前言
这是CSAPP官网上的著名实验，二进制炸弹,
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
实验材料可到我的github仓库 https://github.com/Cheukyin/C... 下载
预备
反汇编：objdump -d bomb > bomb_assembly_32.S
Phase 1
打开bomb_assembly_32.S,定位到·
8048b26: 8b 45 08 mov 0x8(%ebp),%eax
8048b29: 83 c4 f8 add $0xfffffff8,%esp
8048b2c: 68 c0 97 04 08 push $0x80497c0
8048b31: 50 push %eax
8048b32: e8 f9 04 00 00 call 8049030
8048b37: 83 c4 10 add $0x10,%esp
8048b3a: 85 c0 test %eax,%eax
8048b3c: 74 05 je 8048b43
8048b3e: e8 b9 09 00 00 call 80494fc 可以看出,用户输入字串指针保存在0x8(%ebp), 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
待
若equal则进入下一phase,否则
从
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
8048b50: 8b 55 08 mov 0x8(%ebp),%edx
8048b53: 83 c4 f8 add $0xfffffff8,%esp
8048b56: 8d 45 e8 lea -0x18(%ebp),%eax
8048b59: 50 push %eax
8048b5a: 52 push %edx
8048b5b: e8 78 04 00 00 call 8048fd8 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 相当于read_six_numbers( 用户字串指针地址, ebp-0x18 )
现在我们切换到
先来看下面2行:
8048fde: 8b 4d 08 mov 0x8(%ebp),%ecx
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:
8048fe4: 8d 42 14 lea 0x14(%edx),%eaxeax存入了 edx+0x14 这个值
再往下:
8048fe7: 50 push %eax
8048fe8: 8d 42 10 lea 0x10(%edx),%eax
8048feb: 50 push %eax
8048fec: 8d 42 0c lea 0xc(%edx),%eax
8048fef: 50 push %eax
8048ff0: 8d 42 08 lea 0x8(%edx),%eax
8048ff3: 50 push %eax
8048ff4: 8d 42 04 lea 0x4(%edx),%eax
8048ff7: 50 push %eax
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stack frame的 ebp-0x18 这个地址值
8048ff9: 68 1b 9b 04 08 push $0x8049b1b
8048ffe: 51 push %ecx
8048fff: e8 5c f8 ff ff call 8048860 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, 6个数字, 存入stack frame中的(ebp-0x18)中
回到
8048b63: 83 7d e8 01 cmpl $0x1,-0x18(%ebp)
8048b67: 74 05 je 8048b6e
8048b69: e8 8e 09 00 00 call 80494fc 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 01 00 00 00 mov $0x1,%ebx
8048b73: 8d 75 e8 lea -0x18(%ebp),%esi令ebx=1, esi = ebp-18
8048b76: 8d 43 01 lea 0x1(%ebx),%eax
8048b79: 0f af 44 9e fc imul -0x4(%esi,%ebx,4),%eax
8048b7e: 39 04 9e cmp %eax,(%esi,%ebx,4)
8048b81: 74 05 je 8048b88
8048b83: e8 74 09 00 00 call 80494fc
8048b88: 43 inc %ebx
8048b89: 83 fb 05 cmp $0x5,%ebx
8048b8c: 7e e8 jle 8048b76 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a[1]=1, a[n+1] = a[n]*(n+1), n<=61, 2, 6, 24, 120, 720
Phase 3
打开bomb_assembly_32.S,定位到
;; edx stores pointer of user input
8048b9f: 8b 55 08 mov 0x8(%ebp),%edx
8048ba2: 83 c4 f4 add $0xfffffff4,%esp
;; push ebp-4 onto stack
8048ba5: 8d 45 fc lea -0x4(%ebp),%eax
8048ba8: 50 push %eax
;; push ebp-5 onto stack
8048ba9: 8d 45 fb lea -0x5(%ebp),%eax
8048bac: 50 push %eax
;; push ebp-12 onto stack
8048bad: 8d 45 f4 lea -0xc(%ebp),%eax
8048bb0: 50 push %eax
