CMU_CSAPP_LAB2_BOMBLAB

CSAPP第二章的第一个实验BombLab实验记录。

实验内容主要使用 gdb 、objdump 等指令在终端调试程序,利用反汇编查看各个函数运行的汇编代码,了解各个函数的执行过程。

附gdb指令精简手册:http://csapp.cs.cmu.edu/public/students.html


Phase_1

首先打开终端输入 gdb bomb 进入调试界面。

查看 phase_1 的汇编指令:

(gdb) disas phase_1
Dump of assembler code for function phase_1:
   0x0000000000400ee0 <+0>:	sub    $0x8,%rsp
   0x0000000000400ee4 <+4>:	mov    $0x402400,%esi
   0x0000000000400ee9 <+9>:	callq  0x401338 
   0x0000000000400eee <+14>:	test   %eax,%eax
   0x0000000000400ef0 <+16>:	je     0x400ef7 
   0x0000000000400ef2 <+18>:	callq  0x40143a 
   0x0000000000400ef7 <+23>:	add    $0x8,%rsp
   0x0000000000400efb <+27>:	retq   
End of assembler dump.

函数首先接收户输入的字符串,再将立即数 0x402400 输入到寄存器 esi ,接着调用函数,很容易猜到寄存器 esi 中保存了要和用户输入的字符串进行比较的字符串的首地址,即立即数 0x402400 。

(gdb) x/s 0x402400
0x402400:	"Border relations with Canada have never been better."

通过地址直接查看字符串,这就是我们需要输入的内容!

(gdb) run
Starting program: /home/liuyuan/Downloads/CSAPP/bomb/bomb 
Welcome to my fiendish little bomb. You have 6 phases with
which to blow yourself up. Have a nice day!
Border relations with Canada have never been better.
Phase 1 defused. How about the next one?

通过测试!

进一步细致理解函数运行过程:

将函数 phase_1 设置断点,运行程序,随便输入一些内容( "Nice day!" ),程序会停留在进入函数体的过程之前:

(gdb) break phase_1
Breakpoint 1 at 0x400ee0
(gdb) run
Starting program: /home/liuyuan/Downloads/CSAPP/bomb/bomb 
Welcome to my fiendish little bomb. You have 6 phases with
which to blow yourself up. Have a nice day!
Nice day!

Breakpoint 1, 0x0000000000400ee0 in phase_1 ()

此时查看各个寄存器的数据:

(gdb) info registers
rax            0x603780	6305664
rbx            0x0	0
rcx            0x9	9
rdx            0x1	1
rsi            0x603780	6305664
rdi            0x603780	6305664
rbp            0x402210	0x402210 <__libc_csu_init>
rsp            0x7fffffffdd98	0x7fffffffdd98
r8             0x60467a	6309498
r9             0x7ffff7fe5540	140737354028352
r10            0x3	3
r11            0x7ffff7a14890	140737347930256
r12            0x400c90	4197520
r13            0x7fffffffde80	140737488346752
r14            0x0	0
r15            0x0	0
rip            0x400ee0	0x400ee0 
eflags         0x202	[ IF ]
cs             0x33	51
ss             0x2b	43
ds             0x0	0
es             0x0	0
fs             0x0	0
gs             0x0	0

这是并没有出现 0x402400 。猜测刚刚输入的数据保存地址为 0x603780:

(gdb) x/s 0x603780
0x603780 :	"Nice day!"

回到前面的汇编指令,第二条指令将立即数 0x402400 移到寄存器 esi,让程序执行两条汇编指令,再查看寄存器的状态:

(gdb) stepi 2
0x0000000000400ee9 in phase_1 ()
(gdb) i r
rax            0x603780	6305664
rbx            0x0	0
rcx            0x9	9
rdx            0x1	1
rsi            0x402400	4203520
rdi            0x603780	6305664
rbp            0x402210	0x402210 <__libc_csu_init>
rsp            0x7fffffffdd90	0x7fffffffdd90
r8             0x60467a	6309498
r9             0x7ffff7fe5540	140737354028352
r10            0x3	3
r11            0x7ffff7a14890	140737347930256
r12            0x400c90	4197520
r13            0x7fffffffde80	140737488346752
r14            0x0	0
r15            0x0	0
rip            0x400ee9	0x400ee9 
eflags         0x206	[ PF IF ]
cs             0x33	51
ss             0x2b	43
ds             0x0	0
es             0x0	0
fs             0x0	0
gs             0x0	0

看到寄存器 esi (即 rsi 的低 32 位)被更新了!


