CSAPP(Computer Systems A Programmer's Perspective)，中译名为深入理解计算机系统，是一本优秀的计算机教材。该书配套了若干个课后实验，可供读者检验所学知识。其中bomb lab(从该页面下载实验材料)检验的是读者的汇编知识。这个实验笔者久闻大名但一直没有做。最近花时间研究了一番发现果然有趣。

该实验需在linux系统下进行，分六关，每关实验者需要输入符合条件的字符串才能进入下一关。通过六关视为拆弹成功。实验提供了一个可执行文件bomb和bomb.c以及相应的实验说明。

第一关

终端运行bomb

$ ./bomb

终端显示

Welcome to my fiendish little bomb. You have 6 phases with

which to blow yourself up. Have a nice day!

实验者需要输入符合条件的字符串才能进入下一关。

打开bomb.c，main函数代码如下

int main(int argc, char *argv[])

{

char *input;

/* Note to self: remember to port this bomb to Windows and put a

* fantastic GUI on it. */

/* When run with no arguments, the bomb reads its input lines

* from standard input. */

if (argc == 1) {

infile = stdin;

}

/* When run with one argument , the bomb reads from

* until EOF, and then switches to standard input. Thus, as you

* defuse each phase, you can add its defusing string to and

* avoid having to retype it. */

else if (argc == 2) {

if (!(infile = fopen(argv[1], "r"))) {

printf("%s: Error: Couldn't open %s\n", argv[0], argv[1]);

exit(8);

}

}

/* You can't call the bomb with more than 1 command line argument. */

else {

printf("Usage: %s []\n", argv[0]);

exit(8);

}

/* Do all sorts of secret stuff that makes the bomb harder to defuse. */

initialize_bomb();

printf("Welcome to my fiendish little bomb. You have 6 phases with\n");

printf("which to blow yourself up. Have a nice day!\n");

/* Hmm... Six phases must be more secure than one phase! */

input = read_line(); /* Get input */

phase_1(input); /* Run the phase */

phase_defused(); /* Drat! They figured it out!

* Let me know how they did it. */

printf("Phase 1 defused. How about the next one?\n");

/* The second phase is harder. No one will ever figure out

* how to defuse this... */

input = read_line();

phase_2(input);

phase_defused();

printf("That's number 2. Keep going!\n");

/* I guess this is too easy so far. Some more complex code will

* confuse people. */

input = read_line();

phase_3(input);

phase_defused();

printf("Halfway there!\n");

/* Oh yeah? Well, how good is your math? Try on this saucy problem! */

input = read_line();

phase_4(input);

phase_defused();

printf("So you got that one. Try this one.\n");

/* Round and 'round in memory we go, where we stop, the bomb blows! */

input = read_line();

phase_5(input);

phase_defused();

printf("Good work! On to the next...\n");

/* This phase will never be used, since no one will get past the

* earlier ones. But just in case, make this one extra hard. */

input = read_line();

phase_6(input);

phase_defused();

/* Wow, they got it! But isn't something... missing? Perhaps

* something they overlooked? Mua ha ha ha ha! */

return 0;

}

由代码可知函数phase_1~phase_6是关键函数，如果能够知道这些函数的实现，我们就能推导出符合条件的字符串，拆除炸弹。然而实验材料里并不包含phase.c。这时需要逆向工程反推出这些函数的运行机制。

终端输入

$ objdump -d bomb > bomb.S

将反汇编bomb得到的汇编代码存储到bomb.S文件中

phase_1的汇编代码如下所示

0000000000400ee0 :

400ee0: 48 83 ec 08 sub $0x8,%rsp

400ee4: be 00 24 40 00 mov $0x402400,%esi

400ee9: e8 4a 04 00 00 callq 401338

400eee: 85 c0 test %eax,%eax

400ef0: 74 05 je 400ef7

400ef2: e8 43 05 00 00 callq 40143a

400ef7: 48 83 c4 08 add $0x8,%rsp

400efb: c3 retq

从汇编代码可知，phase_1会将用户输入的字符串与内存中的某一特定字符串s做比较，如果相同过关，不同则炸弹爆炸。

如何知道字符串s的内容？借助gdb。命令x/s addr会显示起始地址为addr的字符串

如何知道字符串s的地址？提示:X86-64体系通过寄存器传递函数参数， %rdi存储第一个参数，%rsi存储第二个参数。

第二关

phase_2汇编代码以及注释如下

0000000000400efc :

