做完这些lab，对qiling的作用，这个framework的理解，更加的深，希望看到这里的读者可以自己去做一遍，不要纯看他人的wp，因为每道题都有着很多种解法，但是目标都是成功的hook或Hijack，Qiling还有着一些fuzz or 仿真的 example，我都会去做一遍的！
一
What is Qiling
Qiling是一个先进的二进制仿真框架，具有以下特点：
◆Qiling这个框架对于模拟运行二进制程序时的 hook 非常方便
◆模拟多平台：Windows、MacOS、Linux、Android、BSD、UEFI、DOS、MBR、以太坊虚拟机
◆模拟多架构：8086、X86、X86_64、ARM、ARM64、MIPS、RISCV、PowerPC
◆内置调试器，具有逆向调试功能
◆提供深入的内存、寄存器、操作系统级和文件系统级API
◆细粒度检测：允许在各个级别进行挂钩（指令/基本块/内存访问/异常/系统调用/IO/等）
◆支持跨架构和平台调试能力
◆真正的Python框架，可以轻松地在其上构建定制的安全分析工具
◆等等
在qiling framework的github项目上还有几个示例demo，这里注意到了通过Qiling框架仿真模拟并对二进制程序进行hook可以更加方便的fuzz。
二
Lab
challenge1
要求在0x1337地址上写入0x1337：
| Method | Description |
|---|---|
map | Map a memory region at a certain location so it become available for access |
unmap | Reclaim a mapped memory region |
unmap_all | Reclaim all mapped memory regions |
map_anywhere | Map a memory region in an unspecified location |
protect | Modify access protection bits of a mapped region (rwx) |
find_free_space | Find an available memory region |
is_available | Query whether a memory region is available |
is_mapped | Query whether a memory region is mapped |
Qiling给出了这些Managing memory的方法。
内存在被访问之前必须被映射。map方法将连续的内存区域绑定到指定位置，并设置其访问保护位。可以提供字符串标签以便在映射信息表上轻松识别。
ql.mem.map(addr: int, size: int, perms: int = UC_PROT_ALL, info: Optional[str] = None) -> None
参数：-addr- 请求的映射基地址，应该在页面粒度上；-size- 映射大小（以字节为单位），必须是页面大小；-perms- 保护位图的乘积，定义此内存范围是否可读、可写和/或可执行（可选）；-info- 将字符串标签设置为映射范围以方便识别（可选）
主要也是关注前两个参数，这里显然是执行ql.mem.map(0x1337// 4096 * 4096 , 0x1000);
ql.mem.unmap(addr: int, size: int) -> None:
参数：-addr- 要取消映射的区域基地址 -size- 区域大小（以字节为单位）
如果请求的内存范围未完全映射，则引发：QlMemoryMappedError
address = ql.mem.search(b"\xFF\xFE\xFD\xFC\xFB\xFA", begin= 0x1000, end= 0x2000)
从部分内存范围中搜索字符串，begin和end参数均是可选的，去除则是整个内存范围。
ql.mem.read(address, size)
从内存中读取：
ql.mem.write(address, data)
写入内存：
def challenge1(ql):
ql.mem.map(0x1337//4096*4096 , 0x1000)
ql.mem.write(0x1337, b"\x39\x05")
challenge2
unsigned __int64 __fastcall challenge2(_BYTE *a1)
{
unsigned int v2; // [rsp+10h] [rbp-1D0h]
int v3; // [rsp+14h] [rbp-1CCh]
int v4; // [rsp+18h] [rbp-1C8h]
int v5; // [rsp+1Ch] [rbp-1C4h]
struct utsname name; // [rsp+20h] [rbp-1C0h] BYREF
char s[10]; // [rsp+1A6h] [rbp-3Ah] BYREF
char v8[24]; // [rsp+1B0h] [rbp-30h] BYREF
unsigned __int64 v9; // [rsp+1C8h] [rbp-18h] v9 = __readfsqword(0x28u);
if ( uname(&name) )
{
perror("uname");
}
else
{
strcpy(s, "QilingOS");
s[9] = 0;
strcpy(v8, "ChallengeStart");
v8[15] = 0;
v2 = 0;
v3 = 0;
while ( v4 < strlen(s) )
{
if ( name.sysname[v4] == s[v4] )
++v2;
++v4;
}
while ( v5 < strlen(v8) )
{
if ( name.version[v5] == v8[v5] )
++v3;
++v5;
}
if ( v2 == strlen(s) && v3 == strlen(v8) && v2 > 5 )
*a1 = 1;
}
return __readfsqword(0x28u) ^ v9;
}
要求在uname返回系统信息时的sysname == "QilingOS" and version == "ChallengeStart"。
需要通过劫持结构体
Hijack
