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 (edx) >= the input long int, jump to 0x8048ec5   30   31     8048eb0:    3b 02                    cmp    (%edx),%eax  32   33     8048eb2:    7d 11                    jge    8048ec5 <fun7+0x31>  34   35   36   37     ;; (0x804b320) < eax  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     ;; (edx) >= the input long int  62   63     ;; if (edx) == eax, return eax=0  64   65     8048ec5:    3b 02                    cmp    (%edx),%eax  66   67     8048ec7:    74 17                    je     8048ee0 <fun7+0x4c>  68   69   70   71     ;; (edx) > the input long int  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