该实验给了一个可执行程序,但没有源码,需要利用反汇编,分析汇编代码,依次输入几条正确的字符串,来拆除炸弹。
可以学到的东西有,gdb调试(断点,变量、寄存器、地址空间的查看),分析汇编语言中函数的参数,分析栈中的数组,分析汇编语言的跳转,递归过程,利用下标取数,链表数据结构
objdump -d bomb >bomb.d #反汇编得到assembly代码
查看bomb.c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 <file>, the bomb reads from <file>
* until EOF, and then switches to standard input. Thus, as you
* defuse each phase, you can add its defusing string to <file> 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 [<input_file>]\n", argv[0]);
exit(8);
}
main函数的argc参数允许通过命令行或者gdb输入参数 ;
不输入参数时,argc为1,此时需要通过标准输入stdin读取phase;
输入.txt文件作为参数时,argc为2,从txt文件phase;argc最大为2,main只允许输入一个参数。
从main中可以看到要破解6个phase函数。
gdb bomb #进入调试
layout asm #显示assembly代码
layout reg #显示寄存器值
b main #打断点
b phase1
i b #显示断点
delete #删除所有断点
delete 1#删除标号为1的断点
r #不输入command line参数 直接执行
c #执行到下一断点
si #单步执行汇编
ctrl +L #刷新gdb
r <ans.txt #输入txt文件 开始执行代码
进入phase_1之前,会通过read_line函数读取phase,随便输入一个phase,如123,然后去看phase_1的汇编代码如何执行的。
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> #strings_not_equal返回0则跳转,不爆炸
400ef2: e8 43 05 00 00 callq 40143a <explode_bomb>
400ef7: 48 83 c4 08 add $0x8,%rsp
400efb: c3 retq
x/s $rdi #rdi存放phase_1参数,x/s以string形式查看rdi
(gdb) x/s $rdi
0x603780 <input_strings>: "123" #可以看到是我们输入的123,而rdi的值为0x603780
(gdb) x/4dx 0x603780 #查看0x603780处的4个bytes 正好是123的ascii码和结束符00
0x603780 <input_strings>: 0x31 0x32 0x33 0x00
(gdb) x/dx 0x603780 #字符串的第一个字符在最低的地址
0x603780 <input_strings>: 0x31
0x603781 <input_strings+1>: 0x32
0x603782 <input_strings+2>: 0x33
0x603783 <input_strings+3>: 0x00
继续执行 可以看到phase_1调用了strings_not_equal(),在该函数打个断点,进入后查看其两个参数(分别在rdi和rsi)
(gdb) b strings_not_equal
Breakpoint 3 at 0x401338
(gdb) c
Continuing.
Breakpoint 3, 0x0000000000401338 in strings_not_equal ()
(gdb) x/gx $rdi
0x603780 <input_strings>: "123"
(gdb) x/s $rsi
0x402400: "Border relations with Canada have never been better."
