Today,we will do a experiment about computer system basic knowledge.The experiment is based on Ubuntu system.Before we start our experiment,we should install Ubuntu System.
Step 1:
write program
#include<stdio.h>
int main(void)
{
float x=-0.125;
float y=7.5;
int z=100;
printf("x=%f y=%f z=%d\n",x,y,z);
return 0;
}
Step 2:
compile and run programs
input command:gcc -g -p.c -o p
generate executable file p
Step 3:
view these variables of numbers of machine in the memory.
use command x/nfu<addr>view the memory
Parameter n,which indicates the number of memory unite to view
Parameter f is t,indicating that variables are displayed in binary format
Parameter u is b,indicating that the length of each address unite is one byte
#include<stdio.h>
int main(void)
{
int x=-32768;
short y=522;
unsigned z=65530;
char c='@';
float a=-1.1;
double b=10.5;
return 0;
}
https://blog.csdn.net/u011068702/article/details/53925415
https://blog.csdn.net/maikeerdai/article/details/17530937
使用命令
gcc -S prog1.c
然后prog1.s
使用命令
gcc -c prog1.c
生成文件prog1.o
使用命令
gcc -o prog prog1.c
生成可以执行文件prog
下面查看两个文件的不同：
https://blog.csdn.net/zoomdy/article/details/50563680
使用命令
objdump -d prog1.o
查看prog1.o反汇编之后的源码
使用命令
objdump -d prog
查看prog1反汇编之后的源码
下面要开始认真的做了，嘻嘻嘻
冲鸭
bomb.c源码
/***************************************************************************
* Dr. Evil's Insidious Bomb, Version 1.1
* Copyright 2011, Dr. Evil Incorporated. All rights reserved.
*
* LICENSE:
*
* Dr. Evil Incorporated (the PERPETRATOR) hereby grants you (the
* VICTIM) explicit permission to use this bomb (the BOMB). This is a
* time limited license, which expires on the death of the VICTIM.
* The PERPETRATOR takes no responsibility for damage, frustration,
* insanity, bug-eyes, carpal-tunnel syndrome, loss of sleep, or other
* harm to the VICTIM. Unless the PERPETRATOR wants to take credit,
* that is. The VICTIM may not distribute this bomb source code to
* any enemies of the PERPETRATOR. No VICTIM may debug,
* reverse-engineer, run "strings" on, decompile, decrypt, or use any
* other technique to gain knowledge of and defuse the BOMB. BOMB
* proof clothing may not be worn when handling this program. The
* PERPETRATOR will not apologize for the PERPETRATOR's poor sense of
* humor. This license is null and void where the BOMB is prohibited
* by law.