;; push $0x80497de onto stack
;; gdb x/s 0x80497de: "%d %c %d"
8048bb1: 68 de 97 04 08 push $0x80497de
;; push pointer of user input onto stack
8048bb6: 52 push %edx
8048bb7: e8 a4 fc ff ff call 8048860 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: 83 7d f4 07 cmpl $0x7,-0xc(%ebp)
;; (unsigned)(ebp-12) > 7, jump to 0x8048c88, which will bomb
8048bcd: 0f 87 b5 00 00 00 ja 8048c88
;; jump to *( 0x80497e8 + 4*(the first int) )
8048bd3: 8b 45 f4 mov -0xc(%ebp),%eax
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):
;; bl = 0x71
8048be0: b3 71 mov $0x71,%bl
;; if 0x309==777==the last int,
;; jump to 0x8048c8f, which will compare the char
8048be2: 81 7d fc 09 03 00 00 cmpl $0x309,-0x4(%ebp)
8048be9: 0f 84 a0 00 00 00 je 8048c8f
8048bef: e8 08 09 00 00 call 80494fc 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: 74 05 je 8048c99
8048c94: e8 63 08 00 00 call 80494fc 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
;; edx = pointer of input string
8048ce6: 8b 55 08 mov 0x8(%ebp),%edx
8048ce9: 83 c4 fc add $0xfffffffc,%esp
;; eax = ebp-4
8048cec: 8d 45 fc lea -0x4(%ebp),%eax
;; push ebp-4
8048cef: 50 push %eax
;; push $0x8049808
;; x/s 0x804980: "%d"
8048cf0: 68 08 98 04 08 push $0x8049808
;; push pointer of input string
8048cf5: 52 push %edx
8048cf6: e8 65 fb ff ff call 8048860 就是读入一个整数,存入ebp-4
;; func4( input_number )
8048d11: 8b 45 fc mov -0x4(%ebp),%eax
8048d14: 50 push %eax
8048d15: e8 86 ff ff ff call 8048ca0
8048d1a: 83 c4 10 add $0x10,%esp
;; eax should contain the return value of
;; if eax == 0x37 == 55, defused
8048d1d: 83 f8 37 cmp $0x37,%eax
8048d20: 74 05 je 8048d27
8048d22: e8 d5 07 00 00 call 80494fc 然后比较func4( input_number )==55, 若等于则成功defuse.
接下来看看
;; ebx = input_number
8048ca8: 8b 5d 08 mov 0x8(%ebp),%ebx
;; if input_number<=1, return 1
8048cab: 83 fb 01 cmp $0x1,%ebx
8048cae: 7e 20 jle 8048cd0
8048cb0: 83 c4 f4 add $0xfffffff4,%esp
;; esi == func4( input_number-1 )
8048cb3: 8d 43 ff lea -0x1(%ebx),%eax
8048cb6: 50 push %eax
8048cb7: e8 e4 ff ff ff call 8048ca0
8048cbc: 89 c6 mov %eax,%esi
8048cbe: 83 c4 f4 add $0xfffffff4,%esp
;; esi += func4( input_number-2 )
8048cc1: 8d 43 fe lea -0x2(%ebx),%eax
8048cc4: 50 push %eax
8048cc5: e8 d6 ff ff ff call 8048ca0
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
;; ebx = pointer of input
;; push ebx onto stack
;; call string_length
8048d34: 8b 5d 08 mov 0x8(%ebp),%ebx
8048d37: 83 c4 f4 add $0xfffffff4,%esp
8048d3a: 53 push %ebx
8048d3b: e8 d8 02 00 00 call 8049018
8048d40: 83 c4 10 add $0x10,%esp
;; eax stores the return value of string_length
;; if eax == 6, jump to 0x8048d4d
8048d43: 83 f8 06 cmp $0x6,%eax
8048d46: 74 05 je 8048d4d
8048d48: e8 af 07 00 00 call 80494fc 从上面代码可知,输入需要6个字符.
