csapp bomblab
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实验概述
实验模拟了一个二进制炸弹，有六个“雷管”，输入正确的字符串就可以拆卸。要进行反汇编和gdb调试
/* 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! */
用命令objdump -d bomb>bomb.txt导出汇编到文件bomb.txt
phase1
分析输入时的调用链
rdi：the address of the string that we input
400e3a: e8 a1 00 00 00 callq 400ee0 <phase_1>
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 <strings_not_equal>
调用下面这个函数，后面也会用到
0000000000401338 <strings_not_equal>:
401338: 41 54 push %r12
40133a: 55 push %rbp
40133b: 53 push %rbx
40133c: 48 89 fb mov %rdi,%rbx
40133f: 48 89 f5 mov %rsi,%rbp
401342: e8 d4 ff ff ff callq 40131b <string_length>
401347: 41 89 c4 mov %eax,%r12d
40134a: 48 89 ef mov %rbp,%rdi
40134d: e8 c9 ff ff ff callq 40131b <string_length>
401352: ba 01 00 00 00 mov $0x1,%edx
401357: 41 39 c4 cmp %eax,%r12d
40135a: 75 3f jne 40139b <strings_not_equal+0x63>
40135c: 0f b6 03 movzbl (%rbx),%eax
40135f: 84 c0 test %al,%al
401361: 74 25 je 401388 <strings_not_equal+0x50>
401363: 3a 45 00 cmp 0x0(%rbp),%al
401366: 74 0a je 401372 <strings_not_equal+0x3a>
401368: eb 25 jmp 40138f <strings_not_equal+0x57>
40136a: 3a 45 00 cmp 0x0(%rbp),%al
40136d: 0f 1f 00 nopl (%rax)
401370: 75 24 jne 401396 <strings_not_equal+0x5e>
401372: 48 83 c3 01 add $0x1,%rbx
401376: 48 83 c5 01 add $0x1,%rbp
40137a: 0f b6 03 movzbl (%rbx),%eax
40137d: 84 c0 test %al,%al
40137f: 75 e9 jne 40136a <strings_not_equal+0x32>
401381: ba 00 00 00 00 mov $0x0,%edx
401386: eb 13 jmp 40139b <strings_not_equal+0x63>
401388: ba 00 00 00 00 mov $0x0,%edx
40138d: eb 0c jmp 40139b <strings_not_equal+0x63>
40138f: ba 01 00 00 00 mov $0x1,%edx
401394: eb 05 jmp 40139b <strings_not_equal+0x63>
401396: ba 01 00 00 00 mov $0x1,%edx
40139b: 89 d0 mov %edx,%eax
40139d: 5b pop %rbx
40139e: 5d pop %rbp
40139f: 41 5c pop %r12
4013a1: c3 retq
000000000040131b <string_length>:
40131b: 80 3f 00 cmpb $0x0,(%rdi) //防止rdi输入为空
40131e: 74 12 je 401332 <string_length+0x17>
401320: 48 89 fa mov %rdi,%rdx
401323: 48 83 c2 01 add $0x1,%rdx
401327: 89 d0 mov %edx,%eax
401329: 29 f8 sub %edi,%eax
40132b: 80 3a 00 cmpb $0x0,(%rdx)
40132e: 75 f3 jne 401323 <string_length+0x8>
401330: f3 c3 repz retq
401332: b8 00 00 00 00 mov $0x0,%eax
401337: c3 retq
其实这里rsi存的就是正确答案了
这是个测试字符串长度的程序，地址依次加1进行比较，当最后一个为null时，跳出函数执行，返回字符串长度到eax
然后回到上面第一次调用检查我们输入字符串的长度，放到r12，第二次放入eax后直接和r12比较，如果相等，再进行输入地址是否为空的判断，并且401363这里也进行了一次判断输入和目的地址的第一个字节是否相同的判断，这里40133c指令把输入字符串地址给过rbx了。0x40136a~0x40137f这一段就是借用eax为一个字节的temp不停逐字节的循环判断两个字符串是否相等。为什么要用temp就是源操作数和目的操作数不能同时为内存引用。