***************************************************************************/
#include <stdio.h>
#include <stdlib.h>
#include "support.h"
#include "phases.h"
/*
* Note to self: Remember to erase this file so my victims will have no
* idea what is going on, and so they will all blow up in a
* spectaculary fiendish explosion. -- Dr. Evil
*/
FILE *infile;
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);
}
/* Do all sorts of secret stuff that makes the bomb harder to defuse. */
initialize_bomb();
printf("Welcome to my fiendish little bomb. You have 6 phases with\n");
printf("which to blow yourself up. Have a nice day!\n");
/* 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! */
return 0;
}
看看汇编是啥东西呢？
phase_1
0000000000400ee0 <phase_1>: #函数的名称和函数的起始地址，可见函数最终变成了一些列的指令，没有其他的东西
400ee0: 48 83 ec 08 sub $0x8,%rsp #R[rsp] <- R[rsp]-8 栈指针寄存器往下移动8个字节
400ee4: be 00 24 40 00 mov $0x402400,%esi #R[esi] <- 402400
400ee9: e8 4a 04 00 00 callq 401338 <strings_not_equal> #调用函数，同时把函数的返回地址400eee写入到M[R[rsp]],即M[R[rsp]] <- 400eee
400eee: 85 c0 test %eax,%eax #test测试指令，若EAX是0，则ZF=1,否则ZF=0 eg.test指令对两个操作数进行与与运算，但是不改变操作数的数值
400ef0: 74 05 je 400ef7 <phase_1+0x17> #je = jz ，ZF=1时候执行
400ef2: e8 43 05 00 00 callq 40143a <explode_bomb> #调用函数，同时把函数的返回地址写入M[R[rsp]],覆盖调用函数<string_not_equal>的返回地址，即M[R[rsp]]<-400ef7
400ef7: 48 83 c4 08 add $0x8,%rsp # R[rsp] <- R[rsp]+8
400efb: c3 retq #R[rsp]<-R[rsp]-8 , R[rcs]<-M[R[rsp]] 取出返回地址，并且把返回地址写入RCS寄存器中
<string_length>
000000000040131b <string_length>:
40131b: 80 3f 00 cmpb $0x0,(%rdi) #判断给出的首地址的第一个单元的内容是否为0，如果是0，那么直接跳转到401332，把0放入RAX，然后返回
40131e: 74 12 je 401332 <string_length+0x17>
401320: 48 89 fa mov %rdi,%rdx #使用RDX保存字符串的首地址
401323: 48 83 c2 01 adda $0x1,%rdx #在循环中，RDX不断加一
401327: 89 d0 mov %edx,%eax #R[eax]<-R[edx]
401329: 29 f8 sub %edi,%eax #R[eax]<-R[eax]-R[edi]
40132b: 80 3a 00 cmpb $0x0,(%rdx) #M[R[rdx]]中的内容和0做比较
40132e: 75 f3 jne 401323 <string_length+0x8> #如果M[R[rdx]]中内容不是0，那么继续循环执行，即判断是否到达字符串的结尾
401330: f3 c3 repz retq #
401332: b8 00 00 00 00 mov $0x0,%eax #R[rax]<-0
401337: c3 retq #返回
string_not_equal
0000000000401338 <strings_not_equal>:
401338: 41 54 push %r12 #R[rsp]<-R[rsp]-8 , M[R[rsp]]<-R[r12]
40133a: 55 push %rbp #R[rsp]<-R[rsp]-8,M[R[rsp]]<-R[rbp]
40133b: 53 push %rbx #R[rsp]<-R[rsp]-8,M[R[rsp]]<-R[rbx]
40133c: 48 89 fb mov %rdi,%rbx #R[rbx]<-R[rdi] 使用rbx接受第一个参数，
40133f: 48 89 f5 mov %rsi,%rbp #R[rbp]<-R[rsi]
