CSAPP-bomblab
DO NOT READ THIS ARTICLE. I wrote bullshit in English.
This lab I have finished once, so this article is just for fun.
Phase-1
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
Just scan the memory of address [0x402400] . GDB command is x /s 0x402400 .
(gdb) x /s 0x402400
0x402400: "Border relations with Canada have never been better."
We can see that the answer is Border relations with Canada have never been better. .
Phase-2
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>
400f0a: 83 3c 24 01 cmpl $0x1,(%rsp)
400f0e: 74 20 je 400f30 <phase_2+0x34>
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
400f1a: 01 c0 add %eax,%eax
400f1c: 39 03 cmp %eax,(%rbx)
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
400f29: 48 39 eb cmp %rbp,%rbx
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
400f35: 48 8d 6c 24 18 lea 0x18(%rsp),%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
From this code read_six_number , we can see that we should input six numbers. And the we can know that there is a loop, and the most important line of code is :
400f1a: 01 c0 add %eax,%eax
Now, we can guess that it should be a geometric series, whose ratio is 2. What is the first element of the number series? We can see:
400f0a: 83 3c 24 01 cmpl $0x1,(%rsp)
400f0e: 74 20 je 400f30 <phase_2+0x34>
400f10: e8 25 05 00 00 callq 40143a <explode_bomb>
OK, we can see that if (%rsp) is not equal to 0x1 , it will call function explode_bomb .
Thus, the answer is : 1 2 4 8 16 32 .
Phase-3
0000000000400f43 <phase_3>:
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>
400f65: e8 d0 04 00 00 callq 40143a <explode_bomb>
400f6a: 83 7c 24 08 07 cmpl $0x7,0x8(%rsp)
400f6f: 77 3c ja 400fad <phase_3+0x6a>
400f71: 8b 44 24 08 mov 0x8(%rsp),%eax
400f75: ff 24 c5 70 24 40 00 jmpq *0x402470(,%rax,8)
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
Look at this line of code:
400f51: be cf 25 40 00 mov $0x4025cf,%esi
Run x /s 0x4025cf in GDB, it will print: "%d %d" . Thus, we should input 2 int in this phase.
And we can see that there are many annoying jmp 0x400fb9 instructions in this assembly code, so we can infer that there is a switch-case structure.
The first 2 lines of code below shows that the first number we input (stored in [%rsp + 8]) is less equal than 7, otherwise it will call explode_bomb . The last 2 lines tell us that the program will jump to an address according to our first number.
400f6a: 83 7c 24 08 07 cmpl $0x7,0x8(%rsp)
400f6f: 77 3c ja 400fad <phase_3+0x6a>
400f71: 8b 44 24 08 mov 0x8(%rsp),%eax
400f75: ff 24 c5 70 24 40 00 jmpq *0x402470(,%rax,8)
We know that the first number (now it has been copied to %rax) can be one of number in [0,1,2,3,4,5,6,7]. So scan the address [0x402470 + %rax * 8] :
(gdb) x /x 0x402470 + 8 * 0 => 0x402470: 0x00400f7c => 0 0xcf
(gdb) x /x 0x402470 + 8 * 1 => 0x402478: 0x00400fb9 => 1 0x137
(gdb) x /x 0x402470 + 8 * 2 => 0x402480: 0x00400f83 => 2 0x2c3
(gdb) x /x 0x402470 + 8 * 3 => 0x402488: 0x00400f8a => 3 0x100
(gdb) x /x 0x402470 + 8 * 4 => 0x402490: 0x00400f91 => 4 0x185
(gdb) x /x 0x402470 + 8 * 5 => 0x402498: 0x00400f98 => 5 0xce
(gdb) x /x 0x402470 + 8 * 6 => 0x4024a0: 0x00400f9f => 6 0x2aa
(gdb) x /x 0x402470 + 8 * 7 => 0x4024a8: 0x00400fa6 => 7 0x147
When the first number we input is 0, the program will jump to 0x400f7c. And the code at 0x400f7c is :
400f7c: b8 cf 00 00 00 mov $0xcf,%eax
So the second number is 0xcf = 207. Total answer is 0 207.
