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:

     8048b26:    8b                   mov    0x8(%ebp),%eax

     8048b29:     c4 f8                 add    $0xfffffff8,%esp

     8048b2c:     c0              push   $0x80497c0

     8048b31:                           push   %eax

     8048b32:    e8 f9              call    <strings_not_equal>

     8048b37:     c4                  add    $0x10,%esp

     8048b3a:     c0                    test   %eax,%eax

     8048b3c:                         je     8048b43 <phase_1+0x23>

     8048b3e:    e8 b9              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:

     8048b50:    8b                   mov    0x8(%ebp),%edx

     8048b53:     c4 f8                 add    $0xfffffff8,%esp

     8048b56:    8d  e8                 lea    -0x18(%ebp),%eax

     8048b59:                           push   %eax

     8048b5a:                           push   %edx

     8048b5b:    e8               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:

     8048fde:    8b 4d                  mov    0x8(%ebp),%ecx

     8048fe1:    8b  0c                 mov    0xc(%ebp),%edx

把用户字串指针存入ecx, ebp-0x18存入edx

往下看:

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

eax存入了 edx+0x14 这个值

再往下:

     8048fe7:                           push   %eax

     8048fe8:    8d                   lea    0x10(%edx),%eax

     8048feb:                           push   %eax

     8048fec:    8d  0c                 lea    0xc(%edx),%eax

     8048fef:                           push   %eax

     8048ff0:    8d                   lea    0x8(%edx),%eax

     8048ff3:                           push   %eax

     8048ff4:    8d                   lea    0x4(%edx),%eax

     8048ff7:                           push   %eax

     8048ff8:                           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

     8048ff9:     1b 9b             push   $0x8049b1b

     8048ffe:                           push   %ecx

     8048fff:    e8 5c f8 ff ff           call    <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:

     8048b63:     7d e8               cmpl   $0x1,-0x18(%ebp)

     8048b67:                         je     8048b6e <phase_2+0x26>

     8048b69:    e8 8e              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.

     8048b6e:    bb               mov    $0x1,%ebx

     8048b73:    8d  e8                 lea    -0x18(%ebp),%esi

令ebx=1, esi = ebp-18

     8048b76:    8d                   lea    0x1(%ebx),%eax

     8048b79:    0f af  9e fc           imul   -0x4(%esi,%ebx,),%eax

     8048b7e:      9e                 cmp    %eax,(%esi,%ebx,)

     8048b81:                         je     8048b88 <phase_2+0x40>

     8048b83:    e8               call   80494fc <explode_bomb>

     8048b88:                           inc    %ebx

     8048b89:     fb                  cmp    $0x5,%ebx

     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:

     ;; edx stores pointer of user input

     8048b9f:    8b                   mov    0x8(%ebp),%edx

     8048ba2:     c4 f4                 add    $0xfffffff4,%esp

     ;; push ebp-4 onto stack

     8048ba5:    8d  fc                 lea    -0x4(%ebp),%eax

     8048ba8:                           push   %eax

     ;; push ebp-5 onto stack

     8048ba9:    8d  fb                 lea    -0x5(%ebp),%eax

     8048bac:                           push   %eax

     ;; push ebp-12 onto stack

     8048bad:    8d  f4                 lea    -0xc(%ebp),%eax

     8048bb0:                           push   %eax

     ;; push $0x80497de onto stack

     ;; gdb x/s 0x80497de: "%d %c %d"

     8048bb1:     de              push   $0x80497de

     ;; push pointer of user input onto stack

     8048bb6:                           push   %edx

     8048bb7:    e8 a4 fc ff ff           call    <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:

     ;; (ebp-12) stores the first int, compare to 7

     ;; cmpl takes (ebp-12) as unsigned int

     8048bc9:     7d f4               cmpl   $0x7,-0xc(%ebp)

     ;; (unsigned)(ebp-12) > 7, jump to 0x8048c88, which will bomb

     8048bcd:    0f  b5           ja     8048c88 <phase_3+0xf0>

     ;; jump to *( 0x80497e8 + 4*(the first int) )