401342: e8 d4 ff ff ff callq 40131b <string_length> #计算参数RDI表示的字符串的长度
401347: 41 89 c4 mov %eax,%r12d #R[12d]保存字符串的长度
40134a: 48 89 ef mov %rbp,%rdi #R[rdi]<-R[rbp],此时RDI中的值为0x402400
40134d: e8 c9 ff ff ff callq 40131b <string_length> #计算字符串的长度，具体的字符串的内容可以查看内存中的数据
401352: ba 01 00 00 00 mov $0x1,%edx #R[edx]<-1
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> #如果两个字符串的长度不相同，那么返回1
40135c: 0f b6 03 movzbl (%rbx),%eax #R[eax]<-M[R[rbx]] 这个过程经过0扩展，RBX保存输入的字符串的首地址，所以此时EAX保存第一个字符的数值
40135f: 84 c0 test %al,%al #判断这个字符的低8位是否是0
401361: 74 25 je 401388 <strings_not_equal+0x50> #如果是0,即ZF=1跳转，然后返回0，但是如果第一个字符的数值为0，那么长度也是0
401363: 3a 45 00 cmp 0x0(%rbp),%al #比较RBP这个地址表示的字符串的第一个字符和输入的字符串的第一个
401366: 74 0a je 401372 <strings_not_equal+0x3a> #如果两个字符相同，跳转
401368: eb 25 jmp 40138f <strings_not_equal+0x57> #如果不相同，返回1
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 #R[rbx]<-R[rbx]+1
401376: 48 83 c5 01 add $0x1,%rbp #R[rbp]<-R[rbp]+1
40137a: 0f b6 03 movzbl (%rbx),%eax #把M[R[rbx]]内容放到EAX中，即输入的第二个字符
40137d: 84 c0 test %al,%al #如果是空字符，ZF=1
40137f: 75 e9 jne 40136a <strings_not_equal+0x32> #如果不是空字符，继续判断
401381: ba 00 00 00 00 mov $0x0,%edx #如果两个字符串全部相同，返回0
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 #恢复
40139e: 5d pop %rbp #恢复
40139f: 41 5c pop %r12 #恢复
4013a1: c3 retq #返回
具体的分析过程
<phase_1>在ESI中保存了数据0x402400然后调用函数<strings_not_equal>，然后判断EAX寄存器是否为0.如果为0，那么ZF=1,此时会执行je指令，跳过explode_bomb函数.因此需要EAX为0即可.那么什么时候EAX为0呢？看一下<string_not_equal>函数，发现这个函数使用了一个循环比较ESI和EDI表示的两个字符串，如果相同，返回0，如果不同，返回1。因此，只要输入的字符串满足和程序给出的字符串相同即可。
RAX保存输入的字符串的地址
在调用函数<phase_1>之前，使用RDI保存RAX的值，即第一个参数是输入的字符串的地址
进入函数<phase_1>中时，没有修改RDI的值（RDI仍然是字符串的首地址），同时设置RSI的值，两个寄存器作为两个<string_not_equal>的参数,这两个参数表示两个字符串
把输入的字符串的地址保存的寄存器RBX中
查看内容从地址0x402400开始的字符串,长度为52
使用十六进制显示一下，方便阅读
使用十六进制显示的字符串
这个字符串的长度为54
保存到文件文件
使用C语言读入这段16进制的数，用字符输出
结果：
Border relations with Canada have never been better.
把这个字符串输入，结果如下图所示：
第一个炸弹拆开的截图
第二个炸弹：
<phase_2>
0000000000400efc <phase_2>:
400efc: 55 push %rbp
400efd: 53 push %rbx
400efe: 48 83 ec 28 sub $0x28,%rsp #R[rsp]<-R[rsp]-28
400f02: 48 89 e6 mov %rsp,%rsi #R[rsi]<-R[rsp] ，设置第二个参数
400f05: e8 52 05 00 00 callq 40145c <read_six_numbers>
400f0a: 83 3c 24 01 cmpl $0x1,(%rsp) #比较M[R[rsp]]和1
400f0e: 74 20 je 400f30 <phase_2+0x34> #如果M[R[rsp]]=1
400f10: e8 25 05 00 00 callq 40143a <explode_bomb> #如果不是1，那么爆炸了
400f15: eb 19 jmp 400f30 <phase_2+0x34> #无条件跳转到0x400f30