Phase-4
; first call is func4(edx=a, esi=b, edi=x)
0000000000400fce <func4>:
400fce: 48 83 ec 08 sub $0x8,%rsp
400fd2: 89 d0 mov %edx,%eax ; eax = a
400fd4: 29 f0 sub %esi,%eax ; eax = eax - esi = a - b = t1
400fd6: 89 c1 mov %eax,%ecx ; ecx = eax = a - b
400fd8: c1 e9 1f shr $0x1f,%ecx ; ecx = ecx >> 31 = t1 >> 31 = t2
400fdb: 01 c8 add %ecx,%eax ; eax = eax + ecx = t1 + t2
400fdd: d1 f8 sar %eax ; eax = eax >> 1 = (t1+t2)>>1, t1=(t1+t2)>>1
400fdf: 8d 0c 30 lea (%rax,%rsi,1),%ecx ; ecx = eax + esi = t1 + b, t2=t1+b
400fe2: 39 f9 cmp %edi,%ecx ; ecx <= x ? () : (2 * func4())
400fe4: 7e 0c jle 400ff2 <func4+0x24> ;
400fe6: 8d 51 ff lea -0x1(%rcx),%edx ; edx = ecx - 1 = t2 - 1
400fe9: e8 e0 ff ff ff callq 400fce <func4> ; func4(edx=t2-1, b, x)
400fee: 01 c0 add %eax,%eax ; return 2 * func4()
400ff0: eb 15 jmp 401007 <func4+0x39>
400ff2: b8 00 00 00 00 mov $0x0,%eax ; eax = 0
400ff7: 39 f9 cmp %edi,%ecx ; ecx >= edi ? (return 0) : ()
400ff9: 7d 0c jge 401007 <func4+0x39>
400ffb: 8d 71 01 lea 0x1(%rcx),%esi ; esi = ecx + 1
400ffe: e8 cb ff ff ff callq 400fce <func4> ; func4()
401003: 8d 44 00 01 lea 0x1(%rax,%rax,1),%eax ; return 2 * func4() + 1
401007: 48 83 c4 08 add $0x8,%rsp
40100b: c3 retq
000000000040100c <phase_4>:
40100c: 48 83 ec 18 sub $0x18,%rsp
401010: 48 8d 4c 24 0c lea 0xc(%rsp),%rcx
401015: 48 8d 54 24 08 lea 0x8(%rsp),%rdx
40101a: be cf 25 40 00 mov $0x4025cf,%esi
; scan the address 0x4025cf
40101f: b8 00 00 00 00 mov $0x0,%eax
401024: e8 c7 fb ff ff callq 400bf0 <__isoc99_sscanf@plt>
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)
; the 1st number x should be x <= 15, otherwise explode
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(edx=0xe, esi=0x0, edi=x)
40104d: 85 c0 test %eax,%eax
; the returned val of func4 shouble 0
40104f: 75 07 jne 401058 <phase_4+0x4c>
401051: 83 7c 24 0c 00 cmpl $0x0,0xc(%rsp)
; the 2nd number y should be 0, otherwise explode
; so the total condition is func(0xe,0x0,x) == 0 and y = 0
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
phase_4 call a function func4, which is a recursion. Scan the address 0x4025cf, GDB will print "%d %d", which means we should input 2 int numbers. Assume that our input is x y. The x will be stored in %rsp + 0x8 and y in %rsp + 0xc. Then phase_4 would call func4(edx=0xe, esi=0x0, edi=x), and this function call passes parameters by register. Read the comment above.
The conditions for passing phase_4 is input x y to let func(0xe, 0x0, x) return 0 and x should be less equal than 15 and y must be 0.
Read the code of func4, we can change it into C code:
int func4(int a, int b, int x)
{
x; // edi
int t1 = a - b; // eax
int t2 = ((unsigned int)t1) >> 31; // ecx
t1 = (t1 + t2) >> 1;
t2 = t1 + b;
if (t2 <= x)
{
t1 = 0;
if (t2 < x)
return 2 * func4(a, t2 + 1, x) + 1;
return t1;
}
else
return 2 * func4(t2 - 1, b, x);
}
What we need to pay attention to is this instruction: 400fd8: shr $0x1f,%ecx. The shr implies that it will operation on a unsigned type (read the C code above) .