POSIX 系统调用可以被挂钩以允许用户修改其参数、改变返回值或完全替换其功能。系统调用可以通过其名称或编号进行挂钩，并在一个或多个阶段进行拦截：- QL_INTERCEPT.CALL -：当指定的系统调用即将被调用时，可用于完全替换系统调用功能；- QL_INTERCEPT.ENTER -：在进入系统调用之前；可用于篡改系统调用参数值- QL_INTERCEPT.EXIT -：退出系统调用后，可能被用来篡改返回值。
from qiling import Qiling
from qiling.const import QL_INTERCEPT# customized system calls always use the same arguments list as the original
# ones, but with a Qiling instance on front. The Qiling instance may be used
# to interact with various subsystems, such as the memory or registers
def my_syscall_write(ql: Qiling, fd: int, buf: int, count: int) -> int:
try:
# read data from emulated memory
data = ql.mem.read(buf, count)
# select the emulated file object that corresponds to the requested
# file descriptor
fobj = ql.os.fd[fd]
# write the data into the file object, if it supports write operations
if hasattr(fobj, 'write'):
fobj.write(data)
except:
ret = -1
else:
ret = count
ql.log.info(f'my_syscall_write({fd}, {buf:#x}, {count}) = {ret}')
# return a value to the caller
return ret
if __name__ == "__main__":
ql = Qiling([r'rootfs/arm_linux/bin/arm_hello'], r'rootfs/arm_linux')
# the following call to 'set_syscall' sets 'my_syscall_write' to execute whenever
# the 'write' system call is about to be called. that practically replaces the
# existing implementation with the one in 'my_syscall_write'.
ql.os.set_syscall('write', my_syscall_write, QL_INTERCEPT.CALL)
# note that system calls may be referred to either by their name or number.
# an equivalent alternative that replaces the write syscall by refering its number:
#
#ql.os.set_syscall(4, my_syscall_write)
ql.run()
因此要通过ql.os.set_syscall中的QL_INTERCEPT.EXIT对返回值进行篡改。
def my_uname_ret(ql ,*args):
'''
struct utsname
{
char sysname[65];
char nodename[65];
char release[65];
char version[65];
char machine[65];
char domainname[65];
};
'''
rdi_value=ql.arch.regs.read("rdi")
ql.mem.write(rdi_value,b'QilingOS\x00')
ql.mem.write(rdi_value+65*3,b'ChallengeStart\x00')
return 0def challenge2(ql):
ql.os.set_syscall("uname",my_uname_ret,QL_INTERCEPT.EXIT)
这里有个要注意的点，my_uname_ret的参数要能接受三个参数，不然会报错。
这里看到qiling的一个example：
onexit_hook(self.ql, *self.get_syscall_args())
Qiling都是通过一些函数，来进行hook的。
Qiling的这些hook，都是一些在程序的hook，并不像正常的程序的输入输出，例如这里在退出的地方执行这个oxexit_hook函数（个人理解）
challenge3
unsigned __int64 __fastcall challenge3(_BYTE *a1)
{
int v2; // [rsp+10h] [rbp-60h]
int i; // [rsp+14h] [rbp-5Ch]
int fd; // [rsp+18h] [rbp-58h]
char v5; // [rsp+1Fh] [rbp-51h] BYREF
char buf[32]; // [rsp+20h] [rbp-50h] BYREF
char v7[40]; // [rsp+40h] [rbp-30h] BYREF
unsigned __int64 v8; // [rsp+68h] [rbp-8h] v8 = __readfsqword(0x28u);
fd = open("/dev/urandom", 0);
read(fd, buf, 0x20uLL);
read(fd, &v5, 1uLL);
close(fd);
getrandom((__int64)v7, 32LL, 1LL);
v2 = 0;
for ( i = 0; i <= 31; ++i )
{
if ( buf[i] == v7[i] && buf[i] != v5 )
++v2;
}
if ( v2 == ' ' )
*a1 = 1;
return __readfsqword(0x28u) ^ v8;
}
这里的第一个想法就是hook掉getrandom，控制其函数体，然后将虚拟路径/dev/urandom映射到文件对象下。
以下示例将虚拟路径/dev/urandom映射到托管系统上现有的/dev/urandom文件。当模拟程序访问/dev/random时，将访问映射文件。