400f17: 8b 43 fc mov -0x4(%rbx),%eax #R[eax]<-M[R[rbx]-4]
400f1a: 01 c0 add %eax,%eax #R[eax]<-R[eax]+R[eax]
400f1c: 39 03 cmp %eax,(%rbx) #比较EAX和M[R[rbx]]
400f1e: 74 05 je 400f25 <phase_2+0x29> #如果相同，跳转一下，如果不同，爆炸了
400f20: e8 15 05 00 00 callq 40143a <explode_bomb>
400f25: 48 83 c3 04 add $0x4,%rbx #R[rbx]<-R[rbx]+0x4
400f29: 48 39 eb cmp %rbp,%rbx #比较RBP和RBX
400f2c: 75 e9 jne 400f17 <phase_2+0x1b> #如果不相同，那么循环
400f2e: eb 0c jmp 400f3c <phase_2+0x40> #如果相同，那么可以成功的拆除炸弹了
400f30: 48 8d 5c 24 04 lea 0x4(%rsp),%rbx #R[rbx]<-R[rsp]+0x4
400f35: 48 8d 6c 24 18 lea 0x18(%rsp),%rbp #R[rbp]<-R[rsp]+0x18
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
画一下对应的栈结构，发现<phase_2>通过不断的移动RBX进行相邻值之间的比较,当RBP=RBX时候停止比较
<read_six_numbers>
000000000040145c <read_six_numbers>:
40145c: 48 83 ec 18 sub $0x18,%rsp #R[rsp]<-R[rsp]-0x18
401460: 48 89 f2 mov %rsi,%rdx #R[rdx]<-R[rsi]，R[rsi]保存<phase_2>函数的栈底的地址，从这个地址开始依次存数6个，每个数占4个字节，此时RDX是第一个数的地址
401463: 48 8d 4e 04 lea 0x4(%rsi),%rcx #R[rcx]<-R[rsi]+0x4 第二个数的地址
401467: 48 8d 46 14 lea 0x14(%rsi),%rax#R[rax]<-R[rsi]+0x14 最后一个数的地址
40146b: 48 89 44 24 08 mov %rax,0x8(%rsp) #M[R[rsp]+8]<-R[rax]
401470: 48 8d 46 10 lea 0x10(%rsi),%rax #R[rax]<-R[rsi]+16
401474: 48 89 04 24 mov %rax,(%rsp) #M[R[rsp]]<-R[rax]
401478: 4c 8d 4e 0c lea 0xc(%rsi),%r9 #R[r9]<-R[rsi]+12
40147c: 4c 8d 46 08 lea 0x8(%rsi),%r8 #R[r8]<-[R[rsi]+8
401480: be c3 25 40 00 mov $0x4025c3,%esi #R[esi]<-0x4025c3
401485: b8 00 00 00 00 mov $0x0,%eax #R[eax]<-0
40148a: e8 61 f7 ff ff callq 400bf0 <__isoc99_sscanf@plt>
40148f: 83 f8 05 cmp $0x5,%eax #比较一下
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
第三个炸弹：
<phase_3>
0000000000400f43 <phase_3>:
400f43: 48 83 ec 18 sub $0x18,%rsp #R[rsp]<-R[rsp]-24
400f47: 48 8d 4c 24 0c lea 0xc(%rsp),%rcx #R[rcx]<-R[rsp]+12
400f4c: 48 8d 54 24 08 lea 0x8(%rsp),%rdx #R[rdx]<-R[rsp]+8
400f51: be cf 25 40 00 mov $0x4025cf,%esi #R[esi]<-$0x4025cf
400f56: b8 00 00 00 00 mov $0x0,%eax #R[eax]<-$0x0
400f5b: e8 90 fc ff ff callq 400bf0 <__isoc99_sscanf@plt> #调用输入的函数
400f60: 83 f8 01 cmp $0x1,%eax #EAX和$0x1比较
400f63: 7f 05 jg 400f6a <phase_3+0x27> #如果EAX>1，那么跳转到0x400f6a
400f65: e8 d0 04 00 00 callq 40143a <explode_bomb> #如果小于等于，那么爆炸，说明EAX需要大于1，说明至少输入两个数
400f6a: 83 7c 24 08 07 cmpl $0x7,0x8(%rsp) #比较M[R[rsp]+8]和7
400f6f: 77 3c ja 400fad <phase_3+0x6a> #如果M[R[rsp]+8]>7,那么爆炸，说明M[R[rsp]+8]<=7