;; edx = 0
8048d4d: 31 d2 xor %edx,%edx
;; ecx = ebp-8
8048d4f: 8d 4d f8 lea -0x8(%ebp),%ecx
;; esi = 0x804b220
8048d52: be 20 b2 04 08 mov $0x804b220,%esi
;; edx is a counter from 0 to 5
;; al = (edx + ebx), then al reads a char each time
8048d57: 8a 04 1a mov (%edx,%ebx,1),%al
;; extract the low 4 bit of al
8048d5a: 24 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 04 30 mov (%eax,%esi,1),%al
;; edx + ecx = al,
;; notice that, ecx = ebp-8
;; and edx is a counter from 0 to 5
8048d62: 88 04 0a mov %al,(%edx,%ecx,1)
8048d65: 42 inc %edx
;; loop
8048d66: 83 fa 05 cmp $0x5,%edx
8048d69: 7e ec jle 8048d57
;; ebp-2 = 0, a terminal of string started from ebp-8
8048d6b: c6 45 fe 00 movb $0x0,-0x2(%ebp)
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:
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'获取0x804b220[ input[k] & 0xf ]后,将之copy至 (ebp-8)[k]
继续看:
;; x/s 0x804980b: "giants"
;; push "giants"
8048d72: 68 0b 98 04 08 push $0x804980b
;; push ebp-8
8048d77: 8d 45 f8 lea -0x8(%ebp),%eax
8048d7a: 50 push %eax
;; compare "giants" and the string started from ebp-8
8048d7b: e8 b0 02 00 00 call 8049030
8048d80: 83 c4 10 add $0x10,%esp
8048d83: 85 c0 test %eax,%eax
;; if two strings equal to each other, defused
8048d85: 74 05 je 8048d8c
8048d87: e8 70 07 00 00 call 80494fc 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: 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'因此,
input[0]&0xf = 0xf, input[1]&0xf = 0x0,
input[2]&0xf = 0x5, input[3]&0xf = 0xb,
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:
;; edx = pointer of input
8048da1: 8b 55 08 mov 0x8(%ebp),%edx
;; (ebp-0x34) = $0x804b26c
8048da4: c7 45 cc 6c b2 04 08 movl $0x804b26c,-0x34(%ebp)
8048dab: 83 c4 f8 add $0xfffffff8,%esp
;; read six numbers from input,
;; and storse in the area started from ebp-18
8048dae: 8d 45 e8 lea -0x18(%ebp),%eax
8048db1: 50 push %eax
8048db2: 52 push %edx
8048db3: e8 20 02 00 00 call 8048fd8 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: 31 ff xor %edi,%edi
8048dba: 83 c4 10 add $0x10,%esp
8048dbd: 8d 76 00 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 45 e8 lea -0x18(%ebp),%eax
8048dc3: 8b 04 b8 mov (%eax,%edi,4),%eax
;; eax = six-number[edi]-1
8048dc6: 48 dec %eax
;; if eax <= 5 , continue
8048dc7: 83 f8 05 cmp $0x5,%eax
8048dca: 76 05 jbe 8048dd1
8048dcc: e8 2b 07 00 00 call 80494fc
;; if edi+1 > 5, finish edi loop
8048dd1: 8d 5f 01 lea 0x1(%edi),%ebx
8048dd4: 83 fb 05 cmp $0x5,%ebx
8048dd7: 7f 23 jg 8048dfc
;; (ebp-0x38) = edi*4
8048dd9: 8d 04 bd 00 00 00 00 lea 0x0(,%edi,4),%eax
8048de0: 89 45 c8 mov %eax,-0x38(%ebp)
;; esi = ebp-18 = the beginning address of the six numbers
8048de3: 8d 75 e8 lea -0x18(%ebp),%esi
;; edx = (ebp-0x38) = edi*4
;; inner loops,
;; ebx is the counter from edi+1 to 5
8048de6: 8b 55 c8 mov -0x38(%ebp),%edx
;; eax = edx + esi = six-number[edi]
8048de9: 8b 04 32 mov (%edx,%esi,1),%eax
;; compare six-number[edi] and six-number[edi+ebx]
8048dec: 3b 04 9e cmp (%esi,%ebx,4),%eax
;; if six-number[edi] != six-number[edi+1], continue
8048def: 75 05 jne 8048df6
8048df1: e8 06 07 00 00 call 80494fc
;; ebx++
;; if ebx<=5, jump to 0x8048de6, ebx loops
;; else , finish ebx loop
8048df6: 43 inc %ebx
8048df7: 83 fb 05 cmp $0x5,%ebx
8048dfa: 7e ea jle 8048de6 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
先来看看
;; every time call read_line, ( 0x804b480 )++
;; only with 6 correct answer given ,will the secret phase appear
8049533: 83 3d 80 b4 04 08 06 cmpl $0x6,0x804b480
804953a: 75 63 jne 804959f (0x804b480)是一个计数器,每当调用一次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: 53 push %ebx
;; push ebp-0x54
8049540: 8d 45 ac lea -0x54(%ebp),%eax
8049543: 50 push %eax
;; (gdb) x/s 0x8049d03
;; 0x8049d03: "%d %s"
8049544: 68 03 9d 04 08 push $0x8049d03
;; push the string stores in 0x804b770
;; the address of input of phase 4
8049549: 68 70 b7 04 08 push $0x804b770
804954e: e8 0d f3 ff ff call 8048860
....