后面的话返回值就是0，如果当中任何一个检查没有过返回值就是1了。
400eee: 85 c0 test %eax,%eax
400ef0: 74 05 je 400ef7 <phase_1+0x17>
400ef2: e8 43 05 00 00 callq 40143a <explode_bomb>
400ef7: 48 83 c4 08 add $0x8,%rsp
400efb: c3 retq
000000000040143a <explode_bomb>:
40143a: 48 83 ec 08 sub $0x8,%rsp
40143e: bf a3 25 40 00 mov $0x4025a3,%edi
401443: e8 c8 f6 ff ff callq 400b10 <puts@plt>
401448: bf ac 25 40 00 mov $0x4025ac,%edi
40144d: e8 be f6 ff ff callq 400b10 <puts@plt>
401452: bf 08 00 00 00 mov $0x8,%edi
401457: e8 c4 f7 ff ff callq 400c20 <exit@plt>
直接跳到400ef7，如果eax不为0的话，就会调用引爆炸弹这个函数，这个函数打印字符串后直接exit了，直接退出
那这里的话还会返回到main函数，从400ee0里面return后，下一条指令又是调用4015c4 <phase_defused>函数
00000000004015c4 <phase_defused>:
4015c4: 48 83 ec 78 sub $0x78,%rsp
4015c8: 64 48 8b 04 25 28 00 mov %fs:0x28,%rax
4015cf: 00 00
4015d1: 48 89 44 24 68 mov %rax,0x68(%rsp)
4015d6: 31 c0 xor %eax,%eax
4015d8: 83 3d 81 21 20 00 06 cmpl $0x6,0x202181(%rip) # 603760 <num_input_strings>
4015df: 75 5e jne 40163f <phase_defused+0x7b>
4015e1: 4c 8d 44 24 10 lea 0x10(%rsp),%r8
4015e6: 48 8d 4c 24 0c lea 0xc(%rsp),%rcx
4015eb: 48 8d 54 24 08 lea 0x8(%rsp),%rdx
4015f0: be 19 26 40 00 mov $0x402619,%esi
4015f5: bf 70 38 60 00 mov $0x603870,%edi
4015fa: e8 f1 f5 ff ff callq 400bf0 <__isoc99_sscanf@plt>
4015ff: 83 f8 03 cmp $0x3,%eax
401602: 75 31 jne 401635 <phase_defused+0x71>
401604: be 22 26 40 00 mov $0x402622,%esi
401609: 48 8d 7c 24 10 lea 0x10(%rsp),%rdi
40160e: e8 25 fd ff ff callq 401338 <strings_not_equal>
401613: 85 c0 test %eax,%eax
401615: 75 1e jne 401635 <phase_defused+0x71>
401617: bf f8 24 40 00 mov $0x4024f8,%edi
40161c: e8 ef f4 ff ff callq 400b10 <puts@plt>
401621: bf 20 25 40 00 mov $0x402520,%edi
401626: e8 e5 f4 ff ff callq 400b10 <puts@plt>
40162b: b8 00 00 00 00 mov $0x0,%eax
401630: e8 0d fc ff ff callq 401242 <secret_phase>
401635: bf 58 25 40 00 mov $0x402558,%edi
40163a: e8 d1 f4 ff ff callq 400b10 <puts@plt>
40163f: 48 8b 44 24 68 mov 0x68(%rsp),%rax
401644: 64 48 33 04 25 28 00 xor %fs:0x28,%rax
40164b: 00 00
40164d: 74 05 je 401654 <phase_defused+0x90>
40164f: e8 dc f4 ff ff callq 400b30 <__stack_chk_fail@plt>
401654: 48 83 c4 78 add $0x78,%rsp
401658: c3 retq
401659: 90 nop
40165a: 90 nop
40165b: 90 nop
40165c: 90 nop
40165d: 90 nop
40165e: 90 nop
40165f: 90 nop
这里我写的时候忘记保存了，根据印象应该就是一个恢复环境的，后面也会调用这个函数
Border relations with Canada have never been better.
