CSAPP 炸弹实验解析上

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@plt>
  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 *strconst 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_numbersphase_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@plt>
  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。

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