第4关的输入中多输入一个"austinpowers"即可进入<secret_phase>.

        现在看看<secret_phase>:

     1     8048eef:    e8 08 03 00 00           call   80491fc <read_line>
     2 
     3 
     4 
     5     8048ef4:    6a 00                    push   $0x0
     6 
     7 
     8 
     9     ;; strtol( user input string, 0, 10)
    10 
    11     ;; long int strtol(const char *nptr, char **endptr, int base);
    12 
    13     ;; converts the initial part of the string in nptr to a long integer value according to the given base
    14 
    15     8048ef6:    6a 0a                    push   $0xa
    16 
    17     8048ef8:    6a 00                    push   $0x0
    18 
    19     8048efa:    50                       push   %eax
    20 
    21     8048efb:    e8 f0 f8 ff ff           call   80487f0 <__strtol_internal@plt>

        首先,读入一个字串,并用strtol将之转换为long int.

     1     ;; if fun7( 0x804b320, the input long int )
     2 
     3     ;; x/d 0x804b320: (0x804b320) = 36
     4 
     5     8048f17:    53                       push   %ebx
     6 
     7     8048f18:    68 20 b3 04 08           push   $0x804b320
     8 
     9     8048f1d:    e8 72 ff ff ff           call   8048e94 <fun7>
    10 
    11 
    12 
    13     8048f22:    83 c4 10                 add    $0x10,%esp
    14 
    15     ;; if fun7(0x804b320, the input long int) == 7, defused
    16 
    17     8048f25:    83 f8 07                 cmp    $0x7,%eax
    18 
    19     8048f28:    74 05                    je     8048f2f <secret_phase+0x47>
    20 
    21     8048f2a:    e8 cd 05 00 00           call   80494fc <explode_bomb>

        代码很简单,调用fun7( (void *)0x804b320, 输入的整数 ),若返回值==7, 则成功defused.

        现在看看<fun7>:

        

      1     ;; edx = the first parameter, an address
      2 
      3     8048e9a:    8b 55 08                 mov    0x8(%ebp),%edx
      4 
      5     ;; eax = the input long int
      6 
      7     8048e9d:    8b 45 0c                 mov    0xc(%ebp),%eax
      8 
      9 
     10 
     11     ;; if edx != 0
     12 
     13     8048ea0:    85 d2                    test   %edx,%edx
     14 
     15     8048ea2:    75 0c                    jne    8048eb0 <fun7+0x1c>
     16 
     17 
     18 
     19     8048ea4:    b8 ff ff ff ff           mov    $0xffffffff,%eax
     20 
     21     8048ea9:    eb 37                    jmp    8048ee2 <fun7+0x4e>
     22 
     23     8048eab:    90                       nop
     24 
     25     8048eac:    8d 74 26 00              lea    0x0(%esi,%eiz,1),%esi
     26 
     27 
     28 
     29     ;; if the input number >= (edx), jump to 0x8048ec5 
     30 
     31     8048eb0:    3b 02                    cmp    (%edx),%eax
     32 
     33     8048eb2:    7d 11                    jge    8048ec5 <fun7+0x31>
     34 
     35 
     36 
     37     ;; eax > (edx)
     38 
     39     8048eb4:    83 c4 f8                 add    $0xfffffff8,%esp
     40 
     41     ;; <func7>( (edx+4) ,the input long int )
     42 
     43     8048eb7:    50                       push   %eax
     44 
     45     8048eb8:    8b 42 04                 mov    0x4(%edx),%eax
     46 
     47     8048ebb:    50                       push   %eax
     48 
     49     8048ebc:    e8 d3 ff ff ff           call   8048e94 <fun7>
     50 
     51 
     52 
     53     ;; return eax *= 2, exit
     54 
     55     8048ec1:    01 c0                    add    %eax,%eax
     56 
     57     8048ec3:    eb 1d                    jmp    8048ee2 <fun7+0x4e>
     58 
     59 
     60 
     61     ;; the input number >= (edx)
     62 
     63     ;; if eax == (edx), return eax=0
     64 
     65     8048ec5:    3b 02                    cmp    (%edx),%eax
     66 
     67     8048ec7:    74 17                    je     8048ee0 <fun7+0x4c>
     68 
     69 
     70 
     71     ;; the input number > (edx)
     72 
     73     8048ec9:    83 c4 f8                 add    $0xfffffff8,%esp
     74 
     75     ;; <fun7>( (edx+8) ,the input long int )
     76 
     77     8048ecc:    50                       push   %eax
     78 
     79     8048ecd:    8b 42 08                 mov    0x8(%edx),%eax
     80 
     81     8048ed0:    50                       push   %eax
     82 
     83     8048ed1:    e8 be ff ff ff           call   8048e94 <fun7>
     84 
     85 
     86 
     87     ;; fun7 return 2*eax + 1
     88 
     89     8048ed6:    01 c0                    add    %eax,%eax
     90 
     91     8048ed8:    40                       inc    %eax
     92 
     93     8048ed9:    eb 07                    jmp    8048ee2 <fun7+0x4e>
     94 
     95 
     96 
     97     8048edb:    90                       nop
     98 
     99     8048edc:    8d 74 26 00              lea    0x0(%esi,%eiz,1),%esi
    100 
    101 
    102 
    103     8048ee0:    31 c0                    xor    %eax,%eax

        从上面代码可看出函数原型是:fun7( void *address, long int number ).

        当 number == *(int*)address, fun7( address, number) = 0
    
        当 number > *(int*)address, fun7( address, number) = 2*fun7( address+8, number ) + 1
    
        当 number < *(int*)address, fun7( address, number) = 2*fun7( address+4, number )

        从上面可以看出, 上面的address表示的是棵二叉树(左子树的值<父节点的值, 右子树的值>父节点的值):

     1     struct BST
     2 
     3     {
     4 
     5         int num;
     6 
     7         struct BST *left;
     8 
     9         struct BST *right;
    10 
    11     } *bst;

        则上面的递推式可表示为:

        当 number == bst->num, fun7( bst, number ) = 0;
    
        当 number > bst->num, fun7( bst, number ) = 2*fun7( bst->right, number ) + 1;
    
        当 number < bst->num, fun7( bst, number ) = 2*fun7( bst->left, number );

        鉴于<secret_phase>需要fun7( (struct BST *)0x804b320, number )返回7,一个奇数,所以第一步应该执行第二钟情况,

        又经观察发现以下递推规律:

            fun7( (struct BST *)0x804b320, number )
    
        =   2 * fun7( (struct BST *)0x804b320->right, number ) + 1
    
        =   2 * (2 * fun7( (struct BST *)0x804b320->right->right, number ) + 1) + 1
    
        =   4 * fun7( (struct BST *)0x804b320->right->right, number ) + 3
    
        =   4 * (2 * fun7( (struct BST *)0x804b320->right->right->right, number ) + 1) + 3
    
        =   8 * fun7( (struct BST *)0x804b320->right->right->right, number ) + 7

        因此当 number == (struct BST *)0x804b320->right->right->right->num, fun7便可返回7

        用gdb查看,

        x/wx 0x804b320+8  ==>  0x0804b308 
    
        x/wx 0x804b308+8  ==>  0x0804b2d8
    
        x/wx 0x804b2d8+8  ==>  0x0804b278
    
        x/d  0x0804b278   ==>  1001

        因此应输入1001

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  • 原文地址:https://www.cnblogs.com/cheukyin/p/4508987.html
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