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,%esp12 13 8048b3a: 85 c0 test %eax,%eax14 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 %edx10 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),%ecx2 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 %eax10 11 8048ff0: 8d 42 08 lea 0x8(%edx),%eax12 13 8048ff3: 50 push %eax14 15 8048ff4: 8d 42 04 lea 0x4(%edx),%eax16 17 8048ff7: 50 push %eax18 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 $0x8049b1b2 3 8048ffe: 51 push %ecx4 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,%ebx2 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 %ebx12 13 8048b89: 83 fb 05 cmp $0x5,%ebx14 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),%eax10 11 8048ba8: 50 push %eax12 13 ;; push ebp-5 onto stack14 15 8048ba9: 8d 45 fb lea -0x5(%ebp),%eax16 17 8048bac: 50 push %eax18 19 ;; push ebp-12 onto stack20 21 8048bad: 8d 45 f4 lea -0xc(%ebp),%eax22 23 8048bb0: 50 push %eax24 25 ;; push $0x80497de onto stack26 27 ;; gdb x/s 0x80497de: "%d %c %d"28 29 8048bb1: 68 de 97 04 08 push $0x80497de30 31 ;; push pointer of user input onto stack32 33 8048bb6: 52 push %edx34 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),%eax14 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),%eax10 11 ;; push ebp-412 13 8048cef: 50 push %eax14 15 ;; push $0x804980816 17 ;; x/s 0x804980: "%d"18 19 8048cf0: 68 08 98 04 08 push $0x804980820 21 ;; push pointer of input string22 23 8048cf5: 52 push %edx24 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,%esp12 13 ;; eax should contain the return value of <func4>14 15 ;; if eax == 0x37 == 55, defused16 17 8048d1d: 83 f8 37 cmp $0x37,%eax18 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,%esp14 15 ;; esi == func4( input_number-1 )16 17 8048cb3: 8d 43 ff lea -0x1(%ebx),%eax18 19 8048cb6: 50 push %eax20 21 8048cb7: e8 e4 ff ff ff call 8048ca0 <func4>22 23 8048cbc: 89 c6 mov %eax,%esi24 25 26 27 8048cbe: 83 c4 f4 add $0xfffffff4,%esp28 29 30 31 ;; esi += func4( input_number-2 )32 33 8048cc1: 8d 43 fe lea -0x2(%ebx),%eax34 35 8048cc4: 50 push %eax36 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,%esp10 11 8048d3a: 53 push %ebx12 13 8048d3b: e8 d8 02 00 00 call 8049018 <string_length>14 15 16 17 8048d40: 83 c4 10 add $0x10,%esp18 19 ;; eax stores the return value of string_length20 21 ;; if eax == 6, jump to 0x8048d4d 22 23 8048d43: 83 f8 06 cmp $0x6,%eax24 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 = 0x804b22010 11 8048d52: be 20 b2 04 08 mov $0x804b220,%esi12 13 ;; edx is a counter from 0 to 514 15 ;; al = (edx + ebx), then al reads a char each time16 17 8048d57: 8a 04 1a mov (%edx,%ebx,1),%al18 19 ;; extract the low 4 bit of al20 21 8048d5a: 24 0f and $0xf,%al22 23 ;; sign-extend al to eax24 25 8048d5c: 0f be c0 movsbl %al,%eax26 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),%al36 37 ;; edx + ecx = al,38 39 ;; notice that, ecx = ebp-840 41 ;; and edx is a counter from 0 to 542 43 8048d62: 88 04 0a mov %al,(%edx,%ecx,1)44 45 8048d65: 42 inc %edx46 47 ;; loop48 49 8048d66: 83 fa 05 cmp $0x5,%edx50 51 8048d69: 7e ec jle 8048d57 <phase_5+0x2b>52 53 54 55 ;; ebp-2 = 0, a terminal of string started from ebp-856 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),%eax10 11 8048d7a: 50 push %eax12 13 ;; compare "giants" and the string started from ebp-814 15 8048d7b: e8 b0 02 00 00 call 8049030 <strings_not_equal>16 17 8048d80: 83 c4 10 add $0x10,%esp18 19 8048d83: 85 c0 test %eax,%eax20 21 ;; if two strings equal to each other, defused22 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,%esp10 11 ;; read six numbers from input,12 13 ;; and storse in the area started from ebp-1814 15 8048dae: 8d 45 e8 lea -0x18(%ebp),%eax16 17 8048db1: 50 push %eax18 19 8048db2: 52 push %edx20 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 numbers12 13 ;; edi is a counter from 0 to 514 15 8048dc0: 8d 45 e8 lea -0x18(%ebp),%eax16 17 8048dc3: 8b 04 b8 mov (%eax,%edi,4),%eax18 19 ;; eax = six-number[edi]-120 21 8048dc6: 48 dec %eax22 23 ;; if eax <= 5 , continue24 25 8048dc7: 83 f8 05 cmp $0x5,%eax26 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 loop34 35 8048dd1: 8d 5f 01 lea 0x1(%edi),%ebx36 37 8048dd4: 83 fb 05 cmp $0x5,%ebx38 39 8048dd7: 7f 23 jg 8048dfc <phase_6+0x64>40 41 42 43 ;; (ebp-0x38) = edi*444 45 8048dd9: 8d 04 bd 00 00 00 00 lea 0x0(,%edi,4),%eax46 47 8048de0: 89 45 c8 mov %eax,-0x38(%ebp)48 49 50 51 ;; esi = ebp-18 = the beginning address of the six numbers52 53 8048de3: 8d 75 e8 lea -0x18(%ebp),%esi54 55 ;; edx = (ebp-0x38) = edi*456 57 ;; inner loops,58 59 ;; ebx is the counter from edi+1 to 560 61 8048de6: 8b 55 c8 mov -0x38(%ebp),%edx62 63 ;; eax = edx + esi = six-number[edi]64 65 8048de9: 8b 04 32 mov (%edx,%esi,1),%eax66 67 ;; compare six-number[edi] and six-number[edi+ebx]68 69 8048dec: 3b 04 9e cmp (%esi,%ebx,4),%eax70 71 ;; if six-number[edi] != six-number[edi+1], continue72 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 loops80 81 ;; else , finish ebx loop82 83 8048df6: 43 inc %ebx84 85 8048df7: 83 fb 05 cmp $0x5,%ebx86 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 appear4 5 8049533: 83 3d 80 b4 04 08 06 cmpl $0x6,0x804b4806 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),%eax10 11 8049543: 50 push %eax12 13 ;; (gdb) x/s 0x8049d0314 15 ;; 0x8049d03: "%d %s"16 17 8049544: 68 03 9d 04 08 push $0x8049d0318 19 ;; push the string stores in 0x804b77020 21 ;; the address of input of phase 422 23 8049549: 68 70 b7 04 08 push $0x804b77024 25 804954e: e8 0d f3 ff ff call 8048860 <sscanf@plt>26 27 28 29 ....30 31 32 33 ;; (gdb) x/s 0x8049d0934 35 ;; 0x8049d09: "austinpowers"36 37 804955e: 68 09 9d 04 08 push $0x8049d0938 39 ;; push the %s40 41 8049563: 53 push %ebx42 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 base14 15 8048ef6: 6a 0a push $0xa16 17 8048ef8: 6a 00 push $0x018 19 8048efa: 50 push %eax20 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,%esp14 15 ;; if fun7(0x804b320, the input long int) == 7, defused16 17 8048f25: 83 f8 07 cmp $0x7,%eax18 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),%esi100 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