;; (gdb) x/s 0x8049d09
;; 0x8049d09: "austinpowers"
804955e: 68 09 9d 04 08 push $0x8049d09
;; push the %s
8049563: 53 push %ebx
8049564: e8 c7 fa ff ff call 8049030 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
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
现在看看
8048eef: e8 08 03 00 00 call 80491fc
8048ef4: 6a 00 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 00 push $0x0
8048efa: 50 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: 53 push %ebx
8048f18: 68 20 b3 04 08 push $0x804b320
8048f1d: e8 72 ff ff ff call 8048e94
8048f22: 83 c4 10 add $0x10,%esp
;; if fun7(0x804b320, the input long int) == 7, defused
8048f25: 83 f8 07 cmp $0x7,%eax
8048f28: 74 05 je 8048f2f
8048f2a: e8 cd 05 00 00 call 80494fc 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
;; edx = the first parameter, an address
8048e9a: 8b 55 08 mov 0x8(%ebp),%edx
;; eax = the input long int
8048e9d: 8b 45 0c mov 0xc(%ebp),%eax
;; if edx != 0
8048ea0: 85 d2 test %edx,%edx
8048ea2: 75 0c jne 8048eb0
8048ea4: b8 ff ff ff ff mov $0xffffffff,%eax
8048ea9: eb 37 jmp 8048ee2
8048eab: 90 nop
8048eac: 8d 74 26 00 lea 0x0(%esi,%eiz,1),%esi
;; if the input number >= (edx), jump to 0x8048ec5
8048eb0: 3b 02 cmp (%edx),%eax
8048eb2: 7d 11 jge 8048ec5
;; eax > (edx)
8048eb4: 83 c4 f8 add $0xfffffff8,%esp
;; ( (edx+4) ,the input long int )
8048eb7: 50 push %eax
8048eb8: 8b 42 04 mov 0x4(%edx),%eax
8048ebb: 50 push %eax
8048ebc: e8 d3 ff ff ff call 8048e94
;; return eax *= 2, exit
8048ec1: 01 c0 add %eax,%eax
8048ec3: eb 1d jmp 8048ee2
;; the input number >= (edx)
;; if eax == (edx), return eax=0
8048ec5: 3b 02 cmp (%edx),%eax
8048ec7: 74 17 je 8048ee0
;; the input number > (edx)
8048ec9: 83 c4 f8 add $0xfffffff8,%esp
;; ( (edx+8) ,the input long int )
8048ecc: 50 push %eax
8048ecd: 8b 42 08 mov 0x8(%edx),%eax
8048ed0: 50 push %eax
8048ed1: e8 be ff ff ff call 8048e94
;; fun7 return 2*eax + 1
8048ed6: 01 c0 add %eax,%eax
8048ed8: 40 inc %eax
8048ed9: eb 07 jmp 8048ee2
8048edb: 90 nop
8048edc: 8d 74 26 00 lea 0x0(%esi,%eiz,1),%esi
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):
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 ) = 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