发现phase将0x402400传给了rsi,作为strings_not_equal()的第二个参数
0x400ee4 <phase_1+4> mov $0x402400,%esi
因此,可以猜测strings_not_equal()比较其两个字符串参数是否相同,不同则返回1,相同则返回0;
猜测"Border relations with Canada have never been better."就是我们需要的第一个phase。
研究一下strings_not_equal()
0000000000401338 <strings_not_equal>: #注释的->> -<<是跳转标号
401338: 41 54 push %r12
40133a: 55 push %rbp
40133b: 53 push %rbx
40133c: 48 89 fb mov %rdi,%rbx #存放rdi
40133f: 48 89 f5 mov %rsi,%rbp #存放rsi
401342: e8 d4 ff ff ff callq 40131b <string_length> #计算rdi的str长度 放入r12d
401347: 41 89 c4 mov %eax,%r12d
40134a: 48 89 ef mov %rbp,%rdi
40134d: e8 c9 ff ff ff callq 40131b <string_length> #计算rsi的str长度 放入rax
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> #长度不等,则返回1 (0x1->edx->eax) ->>>4
40135c: 0f b6 03 movzbl (%rbx),%eax #rbx存的rdi 取第一个字符放入eax
40135f: 84 c0 test %al,%al
401361: 74 25 je 401388 <strings_not_equal+0x50> #字符为00,则跳转,并最终返回0 ->>>1
401363: 3a 45 00 cmp 0x0(%rbp),%al #字符不为0 则与rbp的rsi的第一个字符比较
401366: 74 0a je 401372 <strings_not_equal+0x3a> #相等则继续比较 ->>>2
401368: eb 25 jmp 40138f <strings_not_equal+0x57> #不等则跳转 返回1 ->>>5
40136a: 3a 45 00 cmp 0x0(%rbp),%al #比较下一字符 -<<<3
40136d: 0f 1f 00 nopl (%rax)
401370: 75 24 jne 401396 <strings_not_equal+0x5e> #若不等->>>6
401372: 48 83 c3 01 add $0x1,%rbx #地址偏移 -<<<2
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> #下一字符ascii不为00,则继续 ->>>3
401381: ba 00 00 00 00 mov $0x0,%edx
401386: eb 13 jmp 40139b <strings_not_equal+0x63> #->>>4
401388: ba 00 00 00 00 mov $0x0,%edx #-<<<1
40138d: eb 0c jmp 40139b <strings_not_equal+0x63> #->>>4
40138f: ba 01 00 00 00 mov $0x1,%edx #-<<<5
401394: eb 05 jmp 40139b <strings_not_equal+0x63> #->>>4
401396: ba 01 00 00 00 mov $0x1,%edx #-<<<6
40139b: 89 d0 mov %edx,%eax #-<<<4
40139d: 5b pop %rbx
40139e: 5d pop %rbp
40139f: 41 5c pop %r12
4013a1: c3 retq
0000000000400efc <phase_2>:
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 <read_six_numbers>
随便输入123
phase_2()调用了read_six_numbers(),并开辟了长为0x28=40bytes的栈区,栈顶作为参数2(rsi)。
进入到read_six_numbers(),可以看到scanf之前rsi中的地址值处理之后赋给了一系列寄存器。
其中0x7fffffffdb20(rsi+0x10)放入了栈区0x7fffffffdaf0(rsp)
0x7fffffffdb24(rsi+0x14)放入了栈区0x7fffffffdaf8(rsp+8)
000000000040145c <read_six_numbers>: #rdi->"123" rsi "0x7fffffffdb10"
40145c: 48 83 ec 18 sub $0x18,%rsp
401460: 48 89 f2 mov %rsi,%rdx