Therefore, we can find x by enumeration.
int main()
{
int x = 0;
for (x = 0; x <= 15; x++)
{
int t = func4(0xe, 0x0, x);
if (t == 0)
printf("%d ", x);
}
}
The result is:
(x,y) = (0,0), (1,0), (3,0), (7,0)
Phase-5
0000000000401062 <phase_5>:
401062: 53 push %rbx
401063: 48 83 ec 20 sub $0x20,%rsp
; 32 bytes stack for temporary variable
401067: 48 89 fb mov %rdi,%rbx
; rdi is the address of strings we input
40106a: 64 48 8b 04 25 28 00 mov %fs:0x28,%rax
401071: 00 00
; I don't understand here
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>
40107f: 83 f8 06 cmp $0x6,%eax
; we should input a string, whose len is 6
; do while loop begin, eax=0 is the counter
401082: 74 4e je 4010d2 <phase_5+0x70>
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
; t = str[i]
40108f: 88 0c 24 mov %cl,(%rsp)
401092: 48 8b 14 24 mov (%rsp),%rdx
401096: 83 e2 0f and $0xf,%edx
; t = str[i] & 0xf
401099: 0f b6 92 b0 24 40 00 movzbl 0x4024b0(%rdx),%edx
4010a0: 88 54 04 10 mov %dl,0x10(%rsp,%rax,1)
; array[rsp+0x10+rax] = str2[t]
4010a4: 48 83 c0 01 add $0x1,%rax
4010a8: 48 83 f8 06 cmp $0x6,%rax
; eax++ until out of str.length
4010ac: 75 dd jne 40108b <phase_5+0x29>
4010ae: c6 44 24 16 00 movb $0x0,0x16(%rsp)
4010b3: be 5e 24 40 00 mov $0x40245e,%esi
; "flyers"
4010b8: 48 8d 7c 24 10 lea 0x10(%rsp),%rdi
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>
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
4010d7: eb b2 jmp 40108b <phase_5+0x29>
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
This phase_5 want us to input a string str (its length is 6), which is a simple phase.
There is a hash table in the program:
hex index: 0 1 2 3 4 5 6 7 8 9 A B C D E F
hash str : m a d u i e r s n f o t v b y l
phase_5 construct a hash function is: f(x) = hash_str[x & 0xf].
We aim to let f(str) = "flyers".
flyers is equal to character 9, F, E, 4, 5, 6 in the hash table, thus, str[i] & 0xf (i=0,...,5) should be 9, F, E, 4, 5, 6. We can check the ASCII table to get characters:
0x39 0x3F 0x3E 0x34 0x35 0x36
'9' '?' '>' '4' '5' '6'
The string "9?>456" is the key to pass this phase.
If you still do not understand, see the C code:
void phase5(const char *input)
{
const char *hash_str = "maduiersnfotvbyl";
char array[7];
int i = 0;
for (i = 0; i < 6; i++)
array[i] = hash_str[input[i] & 0xf];
if (sting_not_equal(array, "flyers"))
explode();
}
Evaluation
All the answers in a text files sin.ans:
unix > cat sin.ans
Border relations with Canada have never been better.
1 2 4 8 16 32
0 207
7 0 DrEvil
9?>456
4 3 2 1 6 5
20
Execute the bomb binary file:
unix > ./bomb sin.ans
Welcome to my fiendish little bomb. You have 6 phases with
which to blow yourself up. Have a nice day!
Phase 1 defused. How about the next one?
That's number 2. Keep going!
Halfway there!
So you got that one. Try this one.
Good work! On to the next...
Curses, you've found the secret phase!
But finding it and solving it are quite different...
Wow! You've defused the secret stage!
Congratulations! You've defused the bomb!
Summary
好无聊，希望早点开学
CSAPP-bomblab的更多相关文章
- CSAPP 之 BombLab 详解
前言 本篇博客将会展示 CSAPP 之 BombLab 的拆弹过程,粉碎 Dr.Evil 的邪恶阴谋.Dr.Evil 的替身,杀手皇后,总共设置了 6 个炸弹,每个炸弹对应一串字符串,如果字符串错误, ...
- CSAPP：bomblab
BOMBLAB实验总结 CSAPP实验BOMB,很头疼,看不懂,勉强做完了. 答案是这样的: Border relations with Canada have never been better. ...
- 终于要开始做大名鼎鼎的BombLab了！
首先是一些准备工作 lab下载地址:http://csapp.cs.cmu.edu/3e/labs.html 第二个的Bomblab的 self-study handout就是 在做这个lab前,首先 ...
- 在Ubuntu下使用 csapp.h 和 csapp.c
它山之石可以攻玉. 对于<深入理解计算机系统>这本神人写就的神书, 我等凡人就不评论什么啦. 这本书的 第二,三 部分, 真的真的对我理解操作系统有很大的帮助. (当然, 如果你不看第一部 ...
- CSAPP读书随笔之一：为什么汇编器会将call指令中的引用的初始值设置为-4
CSAPP,即<深入理解计算机系统:程序员视角>第三版,是一本好书,但读起来确需要具备相当的基本功.而且,有的表述(中译文)还不太直白. 比如,第463页提到,(对于32位系统)为什么汇编 ...