Phase_2

首先查看 Phase_2 的汇编代码:

(gdb) disas phase_2
Dump of assembler code for function phase_2:
=> 0x0000000000400efc <+0>:	push   %rbp
   0x0000000000400efd <+1>:	push   %rbx
   0x0000000000400efe <+2>:	sub    $0x28,%rsp
   0x0000000000400f02 <+6>:	mov    %rsp,%rsi
   0x0000000000400f05 <+9>:	callq  0x40145c 
   0x0000000000400f0a <+14>:	cmpl   $0x1,(%rsp)
   0x0000000000400f0e <+18>:	je     0x400f30 
   0x0000000000400f10 <+20>:	callq  0x40143a 
   0x0000000000400f15 <+25>:	jmp    0x400f30 
   0x0000000000400f17 <+27>:	mov    -0x4(%rbx),%eax
   0x0000000000400f1a <+30>:	add    %eax,%eax
   0x0000000000400f1c <+32>:	cmp    %eax,(%rbx)
   0x0000000000400f1e <+34>:	je     0x400f25 
   0x0000000000400f20 <+36>:	callq  0x40143a 
   0x0000000000400f25 <+41>:	add    $0x4,%rbx
   0x0000000000400f29 <+45>:	cmp    %rbp,%rbx
   0x0000000000400f2c <+48>:	jne    0x400f17 
   0x0000000000400f2e <+50>:	jmp    0x400f3c 
   0x0000000000400f30 <+52>:	lea    0x4(%rsp),%rbx
   0x0000000000400f35 <+57>:	lea    0x18(%rsp),%rbp
   0x0000000000400f3a <+62>:	jmp    0x400f17 
   0x0000000000400f3c <+64>:	add    $0x28,%rsp
   0x0000000000400f40 <+68>:	pop    %rbx
   0x0000000000400f41 <+69>:	pop    %rbp
   0x0000000000400f42 <+70>:	retq   
End of assembler dump.

函数基本栈帧处理后,调用了一个名为 的函数,顾名思义,应该是读取六个数字,我们也可以反汇编这个函数的机器指令,查看其汇编指令:

(gdb) disas 0x40145c
Dump of assembler code for function read_six_numbers:
   0x000000000040145c <+0>:	sub    $0x18,%rsp
   0x0000000000401460 <+4>:	mov    %rsi,%rdx
   0x0000000000401463 <+7>:	lea    0x4(%rsi),%rcx
   0x0000000000401467 <+11>:	lea    0x14(%rsi),%rax
   0x000000000040146b <+15>:	mov    %rax,0x8(%rsp)
   0x0000000000401470 <+20>:	lea    0x10(%rsi),%rax
   0x0000000000401474 <+24>:	mov    %rax,(%rsp)
   0x0000000000401478 <+28>:	lea    0xc(%rsi),%r9
   0x000000000040147c <+32>:	lea    0x8(%rsi),%r8
   0x0000000000401480 <+36>:	mov    $0x4025c3,%esi
   0x0000000000401485 <+41>:	mov    $0x0,%eax
   0x000000000040148a <+46>:	callq  0x400bf0 <__isoc99_sscanf@plt>
   0x000000000040148f <+51>:	cmp    $0x5,%eax
   0x0000000000401492 <+54>:	jg     0x401499 
   0x0000000000401494 <+56>:	callq  0x40143a 
   0x0000000000401499 <+61>:	add    $0x18,%rsp
   0x000000000040149d <+65>:	retq   
End of assembler dump.

这里是涉及到调用库函数 scanf 时的一些参数配置,笔者也看不太明白,尝试查看了一下调用 scanf 之前配置的寄存器 esi 的立即寻址的内容:

(gdb) x/s 0x4025c3
0x4025c3:	"%d %d %d %d %d %d"

可以看出是逐位读取六个数字。观察寄存器 rsp 在 中的操作,又注意到返回 phase_2 后的第一个指令 cmpl   $0x1,(%rsp) ,猜测 rsp 就是指向我们输入的六个数字的第一个数字,六个数字是顺序保存在程序栈当中的。