from qiling import Qilingif __name__ == "__main__":
ql = Qiling([r'rootfs/x86_linux/bin/x86_fetch_urandom'], r'rootfs/x86_linux')
ql.add_fs_mapper(r'/dev/urandom', r'/dev/urandom')
ql.run()
以下示例将虚拟路径/dev/random映射到用户定义的文件对象，以允许对交互进行更细粒度的控制。请注意，映射对象扩展了QlFsMappedObject。
from qiling import Qiling
from qiling.os.mapper import QlFsMappedObjectclass FakeUrandom(QlFsMappedObject):
def read(self, size: int) -> bytes:
# return a constant value upon reading
return b"\x04"
def fstat(self) -> int:
# return -1 to let syscall fstat ignore it
return -1
def close(self) -> int:
return 0
if __name__ == "__main__":
ql = Qiling([r'rootfs/x86_linux/bin/x86_fetch_urandom'], r'rootfs/x86_linux')
ql.add_fs_mapper(r'/dev/urandom', FakeUrandom())
ql.run()
注意到：
class FakeUrandom(QlFsMappedObject): def read(self, size: int) -> bytes:
# return a constant value upon reading
return b"\x04"
def fstat(self) -> int:
# return -1 to let syscall fstat ignore it
return -1
def close(self) -> int:
return 0
这里其实是一个类，然后定义了一些函数，用self来代替/dev/urandom这个对象。
class FakeUrandom(QlFsMappedObject): def read(self, size=int) -> bytes:
if size==0x20:
return b"\x02"*32
# return a constant value upon reading
return b"\x01"
def fstat(self) -> int:
# return -1 to let syscall fstat ignore it
return -1
def close(self) -> int:
return 0
def my_getrandom_func(ql, buf, count:int, flag:int) ->int:
ql.mem.write(buf,b'\x02'*count)
return count
def challenge3(ql):
ql.add_fs_mapper(r'/dev/urandom', FakeUrandom())
ql.os.set_syscall("getrandom",my_getrandom_func,QL_INTERCEPT.CALL)
challenge4
.text:0000000000000E1D ; __int64 challenge4()
.text:0000000000000E1D public challenge4
.text:0000000000000E1D challenge4 proc near ; CODE XREF: start+18F↓p
.text:0000000000000E1D
.text:0000000000000E1D var_18= qword ptr -18h
.text:0000000000000E1D var_8= dword ptr -8
.text:0000000000000E1D var_4= dword ptr -4
.text:0000000000000E1D
.text:0000000000000E1D ; __unwind {
.text:0000000000000E1D 55 push rbp
.text:0000000000000E1E 48 89 E5 mov rbp, rsp
.text:0000000000000E21 48 89 7D E8 mov [rbp+var_18], rdi
.text:0000000000000E25 C7 45 F8 00 00 00 00 mov [rbp+var_8], 0
.text:0000000000000E2C C7 45 FC 00 00 00 00 mov [rbp+var_4], 0
.text:0000000000000E33 EB 0B jmp short loc_E40
.text:0000000000000E33
.text:0000000000000E35 ; ---------------------------------------------------------------------------
.text:0000000000000E35
.text:0000000000000E35 loc_E35: ; CODE XREF: challenge4+29↓j
.text:0000000000000E35 48 8B 45 E8 mov rax, [rbp+var_18]
.text:0000000000000E39 C6 00 01 mov byte ptr [rax], 1
.text:0000000000000E3C 83 45 FC 01 add [rbp+var_4], 1
.text:0000000000000E3C
.text:0000000000000E40
.text:0000000000000E40 loc_E40: ; CODE XREF: challenge4+16↑j
.text:0000000000000E40 8B 45 F8 mov eax, [rbp+var_8]
.text:0000000000000E43 39 45 FC cmp [rbp+var_4], eax
.text:0000000000000E46 7C ED jl short loc_E35