     8048bd3:    8b  f4                 mov    -0xc(%ebp),%eax

     8048bd6:    ff   e8        jmp    *0x80497e8(,%eax,)

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):

     ;; bl = 0x71

     8048be0:    b3                     mov    $0x71,%bl

     ;; if 0x309==777==the last int,

     ;; jump to 0x8048c8f, which will compare the char

     8048be2:     7d fc         cmpl   $0x309,-0x4(%ebp)

     8048be9:    0f  a0           je     8048c8f <phase_3+0xf7>

     8048bef:    e8               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

     ;; after compare the last int, jump here

     ;; bl = 0x71 = 'q', compare to the char

     ;; if ==, jump to 0x8048c99, and leave this function

     8048c8f:    3a 5d fb                 cmp    -0x5(%ebp),%bl

     8048c92:                         je     8048c99 <phase_3+0x101>

     8048c94:    e8               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:

     ;; edx = pointer of input string

     8048ce6:    8b                   mov    0x8(%ebp),%edx

     8048ce9:     c4 fc                 add    $0xfffffffc,%esp

     ;; eax = ebp-4

     8048cec:    8d  fc                 lea    -0x4(%ebp),%eax

     ;; push ebp-4

     8048cef:                           push   %eax

     ;; push $0x8049808

     ;; x/s 0x804980: "%d"

     8048cf0:                   push   $0x8049808

     ;; push pointer of input string

     8048cf5:                           push   %edx

     8048cf6:    e8  fb ff ff           call    <sscanf@plt>

就是读入一个整数,存入ebp-4

  ;; func4( input_number )

     8048d11:    8b  fc                 mov    -0x4(%ebp),%eax

     8048d14:                           push   %eax

     8048d15:    e8  ff ff ff           call   8048ca0 <func4>

     8048d1a:     c4                  add    $0x10,%esp

     ;; eax should contain the return value of <func4>

     ;; if eax == 0x37 == 55, defused

     8048d1d:     f8                  cmp    $0x37,%eax

     8048d20:                         je     8048d27 <phase_4+0x47>

     8048d22:    e8 d5              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>:

     ;; ebx = input_number

     8048ca8:    8b 5d                  mov    0x8(%ebp),%ebx

     ;; if input_number<=1, <func4> return 1

     8048cab:     fb                  cmp    $0x1,%ebx

     8048cae:    7e                     jle    8048cd0 <func4+0x30>

     8048cb0:     c4 f4                 add    $0xfffffff4,%esp

     ;; esi == func4( input_number-1 )

     8048cb3:    8d  ff                 lea    -0x1(%ebx),%eax

     8048cb6:                           push   %eax

     8048cb7:    e8 e4 ff ff ff           call   8048ca0 <func4>

     8048cbc:     c6                    mov    %eax,%esi

     8048cbe:     c4 f4                 add    $0xfffffff4,%esp

     ;; esi += func4( input_number-2 )

     8048cc1:    8d  fe                 lea    -0x2(%ebx),%eax

     8048cc4:                           push   %eax

     8048cc5:    e8 d6 ff ff ff           call   8048ca0 <func4>

     8048cca:     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:

     ;; ebx = pointer of input

     ;; push ebx onto stack

     ;; call string_length

     8048d34:    8b 5d                  mov    0x8(%ebp),%ebx

     8048d37:     c4 f4                 add    $0xfffffff4,%esp

     8048d3a:                           push   %ebx

     8048d3b:    e8 d8              call    <string_length>

     8048d40:     c4                  add    $0x10,%esp

     ;; eax stores the return value of string_length

     ;; if eax == 6, jump to 0x8048d4d 

     8048d43:     f8                  cmp    $0x6,%eax

     8048d46:                         je     8048d4d <phase_5+0x21>

     8048d48:    e8 af              call   80494fc <explode_bomb>

从上面代码可知,输入需要6个字符.