我们验证一下:在进入指令 cmpl   $0x1,(%rsp) 之前设置断点,查看寄存器 rsp 保存的内存地址数据。

(gdb) break *0x0000000000400f0a
Breakpoint 1 at 0x400f0a
(gdb) run
Starting program: /home/liuyuan/Downloads/CSAPP/bomb/bomb 
Welcome to my fiendish little bomb. You have 6 phases with
which to blow yourself up. Have a nice day!
Border relations with Canada have never been better.
Phase 1 defused. How about the next one?
1 2 3 4 5 6

Breakpoint 1, 0x0000000000400f0a in phase_2 ()
(gdb) i r
rax            0x6	6
rbx            0x0	0
rcx            0x0	0
rdx            0x7fffffffdd74	140737488346484
rsi            0x0	0
rdi            0x7fffffffd6d0	140737488344784
rbp            0x402210	0x402210 <__libc_csu_init>
rsp            0x7fffffffdd60	0x7fffffffdd60
r8             0x0	0
r9             0x0	0
r10            0x7ffff7b82cc0	140737349430464
r11            0x4025d4	4203988
r12            0x400c90	4197520
r13            0x7fffffffde80	140737488346752
r14            0x0	0
r15            0x0	0
rip            0x400f0a	0x400f0a 
eflags         0x202	[ IF ]
cs             0x33	51
ss             0x2b	43
ds             0x0	0
es             0x0	0
fs             0x0	0
gs             0x0	0

查看 0x7fffffffdd60 地址的内存单元保存的 int 型数据,并查看相邻的五个数据(int 型为 32 位,每次偏移 8 字节):

(gdb) x/d 0x7fffffffdd60
0x7fffffffdd60:	1
(gdb) x/d 0x7fffffffdd64
0x7fffffffdd64:	2
(gdb) x/d 0x7fffffffdd68
0x7fffffffdd68:	3
(gdb) x/d 0x7fffffffdd6c
0x7fffffffdd6c:	4
(gdb) x/d 0x7fffffffdd70
0x7fffffffdd70:	5
(gdb) x/d 0x7fffffffdd74
0x7fffffffdd74:	6

和预期一致!接着读 phase_2 的汇编指令,很容易读出,第一位数必须是 1 ,每一位数(除了第一位)必须是相邻的前一位数的 2 倍,所以我们需要输出的是 1 2 4 8 16 32。

(gdb) run
Starting program: /home/liuyuan/Downloads/CSAPP/bomb/bomb 
Welcome to my fiendish little bomb. You have 6 phases with
which to blow yourself up. Have a nice day!
Border relations with Canada have never been better.
Phase 1 defused. How about the next one?
1 2 4 8 16 32
That's number 2.  Keep going!

完成!


Phase_3

还是先找到汇编代码:

(gdb) disas phase_3
Dump of assembler code for function phase_3:
   0x0000000000400f43 <+0>:	sub    $0x18,%rsp
   0x0000000000400f47 <+4>:	lea    0xc(%rsp),%rcx
   0x0000000000400f4c <+9>:	lea    0x8(%rsp),%rdx
   0x0000000000400f51 <+14>:	mov    $0x4025cf,%esi
   0x0000000000400f56 <+19>:	mov    $0x0,%eax
   0x0000000000400f5b <+24>:	callq  0x400bf0 <__isoc99_sscanf@plt>
   0x0000000000400f60 <+29>:	cmp    $0x1,%eax
   0x0000000000400f63 <+32>:	jg     0x400f6a 
   0x0000000000400f65 <+34>:	callq  0x40143a 
   0x0000000000400f6a <+39>:	cmpl   $0x7,0x8(%rsp)
   0x0000000000400f6f <+44>:	ja     0x400fad 
   0x0000000000400f71 <+46>:	mov    0x8(%rsp),%eax
   0x0000000000400f75 <+50>:	jmpq   *0x402470(,%rax,8)
   0x0000000000400f7c <+57>:	mov    $0xcf,%eax
   0x0000000000400f81 <+62>:	jmp    0x400fbe 
   0x0000000000400f83 <+64>:	mov    $0x2c3,%eax
   0x0000000000400f88 <+69>:	jmp    0x400fbe 
   0x0000000000400f8a <+71>:	mov    $0x100,%eax
   0x0000000000400f8f <+76>:	jmp    0x400fbe 
   0x0000000000400f91 <+78>:	mov    $0x185,%eax
   0x0000000000400f96 <+83>:	jmp    0x400fbe 
   0x0000000000400f98 <+85>:	mov    $0xce,%eax
   0x0000000000400f9d <+90>:	jmp    0x400fbe 
   0x0000000000400f9f <+92>:	mov    $0x2aa,%eax
   0x0000000000400fa4 <+97>:	jmp    0x400fbe 
   0x0000000000400fa6 <+99>:	mov    $0x147,%eax
---Type  to continue, or q  to quit---
   0x0000000000400fab <+104>:	jmp    0x400fbe 
   0x0000000000400fad <+106>:	callq  0x40143a 
   0x0000000000400fb2 <+111>:	mov    $0x0,%eax
   0x0000000000400fb7 <+116>:	jmp    0x400fbe 
   0x0000000000400fb9 <+118>:	mov    $0x137,%eax
   0x0000000000400fbe <+123>:	cmp    0xc(%rsp),%eax
   0x0000000000400fc2 <+127>:	je     0x400fc9 
   0x0000000000400fc4 <+129>:	callq  0x40143a 
   0x0000000000400fc9 <+134>:	add    $0x18,%rsp
   0x0000000000400fcd <+138>:	retq   
End of assembler dump.