.text:0000000000000E46
.text:0000000000000E48 90 nop
.text:0000000000000E49 5D pop rbp
.text:0000000000000E4A C3 retn
.text:0000000000000E4A ; } // starts at E1D
.text:0000000000000E4A
.text:0000000000000E4A challenge4 endp
这里会是一个循环，我们想要的是使参数为真，而loc_E35块可以满足我们的要求，但是jl short loc_E35是显然无法满足的，因为-4和-8的位置都被置为0了是相等的，而jl是小于跳转，因此要想办法跳转到loc_E35块上。
读寄存器
◆从字符串“eax”读取
ql.arch.regs.read("EAX")
◆从 Unicorn Engine const 读取
ql.arch.regs.read(UC_X86_REG_EAX)
◆读eax
eax = ql.arch.regs.eax
◆将 0xFF 写入“eax”
ql.arch.regs.write("EAX", 0xFF)
◆通过 Unicorn Engine const 将 0xFF 写入 eax
ql.arch.regs.write(UC_X86_REG_EAX, 0xFF)
◆将 0xFF 写入 eax
ql.arch.regs.eax = 0xFF
还有一些跨架构寄存器的获取方法：
ql.arch.regs.arch_pc
ql.arch.regs.arch_sp #这仅适用于 PC 和 SP。
ql.arch.regs.arch_pc = 0xFF
ql.arch.regs.arch_sp = 0xFF #从当前架构上的 PC/SP 读取，由 ql.arch.type 定义
◆获取当前arch寄存器表列表
ql.arch.regs.register_mapping()
◆在 64 位环境中，这将返回 64
ql.arch.reg_bits("rax")
◆在 64 位环境中，这将返回 32
ql.arch.reg_bits("eax")
Hook
挂钩具体地址。执行指定地址时将调用已注册的回调。
ql.hook_address（回调：可调用，地址：int）
from qiling import Qiling def stop(ql: Qiling) -> None:
ql.log.info('killer switch found, stopping')
ql.emu_stop()
ql = Qiling([r'examples/rootfs/x86_windows/bin/wannacry.bin'], r'examples/rootfs/x86_windows')
# have 'stop' called when execution reaches 0x40819a
ql.hook_address(stop, 0x40819a)
ql.run()
挂钩所有说明。注册的回调将在每个汇编指令执行之前调用：
ql.hook_code（回调：可调用， user_data ：任何=无）
from capstone import Cs
from qiling import Qiling
from qiling.const import QL_VERBOSEdef simple_diassembler(ql: Qiling, address: int, size: int, md: Cs) -> None:
buf = ql.mem.read(address, size)
for insn in md.disasm(buf, address):
ql.log.debug(f':: {insn.address:#x} : {insn.mnemonic:24s} {insn.op_str}')
if __name__ == "__main__":
ql = Qiling([r'examples/rootfs/x8664_linux/bin/x8664_hello'], r'examples/rootfs/x8664_linux', verbose=QL_VERBOSE.DEBUG)
# have 'simple_disassembler' called on each instruction, passing a Capstone disassembler instance bound to
# the underlying architecture as an optional argument
ql.hook_code(simple_diassembler, user_data=ql.arch.disassembler)
ql.run()
还有一些hook，例如挂钩一段代码、挂钩中断号以调用自定义函数、拦截特定类型的指令等等。
我的想法是在：
.text:0000000000000E43 39 45 FC cmp [rbp+var_4], eax
这个地址执行前，在eax里写入1，这样就使得jl条件达成，就成功跳转到成功片段。
def write_eax_1(ql):
ql.arch.regs.write("EAX",1)def challenge4(ql):
base_addr = ql.mem.get_lib_base(ql.path)
ql.hook_address(write_eax_1,base_addr+0x0E43)
challenge5
unsigned __int64 __fastcall challenge5(_BYTE *a1)
{
unsigned int v1; // eax
int i; // [rsp+18h] [rbp-48h]
int j; // [rsp+1Ch] [rbp-44h]
int v5[14]; // [rsp+20h] [rbp-40h]
unsigned __int64 v6; // [rsp+58h] [rbp-8h] v6 = __readfsqword(0x28u);
v1 = time(0LL);
srand(v1);
for ( i = 0; i <= 4; ++i )
{
v5[i] = 0;
v5[i + 8] = rand();
}
for ( j = 0; j <= 4; ++j )
{
if ( v5[j] != v5[j + 8] )
{
*a1 = 0;
return __readfsqword(0x28u) ^ v6;
}
}
*a1 = 1;
return __readfsqword(0x28u) ^ v6;
}
这里就是让rand的返回值为0即可。
