摘要：前言这是官网上的著名实验，二进制炸弹通过和反汇编猜测程序意图，共有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仓库下载预备反汇编打开定位到函数可以看到以下代码可以看出
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·函数,可以看到以下代码:
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), 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
待返回后,测试返回值,
若equal则进入下一phase,否则
从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,定位到函数,留意以下几行:
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,看看这个函数是干什么的:
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),%eax
eax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是的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, 作用就是从用户字串中提取6个数字, 存入 stack frame中的(ebp-0x18)中
回到接着看:
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,定位到函数,可以看到以下代码:
;; 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,定位到函数,可以看到以下代码:
;; 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,定位到函数,可以看到以下代码:
;; 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
首先要找到的入口,经搜索发现入口是在里面.
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: 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