类似之前的方法,先查看调用 scanf 之前配置寄存器 esi 的立即寻址的内容:

(gdb) x/s 0x4025cf
0x4025cf:	"%d %d"

那么这个函数就是需要读取两个整数了,保存的方式应该也是相同的,可以通过 rsp 做偏址寻址找到,我们尝试一下,断点依旧设置在刚刚结束调用 scanf 函数的位置,输入 1 2 两个数,情况如下:

(gdb) x/d 0x7fffffffdd80
0x7fffffffdd80:	0
(gdb) x/d 0x7fffffffdd84
0x7fffffffdd84:	0
(gdb) x/d 0x7fffffffdd88
0x7fffffffdd88:	1
(gdb) x/d 0x7fffffffdd8c
0x7fffffffdd8c:	2

第一个数值就是 rsp 保存的内存地址,进一步比较,可以归纳出调用 scanf 函数的参数情况:一个计数器(eax),一个指定读取数据格式的字符串(esi),以及指定各个数据存储在栈帧当中的地址(用多个寄存器表示)。

继续阅读汇编代码,可以看到要求输入的第一个数字 x1 <= 7(无符号比较),即 0 <= x1 <=7。然后经过一个根据 x1 的值进行跳转的指令,类似于 C 语言当中的 switch 指令,我们逐个测试一下跳转地址:

That's number 2.  Keep going!
0 0

Breakpoint 1, 0x0000000000400f75 in phase_3 ()
(gdb) stepi
0x0000000000400f7c in phase_3 ()

eax = 0xcf

That's number 2.  Keep going!
1 0

Breakpoint 1, 0x0000000000400f75 in phase_3 ()
(gdb) stepi
0x0000000000400fb9 in phase_3 ()

eax = 0x137

That's number 2.  Keep going!
2 0

Breakpoint 1, 0x0000000000400f75 in phase_3 ()
(gdb) stepi
0x0000000000400f83 in phase_3 ()

 eax = 0x2c3

That's number 2.  Keep going!
3 0

Breakpoint 1, 0x0000000000400f75 in phase_3 ()
(gdb) stepi
0x0000000000400f8a in phase_3 ()

eax = 0x100

That's number 2.  Keep going!
4 0

Breakpoint 1, 0x0000000000400f75 in phase_3 ()
(gdb) stepi
0x0000000000400f91 in phase_3 ()

eax = 0x185

That's number 2.  Keep going!
5 0

Breakpoint 1, 0x0000000000400f75 in phase_3 ()
(gdb) stepi
0x0000000000400f98 in phase_3 ()

eax = 0xce

That's number 2.  Keep going!
6 0

Breakpoint 1, 0x0000000000400f75 in phase_3 ()
(gdb) stepi
0x0000000000400f9f in phase_3 ()

eax = 0x2aa

That's number 2.  Keep going!
7 0

Breakpoint 1, 0x0000000000400f75 in phase_3 ()
(gdb) stepi
0x0000000000400fa6 in phase_3 ()

eax = 0x147

这里有 8 个答案,选取第一个,x1 = 0,x2 = 0xcf = 207,输入:

(gdb) run
Starting program: /home/liuyuan/Downloads/CSAPP/bomb/bomb 
Welcome to my fiendish little bomb. You have 6 phases with
which to blow yourself up. Have a nice day!
Border relations with Canada have never been better.
Phase 1 defused. How about the next one?
1 2 4 8 16 32
That's number 2.  Keep going!
0 207
Halfway there!