这里不是系统调用的劫持，而是劫持libc函数，这里看个例子。
与系统调用一样，POSIX libc 函数可以以类似的方式挂钩，从而允许用户控制其功能。
from qiling import Qiling
from qiling.const import QL_INTERCEPT
from qiling.os.const import STRING# customized POSIX libc methods accept a single argument that refers to the active
# Qiling instance. The Qiling instance may be used to interact with various subsystems,
# such as the memory or registers. The customized method may or may not return a value
def my_puts(ql: Qiling):
# Qiling offers a few conviniency methods that abstract away the access to the call
# parameters. specifying the arguments names and types woud allow Qiling to retrieve
# their values and parse them accordingly.
#
# the following call lists a single argument named 's', whose type is 'STRING'.
# a dictionary will be created having the key 's' mapped to the null-terminated
# string read from the memory address pointed by the first argument.
params = ql.os.resolve_fcall_params({'s': STRING})
s = params['s']
ql.log.info(f'my_puts: got "{s}" as an argument')
# emulate puts functionality
print(s)
return len(s)
if __name__ == "__main__":
ql = Qiling([r'rootfs/x8664_linux/bin/x8664_hello'], r'rootfs/x8664_linux')
ql.os.set_api('puts', my_puts, QL_INTERCEPT.CALL)
ql.run()
def rand_rets(ql ,*args):
ql.arch.regs.write("rax",0)def challenge5(ql):
ql.os.set_api('rand', rand_rets)
return
challenge6
.text:0000000000000EF6 public challenge6
.text:0000000000000EF6 challenge6 proc near ; CODE XREF: start+1D7↓p
.text:0000000000000EF6
.text:0000000000000EF6 var_18= qword ptr -18h
.text:0000000000000EF6 var_5= byte ptr -5
.text:0000000000000EF6 var_4= dword ptr -4
.text:0000000000000EF6
.text:0000000000000EF6 ; __unwind {
.text:0000000000000EF6 55 push rbp
.text:0000000000000EF7 48 89 E5 mov rbp, rsp
.text:0000000000000EFA 48 89 7D E8 mov [rbp+var_18], rdi
.text:0000000000000EFE C7 45 FC 00 00 00 00 mov [rbp+var_4], 0
.text:0000000000000F05 C6 45 FB 01 mov [rbp+var_5], 1
.text:0000000000000F09 EB 07 jmp short loc_F12 ; 零扩展
.text:0000000000000F09
.text:0000000000000F0B ; ---------------------------------------------------------------------------
.text:0000000000000F0B
.text:0000000000000F0B loc_F0B: ; CODE XREF: challenge6+22↓j
.text:0000000000000F0B C7 45 FC 01 00 00 00 mov [rbp+var_4], 1
.text:0000000000000F0B
.text:0000000000000F12
.text:0000000000000F12 loc_F12: ; CODE XREF: challenge6+13↑j
.text:0000000000000F12 0F B6 45 FB movzx eax, [rbp+var_5] ; 零扩展
.text:0000000000000F16 84 C0 test al, al
.text:0000000000000F18 75 F1 jnz short loc_F0B
.text:0000000000000F18
.text:0000000000000F1A 48 8B 45 E8 mov rax, [rbp+var_18]
.text:0000000000000F1E C6 00 01 mov byte ptr [rax], 1
.text:0000000000000F21 90 nop
.text:0000000000000F22 5D pop rbp
.text:0000000000000F23 C3 retn
.text:0000000000000F23 ; } // starts at EF6
.text:0000000000000F23
.text:0000000000000F23 challenge6 endp