401463: 48 8d 4e 04 lea 0x4(%rsi),%rcx
401467: 48 8d 46 14 lea 0x14(%rsi),%rax
40146b: 48 89 44 24 08 mov %rax,0x8(%rsp)
401470: 48 8d 46 10 lea 0x10(%rsi),%rax
401474: 48 89 04 24 mov %rax,(%rsp)
401478: 4c 8d 4e 0c lea 0xc(%rsi),%r9
40147c: 4c 8d 46 08 lea 0x8(%rsi),%r8
401480: be c3 25 40 00 mov $0x4025c3,%esi
401485: b8 00 00 00 00 mov $0x0,%eax #rdi(输入的字符串) rsi(格式) rdx rcx r8 r9 rsp存两个
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
查看__isoc99_sscanf@plt的参数rdi rsi
(gdb) x/s $rdi
0x6037d0 <input_strings+80>: "123"
(gdb) x/s $rsi
0x4025c3: "%d %d %d %d %d %d"
可见phase是6个数,以空格隔开,6个数的存储位置为phase2开辟的栈区(scanf("%d %d %d %d %d %d",¶1,¶2....¶6))
重新运行r
,尝试输入1 2 3 4 5 6
(gdb) b *0x40149d
Breakpoint 6 at 0x40149d
(gdb) x/wd 0x7fffffffdb10 #以decimal查看该地址,正好是输入的1 2 3 4 5 6
0x7fffffffdb10: 1
0x7fffffffdb14: 2
0x7fffffffdb18: 3
0x7fffffffdb1c: 4
0x7fffffffdb20: 5
0x7fffffffdb24: 6
结束read_six_numbers后rsp指向0x7fffffffdb10,即数字1的位置
#phase_2后半
400f05: e8 52 05 00 00 callq 40145c <read_six_numbers>
400f0a: 83 3c 24 01 cmpl $0x1,(%rsp) #第一个数与1比较 ,不等则爆炸
400f0e: 74 20 je 400f30 <phase_2+0x34> #相等则跳转 ->>1 表示跳到1
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 #eax存放rbx之前的一个数,记为pre #-<<2
400f1a: 01 c0 add %eax,%eax #pre*=2
400f1c: 39 03 cmp %eax,(%rbx) #判断pre==cur
400f1e: 74 05 je 400f25 <phase_2+0x29> # ->>3
400f20: e8 15 05 00 00 callq 40143a <explode_bomb>
400f25: 48 83 c3 04 add $0x4,%rbx #继续偏移 -<<3
400f29: 48 39 eb cmp %rbp,%rbx #看偏移后是否为rbp,rbp指向为空,若为rbp则终止
400f2c: 75 e9 jne 400f17 <phase_2+0x1b> #->>2
400f2e: eb 0c jmp 400f3c <phase_2+0x40> #终止->>end
400f30: 48 8d 5c 24 04 lea 0x4(%rsp),%rbx #偏移rsp得到下一个数 -<<1
400f35: 48 8d 6c 24 18 lea 0x18(%rsp),%rbp #rsp偏移0x18后就没有数了
400f3a: eb db jmp 400f17 <phase_2+0x1b> # ->>2
400f3c: 48 83 c4 28 add $0x28,%rsp #-<<end
400f40: 5b pop %rbx
400f41: 5d pop %rbp
400f42: c3 retq
阅读可知 其作用是判断当前数字是否为前一个数字的2倍,若不是,则错误
故phase为1 2 4 8 16 32
0000000000400f43 <phase_3>:
400f43: 48 83 ec 18 sub $0x18,%rsp
400f47: 48 8d 4c 24 0c lea 0xc(%rsp),%rcx #num2
400f4c: 48 8d 54 24 08 lea 0x8(%rsp),%rdx #num1
400f51: be cf 25 40 00 mov $0x4025cf,%esi
400f56: b8 00 00 00 00 mov $0x0,%eax #scanf("%d,%d",&num1,&num2)
400f5b: e8 90 fc ff ff callq 400bf0 <__isoc99_sscanf@plt> #这里break 查看rsi 其指明了phase格式