     ;; edx = 0

     8048d4d:     d2                    xor    %edx,%edx

     ;; ecx = ebp-8

     8048d4f:    8d 4d f8                 lea    -0x8(%ebp),%ecx

     ;; esi = 0x804b220

     8048d52:    be  b2             mov    $0x804b220,%esi

     ;; edx is a counter from 0 to 5

     ;; al = (edx + ebx), then al reads a char each time

     8048d57:    8a  1a                 mov    (%edx,%ebx,),%al

     ;; extract the low 4 bit of al

     8048d5a:     0f                    and    $0xf,%al

     ;; sign-extend al to eax

     8048d5c:    0f be c0                 movsbl %al,%eax

     ;; al = ( eax + 0x804b220 )

     ;; x/16c 0x804b220:

     ;; 0x804b220:    105 'i'    115 's'    114 'r'    118 'v'    101 'e'    97 'a'    119 'w'    104 'h'

     ;; 0x804b228:    111 'o'    98 'b'    112 'p'    110 'n'    117 'u'    116 't'    102 'f'    103 'g'

     8048d5f:    8a                   mov    (%eax,%esi,),%al

     ;; edx + ecx = al,

     ;; notice that, ecx = ebp-8

     ;; and edx is a counter from 0 to 5

     8048d62:      0a                 mov    %al,(%edx,%ecx,)

     8048d65:                           inc    %edx

     ;; loop

     8048d66:     fa                  cmp    $0x5,%edx

     8048d69:    7e ec                    jle    8048d57 <phase_5+0x2b>

     ;; ebp-2 = 0, a terminal of string started from ebp-8

     8048d6b:    c6  fe               movb   $0x0,-0x2(%ebp)

     8048d6f:     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:

     0x804b220:     'i'     's'     'r'     'v'     'e'     'a'     'w'     'h'

     0x804b228:     'o'     'b'     'p'     'n'     'u'     't'     'f'     'g'

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

继续看:

     ;; x/s 0x804980b: "giants"

     ;; push "giants"

     8048d72:     0b              push   $0x804980b

     ;; push ebp-8

     8048d77:    8d  f8                 lea    -0x8(%ebp),%eax

     8048d7a:                           push   %eax

     ;; compare "giants" and the string started from ebp-8

     8048d7b:    e8 b0              call    <strings_not_equal>

     8048d80:     c4                  add    $0x10,%esp

     8048d83:     c0                    test   %eax,%eax

     ;; if two strings equal to each other, defused

     8048d85:                         je     8048d8c <phase_5+0x60>

     8048d87:    e8               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 ]

     0x804b220:     'i'     's'     'r'     'v'     'e'     'a'     'w'     'h'

     0x804b228:     'o'     'b'     'p'     'n'     'u'     't'     'f'     'g'

因此,

     input[]&0xf = 0xf, input[]&0xf = 0x0,

     input[]&0xf = 0x5, input[]&0xf = 0xb,

     input[]&0xf = 0xd, input[]&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:

     ;; edx = pointer of input

     8048da1:    8b                   mov    0x8(%ebp),%edx

     ;; (ebp-0x34) = $0x804b26c

     8048da4:    c7  cc 6c b2       movl   $0x804b26c,-0x34(%ebp)

     8048dab:     c4 f8                 add    $0xfffffff8,%esp

     ;; read six numbers from input,

     ;; and storse in the area started from ebp-18

     8048dae:    8d  e8                 lea    -0x18(%ebp),%eax

     8048db1:                           push   %eax

     8048db2:                           push   %edx

     8048db3:    e8               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.