phase_2
直接看400efc
400efc: 55 push %rbp
400efd: 53 push %rbx
400efe: 48 83 ec 28 sub $0x28,%rsp
400f02: 48 89 e6 mov %rsp,%rsi //rsi = rsp
400f05: e8 52 05 00 00 callq 40145c <read_six_numbers>
400f0a: 83 3c 24 01 cmpl $0x1,(%rsp)
000000000040145c <read_six_numbers>:
40145c: 48 83 ec 18 sub $0x18,%rsp
401460: 48 89 f2 mov %rsi,%rdx
401463: 48 8d 4e 04 lea 0x4(%rsi),%rcx //rsp+4
401467: 48 8d 46 14 lea 0x14(%rsi),%rax //rsp+0x14
40146b: 48 89 44 24 08 mov %rax,0x8(%rsp) //rsp + 0x8 = rsp +0x14
401470: 48 8d 46 10 lea 0x10(%rsi),%rax //rsp + 0x10 = rsp +0x14
401474: 48 89 04 24 mov %rax,(%rsp) //(rsp) = rax
401478: 4c 8d 4e 0c lea 0xc(%rsi),%r9 //r9 = rsi +0xc
40147c: 4c 8d 46 08 lea 0x8(%rsi),%r8//r8 = rsi +0x8
401480: be c3 25 40 00 mov $0x4025c3,%esi //esi = 0x40253
401485: b8 00 00 00 00 mov $0x0,%eax //eax=0
40148a: e8 61 f7 ff ff callq 400bf0 <__isoc99_sscanf@plt>
40148f: 83 f8 05 cmp $0x5,%eax
401492: 7f 05 jg 401499 <read_six_numbers+0x3d>
401494: e8 a1 ff ff ff callq 40143a <explode_bomb>
401499: 48 83 c4 18 add $0x18,%rsp
40149d: c3 retq
|
400f0a: 83 3c 24 01 cmpl $0x1,(%rsp)
400f0e: 74 20 je 400f30 <phase_2+0x34>
|
400f30: 48 8d 5c 24 04 lea 0x4(%rsp),%rbx //ebx = &(rsp+4)
400f35: 48 8d 6c 24 18 lea 0x18(%rsp),%rbp //ebp = rsp+0x18
400f3a: eb db jmp 400f17 <phase_2+0x1b>
|
400f17: 8b 43 fc mov -0x4(%rbx),%eax //eax = *rsp temp = a[i]
400f1a: 01 c0 add %eax,%eax //temp= 2a[i]
400f1c: 39 03 cmp %eax,(%rbx)//(rsp) //if (temp == a[i+1])
400f1e: 74 05 je 400f25 <phase_2+0x29>
400f20: e8 15 05 00 00 callq 40143a <explode_bomb>
400f25: 48 83 c3 04 add $0x4,%rbx //判断下一个 i++
400f29: 48 39 eb cmp %rbp,%rbx //判断循环是否终止 if(a[i] == a[5])
400f2c: 75 e9 jne 400f17 <phase_2+0x1b>
400f2e: eb 0c jmp 400f3c <phase_2+0x40>
……………………
400f3c: 48 83 c4 28 add $0x28,%rsp
400f40: 5b pop %rbx
400f41: 5d pop %rbp
400f42: c3 retq
调用sscanf，其中rdi是我们的输入。rsi是格式化字符串%d %d %d %d %d %d，根据scanf给eax值就是对输入进行解析，超过5，也就是6可以跳过401494。这个时候发现输入是每四字节存储到栈里，所以输入的第一个数只能是1，然后跳到400f30
设我们输入的数组是a[6]，所以偏移就是0x4 *6 = 0x18，根据上面的注释，输入是一个首项为1，公比为2的等比数列的前六项。
1 2 4 8 16 32
phase_3
400e6f: 48 89 c7 mov %rax,%rdi
400e72: e8 cc 00 00 00 callq 400f43 <phase_3>
400e77: e8 48 07 00 00 callq 4015c4 <phase_defused>
|
400f43: 48 83 ec 18 sub $0x18,%rsp
400f47: 48 8d 4c 24 0c lea 0xc(%rsp),%rcx
400f4c: 48 8d 54 24 08 lea 0x8(%rsp),%rdx
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 <phase_3+0x27>
|
400f6a: 83 7c 24 08 07 cmpl $0x7,0x8(%rsp)
400f6f: 77 3c ja 400fad <phase_3+0x6a>//jump to bomb
400f71: 8b CSAPP第二個實驗bomblab