完成!这里笔者曾尝试利用指令 jmpq   *0x402470(,%rax,8) ,通过寻址方式计算,得不到结果,最后还是利用逐步调试的方法观察程序的走向,得到了正确的答案。


Plase_4

(gdb) disas phase_4
Dump of assembler code for function phase_4:
   0x000000000040100c <+0>:	sub    $0x18,%rsp
   0x0000000000401010 <+4>:	lea    0xc(%rsp),%rcx
   0x0000000000401015 <+9>:	lea    0x8(%rsp),%rdx
   0x000000000040101a <+14>:	mov    $0x4025cf,%esi
   0x000000000040101f <+19>:	mov    $0x0,%eax
   0x0000000000401024 <+24>:	callq  0x400bf0 <__isoc99_sscanf@plt>
   0x0000000000401029 <+29>:	cmp    $0x2,%eax
   0x000000000040102c <+32>:	jne    0x401035 
   0x000000000040102e <+34>:	cmpl   $0xe,0x8(%rsp)
   0x0000000000401033 <+39>:	jbe    0x40103a 
   0x0000000000401035 <+41>:	callq  0x40143a 
   0x000000000040103a <+46>:	mov    $0xe,%edx
   0x000000000040103f <+51>:	mov    $0x0,%esi
   0x0000000000401044 <+56>:	mov    0x8(%rsp),%edi
   0x0000000000401048 <+60>:	callq  0x400fce 
   0x000000000040104d <+65>:	test   %eax,%eax
   0x000000000040104f <+67>:	jne    0x401058 
   0x0000000000401051 <+69>:	cmpl   $0x0,0xc(%rsp)
   0x0000000000401056 <+74>:	je     0x40105d 
   0x0000000000401058 <+76>:	callq  0x40143a 
   0x000000000040105d <+81>:	add    $0x18,%rsp
   0x0000000000401061 <+85>:	retq   
End of assembler dump.

经过前面几个实验,读这段汇编程序应该很轻松了,翻译过来就是输入两个整数 x1 、x2,要求 0 <= x1 <= 14,x2 = 0。之后是三个参数 edx = 14, esi = 0,edi = x1 穿给一个函数,要求返回的一个值必须为 0。我们可以查看一下函数的汇编代码:

(gdb) disas 0x400fce
Dump of assembler code for function func4:
   0x0000000000400fce <+0>:	sub    $0x8,%rsp
   0x0000000000400fd2 <+4>:	mov    %edx,%eax
   0x0000000000400fd4 <+6>:	sub    %esi,%eax
   0x0000000000400fd6 <+8>:	mov    %eax,%ecx
   0x0000000000400fd8 <+10>:	shr    $0x1f,%ecx
   0x0000000000400fdb <+13>:	add    %ecx,%eax
   0x0000000000400fdd <+15>:	sar    %eax
   0x0000000000400fdf <+17>:	lea    (%rax,%rsi,1),%ecx
   0x0000000000400fe2 <+20>:	cmp    %edi,%ecx
   0x0000000000400fe4 <+22>:	jle    0x400ff2 
   0x0000000000400fe6 <+24>:	lea    -0x1(%rcx),%edx
   0x0000000000400fe9 <+27>:	callq  0x400fce 
   0x0000000000400fee <+32>:	add    %eax,%eax
   0x0000000000400ff0 <+34>:	jmp    0x401007 
   0x0000000000400ff2 <+36>:	mov    $0x0,%eax
   0x0000000000400ff7 <+41>:	cmp    %edi,%ecx
   0x0000000000400ff9 <+43>:	jge    0x401007 
   0x0000000000400ffb <+45>:	lea    0x1(%rcx),%esi
   0x0000000000400ffe <+48>:	callq  0x400fce 
   0x0000000000401003 <+53>:	lea    0x1(%rax,%rax,1),%eax
   0x0000000000401007 <+57>:	add    $0x8,%rsp
   0x000000000040100b <+61>:	retq   
End of assembler dump.

这段代码我懒得看了,有了前面的经验,我觉得这个实验的目的不是精读汇编,而是利用容易得到的信息结合调试的方法得到结果,直接把这段函数当做黑盒,测试一下即可,这里就不贴过程了,注意好断点就行,直接上结果:

1 0
So you got that one.  Try this one.