这里会有一个无限循环，movzx eax, [rbp+var_5]先eax零扩展赋值为1，之后test al, al对al进行逻辑与的运算，结果存入ZF中，而jnz是ZF不为0则跳转，显然会有跳转，之后便是无限循环，我的想法 便是hook rax为0即可。
def write_rax_0(ql):
ql.arch.regs.write("rax",0)def challenge6(ql):
base_addr = ql.mem.get_lib_base(ql.path)
if base_addr is None:
raise ValueError("base_addr is not set correctly")
ql.hook_address(write_rax_0,base_addr+0xF16)
def write_rax_0(ql):
ql.arch.regs.write("rax",0)def challenge6(ql):
base_addr = ql.mem.get_lib_base(ql.path)
if base_addr is None:
raise ValueError("base_addr is not set correctly")
ql.hook_address(write_rax_0,base_addr+0xF16)
challenge7
.text:0000000000000F24 public challenge7
.text:0000000000000F24 challenge7 proc near ; CODE XREF: start+1FB↓p
.text:0000000000000F24
.text:0000000000000F24 var_8= qword ptr -8
.text:0000000000000F24
.text:0000000000000F24 ; __unwind {
.text:0000000000000F24 55 push rbp
.text:0000000000000F25 48 89 E5 mov rbp, rsp
.text:0000000000000F28 48 83 EC 10 sub rsp, 10h
.text:0000000000000F2C 48 89 7D F8 mov [rbp+var_8], rdi
.text:0000000000000F30 48 8B 45 F8 mov rax, [rbp+var_8]
.text:0000000000000F34 C6 00 01 mov byte ptr [rax], 1
.text:0000000000000F37 BF FF FF FF FF mov edi, 0FFFFFFFFh ; seconds
.text:0000000000000F3C E8 0F FB FF FF call _sleep
.text:0000000000000F3C
.text:0000000000000F41 90 nop
.text:0000000000000F42 C9 leave
.text:0000000000000F43 C3 retn
.text:0000000000000F43 ; } // starts at F24
.text:0000000000000F43
.text:0000000000000F43 challenge7 endp
这里有两种想法，第一种便是修改rdi，在call sleep前将rdi改为0，第二种便是hook掉sleep，直接return。
def write_rdi_0(ql):
ql.arch.regs.write("rdi",0)def my_sleep(ql,*args):
return
def challenge7(ql):
base_addr = ql.mem.get_lib_base(ql.path)
if base_addr is None:
raise ValueError("base_addr is not set correctly")
#ql.os.set_api('sleep', my_sleep)
ql.hook_address(write_rdi_0,base_addr+0x00F3C)
challenge8
_DWORD *__fastcall challenge8(__int64 a1)
{
_DWORD *result; // rax
_DWORD *v2; // [rsp+18h] [rbp-8h] v2 = malloc(0x18uLL);
*(_QWORD *)v2 = malloc(0x1EuLL);
v2[2] = 0x539;
v2[3] = 0x3DFCD6EA;
strcpy(*(char **)v2, "Random data");
result = v2;
*((_QWORD *)v2 + 2) = a1;
return result;
}
这一题的target是**Unpack the struct and write at the target address.**解包结构体，然后写入目标地址。
那么我的想法就是在malloc之后hook将rax返回值存入，但是我突然想到，之前我们不是刚学了如何搜索内存的吗，那么我们先通过搜索内存获得地址，然后再写入a1会不会更高端一点。
def search_mem(ql):
MAGIC=0x3DFCD6EA00000539
struct_address = ql.mem.search(p64(MAGIC))
if not struct_address or len(struct_address) < 1:
raise ValueError("struct_address is not properly initialized")
mem_value1, mem_value2 ,mem_value3 = struct.unpack("QQQ", ql.mem.read(struct_address[0]-8,0x18))
print("[*] debug1",hex(mem_value1))
print("[*] debug2",hex(mem_value2))
print("[*] debug3",hex(mem_value3))
ql.mem.write(mem_value3, b"\x01")def challenge8(ql):
base = ql.mem.get_lib_base(ql.path)
ql.hook_address(search_mem, base+0xFB5)
challenge9
unsigned __int64 __fastcall challenge9(bool *a1)
{
char *i; // [rsp+18h] [rbp-58h]