400efc: 55 push %rbp

400efd: 53 push %rbx

400efe: 48 83 ec 28 sub $0x28,%rsp

400f02: 48 89 e6 mov %rsp,%rsi

400f05: e8 52 05 00 00 callq 40145c

400f0a: 83 3c 24 01 cmpl $0x1,(%rsp)

400f0e: 74 20 je 400f30 //first等于1跳至400f30

400f10: e8 25 05 00 00 callq 40143a // 不等于1直接爆炸

400f15: eb 19 jmp 400f30

400f17: 8b 43 fc mov -0x4(%rbx),%eax //%eax = M[%rbx-4]

400f1a: 01 c0 add %eax,%eax //%eax = 2 * %eax

400f1c: 39 03 cmp %eax,(%rbx) //%eax == M[%rbx]

400f1e: 74 05 je 400f25 //等于跳至400f25

400f20: e8 15 05 00 00 callq 40143a //否则爆炸

400f25: 48 83 c3 04 add $0x4,%rbx //%rbx = %rbx + 4

400f29: 48 39 eb cmp %rbp,%rbx //%rbx == %rbp

400f2c: 75 e9 jne 400f17 //不等跳至400f17

400f2e: eb 0c jmp 400f3c //等于phase_2拆弹成功

400f30: 48 8d 5c 24 04 lea 0x4(%rsp),%rbx //%rbx = second_num addr

400f35: 48 8d 6c 24 18 lea 0x18(%rsp),%rbp //%rbp = %rsp + 24

400f3a: eb db jmp 400f17 //跳至400f17

400f3c: 48 83 c4 28 add $0x28,%rsp

400f40: 5b pop %rbx

400f41: 5d pop %rbp

400f42: c3 retq

phase_2调用了函数 read_six_numbers。此函数接收两个参数，在被phase_2调用时read_six_numbers第一个参数是phase_2的第一个参数即实验者输入的字符串地址，第二个参数是%rsp的值。

下面是read_six_numbers的汇编代码

000000000040145c :

40145c: 48 83 ec 18 sub $0x18,%rsp

401460: 48 89 f2 mov %rsi,%rdx //%rdx=%rsi

401463: 48 8d 4e 04 lea 0x4(%rsi),%rcx //%rcx=%rsi+4

401467: 48 8d 46 14 lea 0x14(%rsi),%rax //%rax=%rsi+20

40146b: 48 89 44 24 08 mov %rax,0x8(%rsp) //M[%rsp+8]=%rax

401470: 48 8d 46 10 lea 0x10(%rsi),%rax //%rax=%rsi+16

401474: 48 89 04 24 mov %rax,(%rsp) //M[%rsp]=%rax

401478: 4c 8d 4e 0c lea 0xc(%rsi),%r9 //%r9=%rsi+12

40147c: 4c 8d 46 08 lea 0x8(%rsi),%r8 //%r8=%rsi+8

401480: be c3 25 40 00 mov $0x4025c3,%esi //%esi=char *format

401485: b8 00 00 00 00 mov $0x0,%eax

40148a: e8 61 f7 ff ff callq 400bf0 <__isoc99_sscanf>

40148f: 83 f8 05 cmp $0x5,%eax

401492: 7f 05 jg 401499

401494: e8 a1 ff ff ff callq 40143a

401499: 48 83 c4 18 add $0x18,%rsp

40149d: c3 retq

由代码可知read_six_numbers调用了库函数sscanf从给定字符串中读取数字。那么这6个数字具体类型是什么呢？

$ man sscanf可知sscanf的函数签名为int sscanf(const char *str, const char *format, ...);第二个参数指定读取数据格式。在调用sscanf前有如下指令

mov $0x4025c3,%esi。可见第二个参数const char *format的地址为0x4025c3。借助gdb得到格式字符串的内容为"%d %d %d %d %d %d"。所以这六个数字的类型为int。

由于此处的sscanf要读取六个整型数据，所以一共接收8个参数，前两个参数为const char *str和const char *format，后六个参数为六个int *指针。这八个参数中前六个分别由%rdi，%rsi， %rdx， %rcx， %r8， %r9这六个寄存器传递。后两个参数由内存传递。六个int*指针的值分别为%rsi,%rsi+4, %rsi+8, %rsi+12, %rsi+16, %rsi+20。考虑到read_six_numbers被phase_2调用时，%rsi的值指向phase_2栈帧的顶部。所以这六个数字是以数组形式存储在phase_2的栈帧上，起始地址为phase_2栈帧顶部。