     ;; edi = 0

     8048db8:     ff                    xor    %edi,%edi

     8048dba:     c4                  add    $0x10,%esp

     8048dbd:    8d                   lea    0x0(%esi),%esi

     ;; eax = (ebp-0x18 + 4*edi) = six-number[edi]

     ;; ebp-0x18 = the beginning address of the six numbers

     ;; edi is a counter from 0 to 5

     8048dc0:    8d  e8                 lea    -0x18(%ebp),%eax

     8048dc3:    8b  b8                 mov    (%eax,%edi,),%eax

     ;; eax = six-number[edi]-1

     8048dc6:                           dec    %eax

     ;; if eax <= 5 , continue

     8048dc7:     f8                  cmp    $0x5,%eax

     8048dca:                         jbe    8048dd1 <phase_6+0x39>

     8048dcc:    e8 2b              call   80494fc <explode_bomb>

     ;; if edi+1 > 5, finish edi loop

     8048dd1:    8d 5f                  lea    0x1(%edi),%ebx

     8048dd4:     fb                  cmp    $0x5,%ebx

     8048dd7:    7f                     jg     8048dfc <phase_6+0x64>

     ;; (ebp-0x38) = edi*4

     8048dd9:    8d  bd         lea    0x0(,%edi,),%eax

     8048de0:      c8                 mov    %eax,-0x38(%ebp)

     ;; esi = ebp-18 = the beginning address of the six numbers

     8048de3:    8d  e8                 lea    -0x18(%ebp),%esi

     ;; edx = (ebp-0x38) = edi*4

     ;; inner loops,

     ;; ebx is the counter from edi+1 to 5

     8048de6:    8b  c8                 mov    -0x38(%ebp),%edx

     ;; eax = edx + esi = six-number[edi]

     8048de9:    8b                   mov    (%edx,%esi,),%eax

     ;; compare six-number[edi] and six-number[edi+ebx]

     8048dec:    3b  9e                 cmp    (%esi,%ebx,),%eax

     ;; if six-number[edi] != six-number[edi+1], continue

     8048def:                         jne    8048df6 <phase_6+0x5e>

     8048df1:    e8               call   80494fc <explode_bomb>

     ;; ebx++

     ;; if ebx<=5, jump to 0x8048de6, ebx loops

     ;; else , finish ebx loop

     8048df6:                           inc    %ebx

     8048df7:     fb                  cmp    $0x5,%ebx

     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>:

     ;; every time call read_line, ( 0x804b480 )++

     ;; only with 6 correct answer given ,will the secret phase appear

     :     3d  b4        cmpl   $0x6,0x804b480

     804953a:                         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.

     ;; push ebp-0x50

     804953c:    8d 5d b0                 lea    -0x50(%ebp),%ebx

     804953f:                           push   %ebx

     ;; push ebp-0x54

     :    8d  ac                 lea    -0x54(%ebp),%eax

     :                           push   %eax

     ;; (gdb) x/s 0x8049d03

     ;; 0x8049d03:    "%d %s"

     :      9d             push   $0x8049d03

     ;; push the string stores in 0x804b770

     ;; the address of input of phase 4

     :      b7             push   $0x804b770

     804954e:    e8 0d f3 ff ff           call    <sscanf@plt>

     ....

     ;; (gdb) x/s 0x8049d09

     ;; 0x8049d09:    "austinpowers"

     804955e:      9d             push   $0x8049d09

     ;; push the %s

     :                           push   %ebx

     :    e8 c7 fa ff ff           call    <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>:

     8048eef:    e8               call   80491fc <read_line>

     8048ef4:    6a                     push   $0x0

     ;; strtol( user input string, 0, 10)

     ;; long int strtol(const char *nptr, char **endptr, int base);

     ;; converts the initial part of the string in nptr to a long integer value according to the given base

     8048ef6:    6a 0a                    push   $0xa

     8048ef8:    6a                     push   $0x0

     8048efa:                           push   %eax

     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.

