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Bomb Lab

给了bomb.cbomb二进制可执行目标程序,bomb.c不能直接编译和运行,只是有一些提示,但是程序大致结构是:有6个关卡,每个都需要输入(stdin/文件)一个字符串,运行后判断是否输入了正确的字符串。我们需要反汇编bomb,找到这6个正确的字符串。
我是在Amazon的云服务器上完成的,64位Red Hat。

第一步把汇编代码扔到一个文件中,方便调试。

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objdump -d bomb > bomb.asm

phase 1

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0000000000400ee0 <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>
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

当参数少于7个时, 参数从左到右放入寄存器: rdi, rsi, rdx, rcx, r8, r9。
我们的input作为第一个参数存入rdi,第二个参数0x402400存入rsi,传入函数处理,
调用了401338处的函数strings_not_equal

在400ee4设个断点,看看402400里是啥:

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gdb
file bomb
b *0x400ee4
run
x/s Addr// 显示内存值为字符串

在这里插入图片描述
为了确认这就是我们要的答案,再去看看调用的函数:

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0000000000401338 <strings_not_equal>:
401338: 41 54 push %r12
40133a: 55 push %rbp
40133b: 53 push %rbx
40133c: 48 89 fb mov %rdi,%rbx # 第一个参数(地址)存入rbx
40133f: 48 89 f5 mov %rsi,%rbp # 第二个参数(地址)存入rbp
401342: e8 d4 ff ff ff callq 40131b <string_length> # 用rdi的值调用string_length
401347: 41 89 c4 mov %eax,%r12d # 返回值存入r12d
40134a: 48 89 ef mov %rbp,%rdi # 第二个参数作为入参调用string_length
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 # 比较两个字符串的长度
40135a: 75 3f jne 40139b <strings_not_equal+0x63> # 不相等跳转
40135c: 0f b6 03 movzbl (%rbx),%eax # 将rbx地址中的值(input的第一个字母)存入eax
40135f: 84 c0 test %al,%al
401361: 74 25 je 401388 <strings_not_equal+0x50>
401363: 3a 45 00 cmp 0x0(%rbp),%al # 比较rbp地址中的值(待比较的第一个字母)
401366: 74 0a je 401372 <strings_not_equal+0x3a> # 相等跳转
401368: eb 25 jmp 40138f <strings_not_equal+0x57> # 不相等跳转
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 # 指针+1
401376: 48 83 c5 01 add $0x1,%rbp # 指针+1
40137a: 0f b6 03 movzbl (%rbx),%eax
40137d: 84 c0 test %al,%al
40137f: 75 e9 jne 40136a <strings_not_equal+0x32> # 跳回循环
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
所以这个函数就是比较两个字符串是否相同,先比较长度,再比较每个字符。故我们的第一个key就是Border relations with Canada have never been better.

phase 2

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0000000000400efc <phase_2>:
400efc: 55 push %rbp
400efd: 53 push %rbx
400efe: 48 83 ec 28 sub $0x28,%rsp # 栈指针-40
400f02: 48 89 e6 mov %rsp,%rsi # 栈指针作为第二个参数
400f05: e8 52 05 00 00 callq 40145c <read_six_numbers>
400f0a: 83 3c 24 01 cmpl $0x1,(%rsp) # 检查是否相等
400f0e: 74 20 je 400f30 <phase_2+0x34> # 相等跳转
400f10: e8 25 05 00 00 callq 40143a <explode_bomb>
400f15: eb 19 jmp 400f30 <phase_2+0x34>
400f17: 8b 43 fc mov -0x4(%rbx),%eax # (rbx-4)赋给eax
400f1a: 01 c0 add %eax,%eax
400f1c: 39 03 cmp %eax,(%rbx) # 比较当前数与下一个数
400f1e: 74 05 je 400f25 <phase_2+0x29> # 相等跳转
400f20: e8 15 05 00 00 callq 40143a <explode_bomb>
400f25: 48 83 c3 04 add $0x4,%rbx
400f29: 48 39 eb cmp %rbp,%rbx # 是否比完了6个数
400f2c: 75 e9 jne 400f17 <phase_2+0x1b>
400f2e: eb 0c jmp 400f3c <phase_2+0x40>
400f30: 48 8d 5c 24 04 lea 0x4(%rsp),%rbx # 栈指针+4赋给rbx
400f35: 48 8d 6c 24 18 lea 0x18(%rsp),%rbp # 栈指针+24赋给rbp
400f3a: eb db jmp 400f17 <phase_2+0x1b>
400f3c: 48 83 c4 28 add $0x28,%rsp
400f40: 5b pop %rbx
400f41: 5d pop %rbp
400f42: c3 retq