char dest[32]; // [rsp+20h] [rbp-50h] BYREF
char src[40]; // [rsp+40h] [rbp-30h] BYREF
unsigned __int64 v5; // [rsp+68h] [rbp-8h] v5 = __readfsqword(0x28u);
strcpy(src, "aBcdeFghiJKlMnopqRstuVWxYz");
src[27] = 0;
strcpy(dest, src);
for ( i = dest; *i; ++i )
*i = tolower(*i);
*a1 = strcmp(src, dest) == 0;
return __readfsqword(0x28u) ^ v5;
}
第一个想法就是直接hook掉strcmp函数，或者也可以hook掉tolower函数。
这里尽量hook tolower函数，不然下题就做不了啦。
def my_strcmp(ql,*args) -> int:
ql.arch.regs.write("rax", 0)def my_lower(ql,*args):
ql.arch.regs.rax = ql.arch.regs.rdi
def challenge9(ql):
#ql.os.set_api('strcmp', my_strcmp,QL_INTERCEPT.EXIT)
ql.os.set_api('tolower', my_lower,QL_INTERCEPT.EXIT)
challenge10
unsigned __int64 __fastcall challenge10(_BYTE *a1)
{
int i; // [rsp+10h] [rbp-60h]
int fd; // [rsp+14h] [rbp-5Ch]
ssize_t v4; // [rsp+18h] [rbp-58h]
char buf[72]; // [rsp+20h] [rbp-50h] BYREF
unsigned __int64 v6; // [rsp+68h] [rbp-8h] v6 = __readfsqword(0x28u);
fd = open("/proc/self/cmdline", 0);
if ( fd != -1 )
{
v4 = read(fd, buf, 0x3FuLL);
if ( v4 > 0 )
{
close(fd);
for ( i = 0; v4 > i; ++i )
{
if ( !buf[i] )
buf[i] = ' ';
}
buf[v4] = 0;
if ( !strcmp(buf, "qilinglab") )
*a1 = 1;
}
}
return __readfsqword(0x28u) ^ v6;
}
想法就是hook掉strcmp，但是上个challenge已经成功hook掉strcmp了，因此就劫持cmdline成一个结构体，直接返回qilinglab即可。
class FakeCmdline(QlFsMappedObject): def read(self, size=int) -> bytes:
return b"qilinglab"
def close(self) -> int:
return 0
def challenge10(ql):
ql.add_fs_mapper(r"/proc/self/cmdline",FakeCmdline)
challenge11
.text:0000000000001195 89 75 D0 mov [rbp+var_30], esi
.text:0000000000001198 89 4D CC mov [rbp+var_34], ecx
.text:000000000000119B 89 45 D4 mov [rbp+var_2C], eax
.text:000000000000119E 81 7D D0 51 69 6C 69 cmp [rbp+var_30], 696C6951h
.text:00000000000011A5 75 19 jnz short loc_11C0
.text:00000000000011A5
.text:00000000000011A7 81 7D CC 6E 67 4C 61 cmp [rbp+var_34], 614C676Eh
.text:00000000000011AE 75 10 jnz short loc_11C0
.text:00000000000011AE
.text:00000000000011B0 81 7D D4 62 20 20 20 cmp [rbp+var_2C], 20202062h
.text:00000000000011B7 75 07 jnz short loc_11C0
.text:00000000000011B7
.text:00000000000011B9 48 8B 45 B8 mov rax, [rbp+var_48]
.text:00000000000011BD C6 00 01 mov byte ptr [rax], 1
就是让esi==696C6951h && ecx==614C676Eh && eax=20202062h，就可以解决了，所以直接hook掉1195，使得这些寄存器为相应的值。
def set_regs(ql):
ql.arch.regs.write("esi",0x696C6951)
ql.arch.regs.write("ecx",0x614C676E)
ql.arch.regs.write("eax",0x20202062)def challenge11(ql):
base = ql.mem.get_lib_base(ql.path)
ql.hook_address(set_regs, base+0x1195)
如此便解决了所有的挑战。
exp
from qiling import Qiling
from qiling.const import QL_VERBOSE
from qiling.const import QL_INTERCEPT
from qiling.os.mapper import QlFsMappedObject
from pwn import *
import structdef challenge1(ql):
ql.mem.map(0x1337//4096*4096 , 0x1000)
ql.mem.write(0x1337, b"\x39\x05")
def my_uname_ret(ql ,*args):
'''
struct utsname
{
char sysname[65];
char nodename[65];