Phase_5

(gdb) disas phase_5
Dump of assembler code for function phase_5:
   0x0000000000401062 <+0>:	push   %rbx
   0x0000000000401063 <+1>:	sub    $0x20,%rsp
   0x0000000000401067 <+5>:	mov    %rdi,%rbx
   0x000000000040106a <+8>:	mov    %fs:0x28,%rax
   0x0000000000401073 <+17>:	mov    %rax,0x18(%rsp)
   0x0000000000401078 <+22>:	xor    %eax,%eax
   0x000000000040107a <+24>:	callq  0x40131b 
   0x000000000040107f <+29>:	cmp    $0x6,%eax
   0x0000000000401082 <+32>:	je     0x4010d2 
   0x0000000000401084 <+34>:	callq  0x40143a 
   0x0000000000401089 <+39>:	jmp    0x4010d2 
   0x000000000040108b <+41>:	movzbl (%rbx,%rax,1),%ecx
   0x000000000040108f <+45>:	mov    %cl,(%rsp)
   0x0000000000401092 <+48>:	mov    (%rsp),%rdx
   0x0000000000401096 <+52>:	and    $0xf,%edx
   0x0000000000401099 <+55>:	movzbl 0x4024b0(%rdx),%edx
   0x00000000004010a0 <+62>:	mov    %dl,0x10(%rsp,%rax,1)
   0x00000000004010a4 <+66>:	add    $0x1,%rax
   0x00000000004010a8 <+70>:	cmp    $0x6,%rax
   0x00000000004010ac <+74>:	jne    0x40108b 
   0x00000000004010ae <+76>:	movb   $0x0,0x16(%rsp)
   0x00000000004010b3 <+81>:	mov    $0x40245e,%esi
   0x00000000004010b8 <+86>:	lea    0x10(%rsp),%rdi
   0x00000000004010bd <+91>:	callq  0x401338 
   0x00000000004010c2 <+96>:	test   %eax,%eax
   0x00000000004010c4 <+98>:	je     0x4010d9 
---Type  to continue, or q  to quit---return
   0x00000000004010c6 <+100>:	callq  0x40143a 
   0x00000000004010cb <+105>:	nopl   0x0(%rax,%rax,1)
   0x00000000004010d0 <+110>:	jmp    0x4010d9 
   0x00000000004010d2 <+112>:	mov    $0x0,%eax
   0x00000000004010d7 <+117>:	jmp    0x40108b 
   0x00000000004010d9 <+119>:	mov    0x18(%rsp),%rax
   0x00000000004010de <+124>:	xor    %fs:0x28,%rax
   0x00000000004010e7 <+133>:	je     0x4010ee 
   0x00000000004010e9 <+135>:	callq  0x400b30 <__stack_chk_fail@plt>
   0x00000000004010ee <+140>:	add    $0x20,%rsp
   0x00000000004010f2 <+144>:	pop    %rbx
   0x00000000004010f3 <+145>:	retq   
End of assembler dump.

分析:首先由 24 和 29 ,猜测是输入一个长度为 6 的字符串;然后将 eax 置 0,跳转到 41 到 74 的循环,循环了 6 次进栈操作,然后进入 76 到 98 ,补上字符串尾部的 '\0' ,判断前面依次得到的 6 个字符组成的字符串是否与 0x40245e 上保存的字符串相同,相同则通过。

首先看看我们最终要得到的字符串是什么:

(gdb) x/s 0x40245e
0x40245e:	"flyers"

再仔细分析 6 个字符的压栈过程(进入之前 eax 已置 0):

   0x000000000040108b <+41>:	movzbl (%rbx,%rax,1),%ecx
   0x000000000040108f <+45>:	mov    %cl,(%rsp)
   0x0000000000401092 <+48>:	mov    (%rsp),%rdx
   0x0000000000401096 <+52>:	and    $0xf,%edx
   0x0000000000401099 <+55>:	movzbl 0x4024b0(%rdx),%edx
   0x00000000004010a0 <+62>:	mov    %dl,0x10(%rsp,%rax,1)
   0x00000000004010a4 <+66>:	add    $0x1,%rax
   0x00000000004010a8 <+70>:	cmp    $0x6,%rax
   0x00000000004010ac <+74>:	jne    0x40108b 