400f60: 83 f8 01 cmp $0x1,%eax #输入大于等于2个数时,rax为1,输入一个数时rax为1
400f63: 7f 05 jg 400f6a <phase_3+0x27> # ->>1
400f65: e8 d0 04 00 00 callq 40143a <explode_bomb>
400f6a: 83 7c 24 08 07 cmpl $0x7,0x8(%rsp) # -<<1
400f6f: 77 3c ja 400fad <phase_3+0x6a> #num1 > 7 则爆炸 ->>2
400f71: 8b 44 24 08 mov 0x8(%rsp),%eax #num1->rax
400f75: ff 24 c5 70 24 40 00 jmpq *0x402470(,%rax,8) #以num1做偏移然后跳转,可见0x402470地址的值很重要
400f7c: b8 cf 00 00 00 mov $0xcf,%eax # rax存放0xcf= 207 -<< num1==0 跳转位置
400f81: eb 3b jmp 400fbe <phase_3+0x7b> #->> all
400f83: b8 c3 02 00 00 mov $0x2c3,%eax # rax=0x2c3=707 -<<num2==2
400f88: eb 34 jmp 400fbe <phase_3+0x7b> #->>all
400f8a: b8 00 01 00 00 mov $0x100,%eax # rax=0x100 -<<num2==3
400f8f: eb 2d jmp 400fbe <phase_3+0x7b> #->>all
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> #-<<2
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 # rax=0x137=311 -<< num==1 跳转位置
400fbe: 3b 44 24 0c cmp 0xc(%rsp),%eax #num2==rax ? -<< all
400fc2: 74 05 je 400fc9 <phase_3+0x86> # ->>end
400fc4: e8 71 04 00 00 callq 40143a <explode_bomb>
400fc9: 48 83 c4 18 add $0x18,%rsp # -<<end
400fcd: c3 retq
(gdb) b *0x400f5b
Breakpoint 8 at 0x400f5b
(gdb) x/s $rdi
0x603820 <input_strings+160>: "1 10" #随便输入的1 10
(gdb) x/s $rsi
0x4025cf: "%d %d"
查看rsi知道要输入两个数
查看0x402470,其保存了跳转方向
(gdb) x/8gx 0x402470
0x402470: 0x0000000000400f7c 0x0000000000400fb9
0x402480: 0x0000000000400f83 0x0000000000400f8a
0x402490: 0x0000000000400f91 0x0000000000400f98
0x4024a0: 0x0000000000400f9f 0x0000000000400fa6
如phase_2注释所示,num1是跳转地址相对0x402470的偏移量,跳转后,对rax赋值,然后判断num2和rax的值是否相等。
因此,有8组答案
0 207
1 311 ...
000000000040100c <phase_4>:
40100c: 48 83 ec 18 sub $0x18,%rsp
401010: 48 8d 4c 24 0c lea 0xc(%rsp),%rcx #num2地址
401015: 48 8d 54 24 08 lea 0x8(%rsp),%rdx #num1地址
40101a: be cf 25 40 00 mov $0x4025cf,%esi
40101f: b8 00 00 00 00 mov $0x0,%eax
401024: e8 c7 fb ff ff callq 400bf0 <__isoc99_sscanf@plt> #scanf(rdi,rsi,rdx,rcx)
401029: 83 f8 02 cmp $0x2,%eax
40102c: 75 07 jne 401035 <phase_4+0x29>
40102e: 83 7c 24 08 0e cmpl $0xe,0x8(%rsp) #num1要小于0xe=13
401033: 76 05 jbe 40103a <phase_4+0x2e>
401035: e8 00 04 00 00 callq 40143a <explode_bomb>
40103a: ba 0e 00 00 00 mov $0xe,%edx
40103f: be 00 00 00 00 mov $0x0,%esi
401044: 8b 7c 24 08 mov 0x8(%rsp),%edi
401048: e8 81 ff ff ff callq 400fce <func4> #func4(edi,esi,edx) 3个参数