400f71: 8b 44 24 08 mov 0x8(%rsp),%eax #R[eax]<-M[R[rsp]+8]
400f75: ff 24 c5 70 24 40 00 jmpq *0x402470(,%rax,8) #这是一个间接跳转的指令，地址=*（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 #比较 EAX 和 M[R[rsp]+12]
400fc2: 74 05 je 400fc9 <phase_3+0x86> #如果相等，那么成功拆掉炸弹，否则继续执行（继续执行会爆炸）
400fc4: e8 71 04 00 00 callq 40143a <explode_bomb>
400fc9: 48 83 c4 18 add $0x18,%rsp #把栈增长的部分收回
400fcd: c3 retq #返回
函数在执行的过程中有一个间接跳转指令，跳转地址是M[0x402470+R[rax]*8 ] ,于是查看一下内存，发现根据第一个参数的不同（第一个参数需要小于等于7），对应的跳转的地址不同，因此只要确定第一个参数，然后根据跳转之后的指令可以确定第二个参数的值
输入文件的内容
第三个炸弹拆除的截图
作业完成
第四个炸弹：
<phase_4>函数执行的流程：
- 读入两个参数，地址为RSP+8,RSP+12(RSP+0xc)
- 如果没有成功读取两个参数或者第一个参数大于0xe，那么爆炸，否则执行3
- 调用函数<func4>,如果函数返回值不是0，爆炸，否则执行4
- 判断给出的第二个参数是否是0，如果不是0则爆炸，如果是0，那么这个炸弹可以成功拆除.
<phase_4>
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
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)
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>
40104d: 85 c0 test %eax,%eax
40104f: 75 07 jne 401058 <phase_4+0x4c>
401051: 83 7c 24 0c 00 cmpl $0x0,0xc(%rsp)
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
<func4>
0000000000400fce <func4>:
初始值：ESI=0x0,EDX=0xe,EDI=第一个变量的地址
400fce: 48 83 ec 08 sub $0x8,%rsp
400fd2: 89 d0 mov %edx,%eax #R[eax]<-R[edx],即EAX=0xe
400fd4: 29 f0 sub %esi,%eax #R[eax]<-R[eax]-R[esi],即EAX=0xe-0=0xe
400fd6: 89 c1 mov %eax,%ecx #R[ecx]<-R[eax],即ECX=0xe
400fd8: c1 e9 1f shr $0x1f,%ecx #R[ecx]<-R[ecx]>>1，逻辑右移，ECX=7
400fdb: 01 c8 add %ecx,%eax #R[eax]<-R[eax]+R[ecx] EAX=22
400fdd: d1 f8 sar %eax #R[eax]<-R[eax]>>1 ,算术右移 EAX=11
400fdf: 8d 0c 30 lea (%rax,%rsi,1),%ecx #R[ecx]<-R[rax]+R[rsi]
400fe2: 39 f9 cmp %edi,%ecx #比较EDI和ECX
400fe4: 7e 0c jle 400ff2 <func4+0x24> #如果ECX小于等于EDI，跳转0x400ff2
400fe6: 8d 51 ff lea -0x1(%rcx),%edx #R[edx]<-R[rcx]-1
400fe9: e8 e0 ff ff ff callq 400fce <func4> #递归调用函数
400fee: 01 c0 add %eax,%eax #R[eax]<-R[eax]+R[eax]
400ff0: eb 15 jmp 401007 <func4+0x39> #跳转一下，然后回收增长的空间，之后返回上一层函数
400ff2: b8 00 00 00 00 mov $0x0,%eax #R[eax]<-0
400ff7: 39 f9 cmp %edi,%ecx #比较EDI和ECX
400ff9: 7d 0c jge 401007 <func4+0x39> #如果ECX>=EDI,跳转到0x401007，直接函数递归结束，递归的终止条件
400ffb: 8d 71 01 lea 0x1(%rcx),%esi #R[esi]<-R[rcx]+1
400ffe: e8 cb ff ff ff callq 400fce <func4> #递归
401003: 8d 44 00 01 lea 0x1(%rax,%rax,1),%eax #R[eax]<-R[rax]+R[rax]+1
401007: 48 83 c4 08 add $0x8,%rsp
40100b: c3 retq
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