回到phase_2的汇编代码，在调用read_six_numbers后有如下指令

400f0a: 83 3c 24 01 cmpl $0x1,(%rsp)

400f0e: 74 20 je 400f30 //first等于1跳至400f30

400f10: e8 25 05 00 00 callq 40143a

可知第一个数字必须为1。

阅读接下来的代码，易知此数组相邻数字间满足固定的比例关系。至此答案已呼之欲出。

第三关

0000000000400f43 : //sscanf(str, "%d, %d", &a, &b);

400f43: 48 83 ec 18 sub $0x18,%rsp

400f47: 48 8d 4c 24 0c lea 0xc(%rsp),%rcx // %rcx = %rsp+12

400f4c: 48 8d 54 24 08 lea 0x8(%rsp),%rdx // %rdx = %rsp+8

400f51: be cf 25 40 00 mov $0x4025cf,%esi

400f56: b8 00 00 00 00 mov $0x0,%eax

400f5b: e8 90 fc ff ff callq 400bf0 <__isoc99_sscanf>

400f60: 83 f8 01 cmp $0x1,%eax

400f63: 7f 05 jg 400f6a //返回值>1,jump 0x400f6a

400f65: e8 d0 04 00 00 callq 40143a //否则爆炸

400f6a: 83 7c 24 08 07 cmpl $0x7,0x8(%rsp) //a == 7

400f6f: 77 3c ja 400fad // a>7 爆炸

400f71: 8b 44 24 08 mov 0x8(%rsp),%eax //%eax = a

400f75: ff 24 c5 70 24 40 00 jmpq *0x402470(,%rax,8) //jump M[0x402470+8*%rax]

400f7c: b8 cf 00 00 00 mov $0xcf,%eax //%eax = 207

400f81: eb 3b jmp 400fbe //jump 0x400fbe

400f83: b8 c3 02 00 00 mov $0x2c3,%eax //%eax = 707

400f88: eb 34 jmp 400fbe //jump 0x400fbe

400f8a: b8 00 01 00 00 mov $0x100,%eax //%eax = 256

400f8f: eb 2d jmp 400fbe

400f91: b8 85 01 00 00 mov $0x185,%eax

400f96: eb 26 jmp 400fbe

400f98: b8 ce 00 00 00 mov $0xce,%eax

400f9d: eb 1f jmp 400fbe

400f9f: b8 aa 02 00 00 mov $0x2aa,%eax

400fa4: eb 18 jmp 400fbe

400fa6: b8 47 01 00 00 mov $0x147,%eax

400fab: eb 11 jmp 400fbe

400fad: e8 88 04 00 00 callq 40143a

400fb2: b8 00 00 00 00 mov $0x0,%eax

400fb7: eb 05 jmp 400fbe

400fb9: b8 37 01 00 00 mov $0x137,%eax

400fbe: 3b 44 24 0c cmp 0xc(%rsp),%eax

400fc2: 74 05 je 400fc9 //if b==%eax defused

400fc4: e8 71 04 00 00 callq 40143a //否则爆炸

400fc9: 48 83 c4 18 add $0x18,%rsp

400fcd: c3 retq

阅读0x400f43-0x400f6f的汇编代码易知phase_3通过sscanf扫描用户输入的字符串从中提取出两个无符号整数设为a, b，其中a小等于7。

关键代码为如下两行

400f71: 8b 44 24 08 mov 0x8(%rsp),%eax //%eax = a

400f75: ff 24 c5 70 24 40 00 jmpq *0x402470(,%rax,8) //jump M[0x402470+8*%rax]

其中jmpq *0x402470(,%rax,8)为indirect jump。该指令没有直接给出跳转地址，而是指定一个内存地址，此内存地址对应的内存区域存储着跳转地址。CPU先访问内存获取跳转地址再进行跳转。由代码可知跳转地址的内存地址为0x402470+8*%rax，其中%rax=a。可见该跳转指令的跳转地址由用户输入的第一个数字决定。

在看接下来的代码，发现他们都具有同一特征：给%eax赋值，然后跳转至0x400fbe。

400fbe: 3b 44 24 0c cmp 0xc(%rsp),%eax

400fc2: 74 05 je 400fc9 //if b==%eax defused

400fc4: e8 71 04 00 00 callq 40143a //否则爆炸

阅读这三行代码可知，b必需等于%eax。问题是如何知道%eax的值，借助gdb输入合法的a，自然就能知道对应的%eax的值，从而得到对应的b。至此第三关解析完毕。