40104d: 85 c0 test %eax,%eax
40104f: 75 07 jne 401058 <phase_4+0x4c>
401051: 83 7c 24 0c 00 cmpl $0x0,0xc(%rsp) #num2要等于0 才跳转
401056: 74 05 je 40105d <phase_4+0x51>
401058: e8 dd 03 00 00 callq 40143a <explode_bomb>
40105d: 48 83 c4 18 add $0x18,%rsp
401061: c3 retq
b *401024
查看__isoc99_sscanf@plt
参数个数(esi指示了要输入的数的格式)
(gdb) x/s $rdi
0x603870 <input_strings+240>: "1 2"
(gdb) x/s $rsi
0x4025cf: "%d %d"
于是知道该题要输入两个数,若不是两个则explode_bomb;随便输入 1 2。
之后调用了func4 要求func4返回0。
根据 401051: 83 7c 24 0c 00 cmpl $0x0,0xc(%rsp)
,num2显然要为0 根据
func4(num1,0,13) 代码如下: 研究一下num1即可
0000000000400fce <func4>: #3个参数 edi=num1 esi=0 edx=13
400fce: 48 83 ec 08 sub $0x8,%rsp
400fd2: 89 d0 mov %edx,%eax #edx->eax
400fd4: 29 f0 sub %esi,%eax #eax=eax-esi=edx-esi
400fd6: 89 c1 mov %eax,%ecx #ecx=eax
400fd8: c1 e9 1f shr $0x1f,%ecx #ecx=>>31
400fdb: 01 c8 add %ecx,%eax #eax=ecx+eax
400fdd: d1 f8 sar %eax #eax=>>1 综上 eax=((edx-esi)>>31+edx-esi)>>1
400fdf: 8d 0c 30 lea (%rax,%rsi,1),%ecx #ecx=rax+rsi=((edx-esi)>>31+edx-esi)>>1+esi
400fe2: 39 f9 cmp %edi,%ecx
400fe4: 7e 0c jle 400ff2 <func4+0x24> #ecx<=edi ->>1
400fe6: 8d 51 ff lea -0x1(%rcx),%edx
400fe9: e8 e0 ff ff ff callq 400fce <func4>
400fee: 01 c0 add %eax,%eax
400ff0: eb 15 jmp 401007 <func4+0x39> #return
400ff2: b8 00 00 00 00 mov $0x0,%eax #-<<1
400ff7: 39 f9 cmp %edi,%ecx
400ff9: 7d 0c jge 401007 <func4+0x39> #ecx>=edi return
400ffb: 8d 71 01 lea 0x1(%rcx),%esi
400ffe: e8 cb ff ff ff callq 400fce <func4>
401003: 8d 44 00 01 lea 0x1(%rax,%rax,1),%eax
401007: 48 83 c4 08 add $0x8,%rsp
40100b: c3 retq
ecx<=edi # 400fe4: 7e 0c jle 400ff2 <func4+0x24>
ecx>=edi return 0 # 400ff9: 7d 0c jge 401007 <func4+0x39>
ecx<edi
esi=ecx+1
return 2*func(edi,esi,edx)+1
ecx>edi
edx=rcx-1
return 2*func(edi,esi,edx) # 400fe9: e8 e0 ff ff ff callq 400fce <func4>
易知当ecx==edi时,func4返回0;而ecx=((edx-esi)>>31+edx-esi)>>1+esi=7
故答案 7 0
0000000000401062 <phase_5>: #rdi string
401062: 53 push %rbx
401063: 48 83 ec 20 sub $0x20,%rsp
401067: 48 89 fb mov %rdi,%rbx #rbx保存了rdi
40106a: 64 48 8b 04 25 28 00 mov %fs:0x28,%rax
401071: 00 00
401073: 48 89 44 24 18 mov %rax,0x18(%rsp)
401078: 31 c0 xor %eax,%eax
40107a: e8 9c 02 00 00 callq 40131b <string_length> #rdi是phase5输入的字符串
40107f: 83 f8 06 cmp $0x6,%eax #字符串长度是否为6 不是则错误
