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