分析:rax 其实就是一个计数器,每完成一个字符的进栈就 +1,直至等于 6;rbx 存放的是我们输入的 6 个字符组成的字符串的首地址,rax 在这里又作为索引,依次取其中一个字符,传给 rcx ,又传给 rdx ;rdx 截取其中的低 4 位,又作为索引,找到 0x4024b0 保存的字符串的相应位置的字符,写入栈中 0x10(%rsp) 到 0x15(%rsp) 的位置上。

那么我们看看0x4024b0 保存的字符串是什么:

(gdb) x/s 0x4024b0
0x4024b0 :	"maduiersnfotvbylSo you think you can stop the bomb with ctrl-c, do you?"

对应 "flyers" 的索引分别是 +9、+15、+14、+5、+6、+7。二进制数分别为 1001、 1111、 1110 、0101、 0110、 0111,将高 4 位都置为 0110 ,组成的 ascii 码代表字母分别为:i o n e f g,即为我们要输入的字符串:

ionefg
Good work!  On to the next...

Phase_6

这段汇编代码是真的长,分段贴:

(gdb) disas phase_6
Dump of assembler code for function phase_6:
   0x00000000004010f4 <+0>:	push   %r14
   0x00000000004010f6 <+2>:	push   %r13
   0x00000000004010f8 <+4>:	push   %r12
   0x00000000004010fa <+6>:	push   %rbp
   0x00000000004010fb <+7>:	push   %rbx
   0x00000000004010fc <+8>:	sub    $0x50,%rsp
   0x0000000000401100 <+12>:	mov    %rsp,%r13
   0x0000000000401103 <+15>:	mov    %rsp,%rsi
   0x0000000000401106 <+18>:	callq  0x40145c 

这里对一些寄存器做了数据保存,然后是输入 6 个整数,和第二个实验调用的函数相同。

   0x000000000040110b <+23>:	mov    %rsp,%r14
   0x000000000040110e <+26>:	mov    $0x0,%r12d
   0x0000000000401114 <+32>:	mov    %r13,%rbp
   0x0000000000401117 <+35>:	mov    0x0(%r13),%eax
   0x000000000040111b <+39>:	sub    $0x1,%eax
   0x000000000040111e <+42>:	cmp    $0x5,%eax
   0x0000000000401121 <+45>:	jbe    0x401128 
   0x0000000000401123 <+47>:	callq  0x40143a 
   0x0000000000401128 <+52>:	add    $0x1,%r12d
   0x000000000040112c <+56>:	cmp    $0x6,%r12d
   0x0000000000401130 <+60>:	je     0x401153 
   0x0000000000401132 <+62>:	mov    %r12d,%ebx
   0x0000000000401135 <+65>:	movslq %ebx,%rax
   0x0000000000401138 <+68>:	mov    (%rsp,%rax,4),%eax
   0x000000000040113b <+71>:	cmp    %eax,0x0(%rbp)
   0x000000000040113e <+74>:	jne    0x401145 
   0x0000000000401140 <+76>:	callq  0x40143a 
---Type  to continue, or q  to quit---continue
   0x0000000000401145 <+81>:	add    $0x1,%ebx
   0x0000000000401148 <+84>:	cmp    $0x5,%ebx
   0x000000000040114b <+87>:	jle    0x401135 
   0x000000000040114d <+89>:	add    $0x4,%r13
   0x0000000000401151 <+93>:	jmp    0x401114 

简单说就是要求输入的 6 个整数都大于 0 且小于等于 6,并且 6 个数彼此互不相等。

   0x0000000000401158 <+100>:	mov    %r14,%rax
   0x000000000040115b <+103>:	mov    $0x7,%ecx
   0x0000000000401160 <+108>:	mov    %ecx,%edx
   0x0000000000401162 <+110>:	sub    (%rax),%edx
   0x0000000000401164 <+112>:	mov    %edx,(%rax)
   0x0000000000401166 <+114>:	add    $0x4,%rax
   0x000000000040116a <+118>:	cmp    %rsi,%rax
   0x000000000040116d <+121>:	jne    0x401160 