401082: 74 4e je 4010d2 <phase_5+0x70> #--->>1
401084: e8 b1 03 00 00 callq 40143a <explode_bomb>
401089: eb 47 jmp 4010d2 <phase_5+0x70>
40108b: 0f b6 0c 03 movzbl (%rbx,%rax,1),%ecx #(rax是偏移量) 取rdi的第rax个字符 #-----<<2
40108f: 88 0c 24 mov %cl,(%rsp)
401092: 48 8b 14 24 mov (%rsp),%rdx
401096: 83 e2 0f and $0xf,%edx #取字符值ascii值的后4位 写入edx
401099: 0f b6 92 b0 24 40 00 movzbl 0x4024b0(%rdx),%edx #以0x4024b0作为基址 edx作为偏移 取一个字节存入edx
4010a0: 88 54 04 10 mov %dl,0x10(%rsp,%rax,1) #把edx中的值写入(%rsp)+0x10+(%rax)的栈处
4010a4: 48 83 c0 01 add $0x1,%rax #rax加1 知道其为6 依次取出了edi的6个字符
4010a8: 48 83 f8 06 cmp $0x6,%rax
4010ac: 75 dd jne 40108b <phase_5+0x29> #----->>2
4010ae: c6 44 24 16 00 movb $0x0,0x16(%rsp)
4010b3: be 5e 24 40 00 mov $0x40245e,%esi
4010b8: 48 8d 7c 24 10 lea 0x10(%rsp),%rdi #(%rsp)+0x10是我们前面构造的字符串的起点
4010bd: e8 76 02 00 00 callq 401338 <strings_not_equal>
4010c2: 85 c0 test %eax,%eax
4010c4: 74 13 je 4010d9 <phase_5+0x77> #eax==0 phase_5正常return
4010c6: e8 6f 03 00 00 callq 40143a <explode_bomb>
4010cb: 0f 1f 44 00 00 nopl 0x0(%rax,%rax,1)
4010d0: eb 07 jmp 4010d9 <phase_5+0x77>
4010d2: b8 00 00 00 00 mov $0x0,%eax #---<<1
4010d7: eb b2 jmp 40108b <phase_5+0x29> #------>>2
4010d9: 48 8b 44 24 18 mov 0x18(%rsp),%rax
4010de: 64 48 33 04 25 28 00 xor %fs:0x28,%rax
4010e5: 00 00
4010e7: 74 05 je 4010ee <phase_5+0x8c>
4010e9: e8 42 fa ff ff callq 400b30 <__stack_chk_fail@plt>
4010ee: 48 83 c4 20 add $0x20,%rsp
4010f2: 5b pop %rbx
4010f3: c3 retq
如下截图 需要(%rsp)+0x10
处开始的字符串为flyers
找到flyers
而(%rsp)+0x10
是从下面构造的
401099: 0f b6 92 b0 24 40 00 movzbl 0x4024b0(%rdx),%edx #以0x4024b0作为基址 edx作为偏移 取一个字节存入edx
rbx是输入的字符串参数的指针 rax是其下标i
上述汇编做的运算为取char的后4位,然后作为下标 , 在“maduier...”中寻找字符
flyers在"maduier..."中的下标为:
f:9
l:15 需要char字符后四位为f(15) 0x4f->O
y:14 需要char字符后四位为e(14) 0x4e->N
e:5 0x35->5
r:6
s 7
故答案可以是 9ON567 (不唯一)
phase6的汇编稍微有点长
首先看到 像phase2一样,要求写入六个数
我们查看一下这段代码
注意rsp存放了我们的6个数 每个数占4个字节
401100: 49 89 e5 mov %rsp,%r13 #rsp->r13
401103: 48 89 e6 mov %rsp,%rsi
401106: e8 51 03 00 00 callq 40145c <read_six_numbers>
40110b: 49 89 e6 mov %rsp,%r14 #rsp->r14
40110e: 41 bc 00 00 00 00 mov $0x0,%r12d #r12d=0
401114: 4c 89 ed mov %r13,%rbp #r13->rbp #-<<<4
401117: 41 8b 45 00 mov 0x0(%r13),%eax #rax=(rsp)
40111b: 83 e8 01 sub $0x1,%eax #rax-1
40111e: 83 f8 05 cmp $0x5,%eax