- 深入理解计算机中的 csapp.h和csapp.c
csapp.h其实就是一堆头文件的打包,在http://csapp.cs.cmu.edu/public/code.html 这里可以下载.这是<深入理解计算机系统>配套网站. 在头文件的# ...
- CSAPP(前言)
很久之前就听过有过CSAPP这本书的传闻了,今天终于决定上手这本神作:既然是神作,就要仔细拜读一下,今天看了一下前言部分还真的令人耳目一新,单单是前言部分就让我学习到几个新的知识点: 1.c和Java ...
- CSAPP缓冲区溢出攻击实验(上)
CSAPP缓冲区溢出攻击实验(上) 下载实验工具.最新的讲义在这. 网上能找到的实验材料有些旧了,有的地方跟最新的handout对不上.只是没有关系,大体上仅仅是程序名(sendstring)或者參数 ...
- 读完了csapp(中文名：深入理解计算机系统)
上个星期最终把csapp看完了. 我买的是中文版的,由于除了貌似评价不错以外,由于涉及到些自己不了解的底层东西,怕是看英文会云里雾里.如今看来,大概不能算是个长处,可是的确可以加快我的看书速度,否则一 ...
- CSAPP Lab2: Binary Bomb
著名的CSAPP实验:二进制炸弹 就是通过gdb和反汇编猜测程序意图,共有6关和一个隐藏关卡 只有输入正确的字符串才能过关,否则会程序会bomb终止运行 隐藏关卡需要输入特定字符串方会开启 实验材料下 ...
随机推荐
- 一道二叉树题的n步优化——LeetCode98validate binary search tree(草稿)
树的题目,往往可以用到三种遍历.以及递归,因为其结构上天然地可以往深处递归,且判断条件也往往不复杂(左右子树都是空的). LeetCode 98题讲的是,判断一棵树是不是二叉搜索树. 题目中给的是标准 ...
- Golang/Python/PHP带你彻底学会gRPC
目录 一.gRPC是什么? 二.Protocol Buffers是什么? 三.需求:开发健身房服务 四.最佳实践 Golang 1. 安装protoc工具 2. 安装protoc-gen-go 3. ...
- 压力测试（三）-自定义变量和CSV可变参数实操
1.Jmeter用户自定义变量实战 简介:什么是用户自定义变量,怎样使用 为什么使用:很多变量在全局中都有使用,或者测试数据更改,可以在一处定义,四处使用 比如服务器地址 1.线程组->add ...
- 基于netty的群聊
基于netty的群聊 学了一段时间的netty知识,现在通过这个基于console的程序来对netty的相关接口做个简单的应用. 准备 依赖 <dependency> <groupI ...
- windows下tensorflow/objectdetection API环境搭建（基于tensorflow1.14和python3.6）
此前就听闻室友说tensorflow在windows下坑很多,这次终于亲身领会到了.以下是参考网上大佬的教程以及自己的踩坑史总结出的有效步骤(亲测有效) 1.下载objectdetection所在的m ...
- JZOJ 1776. 经济编码 (Standard IO)
1776. 经济编码 (Standard IO) Time Limits: 1000 ms Memory Limits: 128000 KB Description 为降低资料储存的空间或增加资料传送 ...
- 学以致用：手把手教你撸一个工具库并打包发布，顺便解决JS浮点数计算精度问题
本文讲解的是怎么实现一个工具库并打包发布到npm给大家使用.本文实现的工具是一个分数计算器,大家考虑如下情况: \[ \sqrt{(((\frac{1}{3}+3.5)*\frac{2}{9}-\fr ...
- nes 红白机模拟器 第7篇 编译使用方法
模拟器,基于 InfoNES ,作者添加修改以下功能: 1, joypad 真实手柄驱动程序(字符型设备驱动) 2,原始图像只有256*240 ,添加 图像放大算法,这里实现了2种,a, 最近邻插值 ...
- Spring源码阅读笔记05：自定义xml标签解析
在上篇文章中,提到了在Spring中存在默认标签与自定义标签两种,并且详细分析了默认标签的解析,本文就来分析自定义标签的解析,像Spring中的AOP就是通过自定义标签来进行配置的,这里也是为后面学习 ...
- 双列集合Map相关面试题
一.了解Map集合吗?Map集合都有哪些实现 HashMap HashTable LinkedHashMap TreeMap ConcurrentHashMap 二.HashMap和HashTable ...