将原来的 6 个整数分别用 7 减并替换,如原来是 3 则替换为 4。

   0x000000000040116f <+123>:	mov    $0x0,%esi
   0x0000000000401174 <+128>:	jmp    0x401197 
   0x0000000000401176 <+130>:	mov    0x8(%rdx),%rdx
   0x000000000040117a <+134>:	add    $0x1,%eax
   0x000000000040117d <+137>:	cmp    %ecx,%eax
   0x000000000040117f <+139>:	jne    0x401176 
   0x0000000000401181 <+141>:	jmp    0x401188 
   0x0000000000401183 <+143>:	mov    $0x6032d0,%edx
   0x0000000000401188 <+148>:	mov    %rdx,0x20(%rsp,%rsi,2)
   0x000000000040118d <+153>:	add    $0x4,%rsi
   0x0000000000401191 <+157>:	cmp    $0x18,%rsi
   0x0000000000401195 <+161>:	je     0x4011ab 
   0x0000000000401197 <+163>:	mov    (%rsp,%rsi,1),%ecx
---Type  to continue, or q  to quit---continue
   0x000000000040119a <+166>:	cmp    $0x1,%ecx
   0x000000000040119d <+169>:	jle    0x401183 
   0x000000000040119f <+171>:	mov    $0x1,%eax
   0x00000000004011a4 <+176>:	mov    $0x6032d0,%edx
   0x00000000004011a9 <+181>:	jmp    0x401176 

这一段是创建了一个链表,直接查看其中立即数直接赋值的地址:

(gdb) x/24w 0x6032d0
0x6032d0 :	0x0000014c	0x00000001	0x006032e0	0x00000000
0x6032e0 :	0x000000a8	0x00000002	0x006032f0	0x00000000
0x6032f0 :	0x0000039c	0x00000003	0x00603300	0x00000000
0x603300 :	0x000002b3	0x00000004	0x00603310	0x00000000
0x603310 :	0x000001dd	0x00000005	0x00603320	0x00000000
0x603320 :	0x000001bb	0x00000006	0x00000000	0x00000000

 接着往下看:

   0x00000000004011ab <+183>:	mov    0x20(%rsp),%rbx
   0x00000000004011b0 <+188>:	lea    0x28(%rsp),%rax
   0x00000000004011b5 <+193>:	lea    0x50(%rsp),%rsi
   0x00000000004011ba <+198>:	mov    %rbx,%rcx
   0x00000000004011bd <+201>:	mov    (%rax),%rdx
   0x00000000004011c0 <+204>:	mov    %rdx,0x8(%rcx)
   0x00000000004011c4 <+208>:	add    $0x8,%rax
   0x00000000004011c8 <+212>:	cmp    %rsi,%rax
   0x00000000004011cb <+215>:	je     0x4011d2 
   0x00000000004011cd <+217>:	mov    %rdx,%rcx
   0x00000000004011d0 <+220>:	jmp    0x4011bd 
   0x00000000004011d2 <+222>:	movq   $0x0,0x8(%rdx)
   0x00000000004011da <+230>:	mov    $0x5,%ebp
   0x00000000004011df <+235>:	mov    0x8(%rbx),%rax
   0x00000000004011e3 <+239>:	mov    (%rax),%eax
   0x00000000004011e5 <+241>:	cmp    %eax,(%rbx)
   0x00000000004011e7 <+243>:	jge    0x4011ee 
   0x00000000004011e9 <+245>:	callq  0x40143a 
   0x00000000004011ee <+250>:	mov    0x8(%rbx),%rbx
   0x00000000004011f2 <+254>:	sub    $0x1,%ebp
   0x00000000004011f5 <+257>:	jne    0x4011df 

配置链表各个节点的 next 域,将各个节点关联起来,并将链表按照数值部分降序排序,根据第一列的数值降序排序重组链表顺序,得到的第二值顺序是 3 4 5 6 1 2 :

(gdb) x/24w 0x6032d0
0x6032d0 :	0x0000014c	0x00000001	0x006032e0	0x00000000
0x6032e0 :	0x000000a8	0x00000002	0x00000000	0x00000000
0x6032f0 :	0x0000039c	0x00000003	0x00603300	0x00000000
0x603300 :	0x000002b3	0x00000004	0x00603310	0x00000000
0x603310 :	0x000001dd	0x00000005	0x00603320	0x00000000
0x603320 :	0x000001bb	0x00000006	0x006032d0	0x00000000

即我们要输入的值为 4 3 2 1 6 5:

4 3 2 1 6 5
Congratulations! You've defused the bomb!

完成,还有隐藏任务,这里就先不做了。

你可能感兴趣的:(计算机专业课程学习记录)