401121: 76 05 jbe 401128 <phase_6+0x34> #需要rax<=5 ->>>1
401123: e8 12 03 00 00 callq 40143a <explode_bomb>
401128: 41 83 c4 01 add $0x1,%r12d #r12d++ -<<<1
40112c: 41 83 fc 06 cmp $0x6,%r12d
401130: 74 21 je 401153 <phase_6+0x5f> #r12d==6 则跳 ->>>2
401132: 44 89 e3 mov %r12d,%ebx #rbx用于计算偏移 -<<<3
401135: 48 63 c3 movslq %ebx,%rax
401138: 8b 04 84 mov (%rsp,%rax,4),%eax #rbx*4作为偏移 即取下一个数
40113b: 39 45 00 cmp %eax,0x0(%rbp) #需要下一个数不等于rbp指向的数
40113e: 75 05 jne 401145 <phase_6+0x51>
401140: e8 f5 02 00 00 callq 40143a <explode_bomb>
401145: 83 c3 01 add $0x1,%ebx #rbx++
401148: 83 fb 05 cmp $0x5,%ebx
40114b: 7e e8 jle 401135 <phase_6+0x41> #rbx<=5 ->>>3
40114d: 49 83 c5 04 add $0x4,%r13 #rbx=6时 r13+4
401151: eb c1 jmp 401114 <phase_6+0x20> #->>>4
401153: 48 8d 74 24 18 lea 0x18(%rsp),%rsi #-<<<2
上述代码实现的效果相当于
r13=(%rsp)
for(r12d=1;r12d++;r12d<6){
if ((r13)-1>5) explode;
for(rbx=0;rbx++;rbxx<=5){
if((r13d+rbx*4)==(r13d)) explode;
}
}
即6个数不能相等 且小于等于6
我们随便输入6 3 2 5 1 4
执行完│ 0x401106 <phase_6+18> callq 0x40145c <read_six_numbers>
之后,查看栈区,正好是我们写入的6的数
0x7fffffffdab0: 6 3 2 5
0x7fffffffdac0: 1 4
查看下面代码
注意r14 rsi此时存的是rsp
401153: 48 8d 74 24 18 lea 0x18(%rsp),%rsi
401158: 4c 89 f0 mov %r14,%rax
40115b: b9 07 00 00 00 mov $0x7,%ecx
401160: 89 ca mov %ecx,%edx
401162: 2b 10 sub (%rax),%edx
401164: 89 10 mov %edx,(%rax)
401166: 48 83 c0 04 add $0x4,%rax
40116a: 48 39 f0 cmp %rsi,%rax
40116d: 75 f1 jne 401160 <phase_6+0x6c>
易知其效果是:用7减去这6个数 得到新的数
我们的输入为6 3 2 5 1 4 ;此时变成
(gdb) x/6wd $rsp
0x7fffffffdab0: 1 4 5 2
0x7fffffffdac0: 6 3
继续查看汇编
40116f: be 00 00 00 00 mov $0x0,%esi #rsi用于取输入的数 1 4 5 2 6 3
401174: eb 21 jmp 401197 <phase_6+0xa3> #->>>1
401176: 48 8b 52 08 mov 0x8(%rdx),%rdx #链表跳转 取到下一个node的地址 -<<<3
40117a: 83 c0 01 add $0x1,%eax #eax++
40117d: 39 c8 cmp %ecx,%eax
40117f: 75 f5 jne 401176 <phase_6+0x82> #eax!=ecx 直到相等 ->>>3
401181: eb 05 jmp 401188 <phase_6+0x94> #->>>>4
401183: ba d0 32 60 00 mov $0x6032d0,%edx #->>>2
401188: 48 89 54 74 20 mov %rdx,0x20(%rsp,%rsi,2) #edx存了个地址$0x6032d0 将其放入(%rsp)=+20 #-<<<4
40118d: 48 83 c6 04 add $0x4,%rsi #rsi+=4
401191: 48 83 fe 18 cmp $0x18,%rsi
401195: 74 14 je 4011ab <phase_6+0xb7>
401197: 8b 0c 34 mov (%rsp,%rsi,1),%ecx #取栈顶rsp处的数放入ecx -<<<1
40119a: 83 f9 01 cmp $0x1,%ecx
40119d: 7e e4 jle 401183 <phase_6+0x8f> #ecx<=1 ->>>2