     ;; if fun7( 0x804b320, the input long int )

     ;; x/d 0x804b320: (0x804b320) = 36

     8048f17:                           push   %ebx

     8048f18:      b3             push   $0x804b320

     8048f1d:    e8  ff ff ff           call   8048e94 <fun7>

     8048f22:     c4                  add    $0x10,%esp

     ;; if fun7(0x804b320, the input long int) == 7, defused

     8048f25:     f8                  cmp    $0x7,%eax

     8048f28:                         je     8048f2f <secret_phase+0x47>

     8048f2a:    e8 cd              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>:

     ;; edx = the first parameter, an address

     8048e9a:    8b                   mov    0x8(%ebp),%edx

     ;; eax = the input long int

     8048e9d:    8b  0c                 mov    0xc(%ebp),%eax

     ;; if edx != 0

     8048ea0:     d2                    test   %edx,%edx

     8048ea2:     0c                    jne    8048eb0 <fun7+0x1c>

     8048ea4:    b8 ff ff ff ff           mov    $0xffffffff,%eax

     8048ea9:    eb                     jmp    8048ee2 <fun7+0x4e>

     8048eab:                           nop

     8048eac:    8d                 lea    0x0(%esi,%eiz,),%esi

     ;; if the input number >= (edx), jump to 0x8048ec5 

     8048eb0:    3b                     cmp    (%edx),%eax

     8048eb2:    7d                     jge    8048ec5 <fun7+0x31>

     ;; eax > (edx)

     8048eb4:     c4 f8                 add    $0xfffffff8,%esp

     ;; <func7>( (edx+4) ,the input long int )

     8048eb7:                           push   %eax

     8048eb8:    8b                   mov    0x4(%edx),%eax

     8048ebb:                           push   %eax

     8048ebc:    e8 d3 ff ff ff           call   8048e94 <fun7>

     ;; return eax *= 2, exit

     8048ec1:     c0                    add    %eax,%eax

     8048ec3:    eb 1d                    jmp    8048ee2 <fun7+0x4e>

     ;; the input number >= (edx)

     ;; if eax == (edx), return eax=0

     8048ec5:    3b                     cmp    (%edx),%eax

     8048ec7:                         je     8048ee0 <fun7+0x4c>

     ;; the input number > (edx)

     8048ec9:     c4 f8                 add    $0xfffffff8,%esp

     ;; <fun7>( (edx+8) ,the input long int )

     8048ecc:                           push   %eax

     8048ecd:    8b                   mov    0x8(%edx),%eax

     8048ed0:                           push   %eax

     8048ed1:    e8 be ff ff ff           call   8048e94 <fun7>

     ;; fun7 return 2*eax + 1

     8048ed6:     c0                    add    %eax,%eax

     8048ed8:                           inc    %eax

     8048ed9:    eb                     jmp    8048ee2 <fun7+0x4e>

     8048edb:                           nop

     8048edc:    8d                 lea    0x0(%esi,%eiz,),%esi

     8048ee0:     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) = 

    当 number > *(int*)address, fun7( address, number) = *fun7( address+, number ) + 

    当 number < *(int*)address, fun7( address, number) = *fun7( address+, 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):

     struct BST

     {

         int num;

         struct BST *left;

         struct BST *right;

     } *bst;

则上面的递推式可表示为:

    当 number == bst->num, fun7( bst, number ) = ;

    当 number > bst->num, fun7( bst, number ) = *fun7( bst->right, number ) + ;

    当 number < bst->num, fun7( bst, number ) = *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 )

    =    * fun7( (struct BST *)0x804b320->right, number ) + 

    =    * ( * fun7( (struct BST *)0x804b320->right->right, number ) + ) + 

    =    * fun7( (struct BST *)0x804b320->right->right, number ) + 

    =    * ( * fun7( (struct BST *)0x804b320->right->right->right, number ) + ) + 

    =    * fun7( (struct BST *)0x804b320->right->right->right, number ) + 

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

用gdb查看,

    x/wx 0x804b320+  ==>  0x0804b308 

    x/wx 0x804b308+  ==>  0x0804b2d8

    x/wx 0x804b2d8+  ==>  0x0804b278

    x/d  0x0804b278   ==>  

因此应输入1001