第一次是1,第二次是2,4,8,16,32

phase 3

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0000000000400f43 <phase_3>:
400f43: 48 83 ec 18 sub $0x18,%rsp # 栈指针-24
400f47: 48 8d 4c 24 0c lea 0xc(%rsp),%rcx # 栈指针+12赋给rcx
400f4c: 48 8d 54 24 08 lea 0x8(%rsp),%rdx # 栈指针+8赋给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> # 返回值存入eax
400f60: 83 f8 01 cmp $0x1,%eax
400f63: 7f 05 jg 400f6a <phase_3+0x27> # 大于跳转
400f65: e8 d0 04 00 00 callq 40143a <explode_bomb> # 否则爆炸
400f6a: 83 7c 24 08 07 cmpl $0x7,0x8(%rsp) #(rsp+8)即输入的第一个整数与7比较
400f6f: 77 3c ja 400fad <phase_3+0x6a> # 大于爆炸
400f71: 8b 44 24 08 mov 0x8(%rsp),%eax # 输入的第一个整数赋给eax
400f75: ff 24 c5 70 24 40 00 jmpq *0x402470(,%rax,8) # 跳转8*rax+0x402470
400f7c: b8 cf 00 00 00 mov $0xcf,%eax
400f81: eb 3b jmp 400fbe <phase_3+0x7b>
400f83: b8 c3 02 00 00 mov $0x2c3,%eax
400f88: eb 34 jmp 400fbe <phase_3+0x7b>
400f8a: b8 00 01 00 00 mov $0x100,%eax
400f8f: eb 2d jmp 400fbe <phase_3+0x7b>
400f91: b8 85 01 00 00 mov $0x185,%eax
400f96: eb 26 jmp 400fbe <phase_3+0x7b>
400f98: b8 ce 00 00 00 mov $0xce,%eax
400f9d: eb 1f jmp 400fbe <phase_3+0x7b>
400f9f: b8 aa 02 00 00 mov $0x2aa,%eax
400fa4: eb 18 jmp 400fbe <phase_3+0x7b>
400fa6: b8 47 01 00 00 mov $0x147,%eax
400fab: eb 11 jmp 400fbe <phase_3+0x7b>
400fad: e8 88 04 00 00 callq 40143a <explode_bomb>
400fb2: b8 00 00 00 00 mov $0x0,%eax
400fb7: eb 05 jmp 400fbe <phase_3+0x7b>
400fb9: b8 37 01 00 00 mov $0x137,%eax
400fbe: 3b 44 24 0c cmp 0xc(%rsp),%eax
400fc2: 74 05 je 400fc9 <phase_3+0x86>
400fc4: e8 71 04 00 00 callq 40143a <explode_bomb>
400fc9: 48 83 c4 18 add $0x18,%rsp
400fcd: c3 retq

400f51将地址0x4025cf赋给esi,作为第二个参数,看下这个地址有啥:为了方便,我们把前面问题的答案仍在一个文件ans.txt中,
在这里插入图片描述
输入是2个整数,第一个不能大于7,基于第一个整数(0-7)跳转......