40119f: b8 01 00 00 00 mov $0x1,%eax
4011a4: ba d0 32 60 00 mov $0x6032d0,%edx
4011a9: eb cb jmp 401176 <phase_6+0x82> #->>>3
我们查看一下```x/gx 0x6032d0``,可以看到这是个链表结构体,nodei存放的是值,nodei+8存放的是下一个node的地址。
且其指向是node1->node2->3->4->5->6
x/gx 0x6032d0
0x6032d0 <node1>: 0x000000010000014c
0x6032d8 <node1+8>: 0x00000000006032e0 # &node2
0x6032e0 <node2>: 0x00000002000000a8
0x6032e8 <node2+8>: 0x00000000006032f0
0x6032f0 <node3>: 0x000000030000039c
0x6032f8 <node3+8>: 0x0000000000603300
0x603300 <node4>: 0x00000004000002b3
0x603308 <node4+8>: 0x0000000000603310
0x603310 <node5>: 0x00000005000001dd
0x603318 <node5+8>: 0x0000000000603320
0x603320 <node6>: 0x00000006000001bb
0x603328 <node6+8>: 0x0000000000000000
struct node{
long num;
node* next;
}
而上述汇编的效果是,将nodei的地址按照1 4 5 2 6 3的顺序放入以(%rsp)+0x20开始的栈区。查看一下该栈区
(gdb) x/gx 0x7fffffffdad0
0x7fffffffdad0: 0x00000000006032d0 #node1
0x7fffffffdad8: 0x0000000000603300 #node4
0x7fffffffdae0: 0x0000000000603310 #node5
0x7fffffffdae8: 0x00000000006032e0 #node2
0x7fffffffdaf0: 0x0000000000603320 #node6
0x7fffffffdaf8: 0x00000000006032f0 #node3
继续分析汇编代码
4011ab: 48 8b 5c 24 20 mov 0x20(%rsp),%rbx #rbx存放栈区基址rsp+0x20=0x7fffffffdad0处的node地址0x6032d0
4011b0: 48 8d 44 24 28 lea 0x28(%rsp),%rax #rax 是栈区第二个node地址 0x7fffffffdad8
4011b5: 48 8d 74 24 50 lea 0x50(%rsp),%rsi #rsi 存放栈区末尾地址 0x7fffffffdb00
4011ba: 48 89 d9 mov %rbx,%rcx
4011bd: 48 8b 10 mov (%rax),%rdx
4011c0: 48 89 51 08 mov %rdx,0x8(%rcx) #改变node的指向
4011c4: 48 83 c0 08 add $0x8,%rax
4011c8: 48 39 f0 cmp %rsi,%rax
4011cb: 74 05 je 4011d2 <phase_6+0xde>
4011cd: 48 89 d1 mov %rdx,%rcx
4011d0: eb eb jmp 4011bd <phase_6+0xc9>
上述汇编使得 链表重连 1->4->5->2->6->3
(gdb) x/gx 0x6032d0
0x6032d0 <node1>: 0x000000010000014c
0x6032d8 <node1+8>: 0x0000000000603300 # &node4
0x6032e0 <node2>: 0x00000002000000a8
0x6032e8 <node2+8>: 0x0000000000603320
0x6032f0 <node3>: 0x000000030000039c
0x6032f8 <node3+8>: 0x0000000000603300
0x603300 <node4>: 0x00000004000002b3
0x603308 <node4+8>: 0x0000000000603310 # &node5
0x603310 <node5>: 0x00000005000001dd
0x603318 <node5+8>: 0x00000000006032e0 # &node2
0x603320 <node6>: 0x00000006000001bb
0x603328 <node6+8>: 0x00000000006032f0
后续汇编要求重连后node连接由大到小 即node3 4 5 6 1 2(node值由大到小排列)
故输入应数字为4 3 2 1 6 5,其被7减后变为3 4 5 6 1 2 这就是我们需要的node指向