char release[65];
char version[65];
char machine[65];
char domainname[65];
};
'''
rdi_value=ql.arch.regs.read("rdi")
ql.mem.write(rdi_value,b'QilingOS\x00')
ql.mem.write(rdi_value+65*3,b'ChallengeStart\x00')
return 0
def challenge2(ql):
ql.os.set_syscall("uname",my_uname_ret,QL_INTERCEPT.EXIT)
class FakeUrandom(QlFsMappedObject):
def read(self, size=int) -> bytes:
if size==0x20:
return b"\x02"*32
# return a constant value upon reading
return b"\x01"
def fstat(self) -> int:
# return -1 to let syscall fstat ignore it
return -1
def close(self) -> int:
return 0
def my_getrandom_func(ql, buf, count:int, flag:int) ->int:
ql.mem.write(buf,b'\x02'*count)
return count
def challenge3(ql):
ql.add_fs_mapper(r'/dev/urandom', FakeUrandom())
ql.os.set_syscall("getrandom",my_getrandom_func,QL_INTERCEPT.CALL)
def write_rax_1(ql):
ql.arch.regs.write("rax",1)
def challenge4(ql):
base_addr = ql.mem.get_lib_base(ql.path)
if base_addr is None:
raise ValueError("base_addr is not set correctly")
ql.hook_address(write_rax_1,base_addr+0x0E43)
def rand_rets(ql ,*args):
ql.arch.regs.write("rax",0)
def win(ql: Qiling):
print('[*] win')
def challenge5(ql):
ql.os.set_api('rand', rand_rets)
return
def write_rax_0(ql):
ql.arch.regs.write("rax",0)
def challenge6(ql):
base_addr = ql.mem.get_lib_base(ql.path)
if base_addr is None:
raise ValueError("base_addr is not set correctly")
ql.hook_address(write_rax_0,base_addr+0xF16)
def write_rdi_0(ql):
ql.arch.regs.write("rdi",0)
def my_sleep(ql,*args):
return
def challenge7(ql):
base_addr = ql.mem.get_lib_base(ql.path)
if base_addr is None:
raise ValueError("base_addr is not set correctly")
#ql.os.set_api('sleep', my_sleep)
ql.hook_address(write_rdi_0,base_addr+0x00F3C)
def search_mem(ql):
MAGIC=0x3DFCD6EA00000539
struct_address = ql.mem.search(p64(MAGIC))
if not struct_address or len(struct_address) < 1:
raise ValueError("struct_address is not properly initialized")
mem_value1, mem_value2 ,mem_value3 = struct.unpack("QQQ", ql.mem.read(struct_address[0]-8,0x18))
print("[*] debug1",hex(mem_value1))
print("[*] debug2",hex(mem_value2))
print("[*] debug3",hex(mem_value3))
ql.mem.write(mem_value3, b"\x01")
def challenge8(ql):
base = ql.mem.get_lib_base(ql.path)
ql.hook_address(search_mem, base+0xFB5)
def my_strcmp(ql,*args) -> int:
ql.arch.regs.write("rax", 0)
def my_lower(ql,*args):
ql.arch.regs.rax = ql.arch.regs.rdi
def challenge9(ql):
#ql.os.set_api('strcmp', my_strcmp,QL_INTERCEPT.EXIT)
ql.os.set_api('tolower', my_lower,QL_INTERCEPT.EXIT)
class FakeCmdline(QlFsMappedObject):
def read(self, size=int) -> bytes:
return b"qilinglab"
def close(self) -> int:
return 0
def challenge10(ql):
ql.add_fs_mapper(r"/proc/self/cmdline",FakeCmdline)
def set_regs(ql):
ql.arch.regs.write("esi",0x696C6951)
ql.arch.regs.write("ecx",0x614C676E)
ql.arch.regs.write("eax",0x20202062)
def challenge11(ql):
base = ql.mem.get_lib_base(ql.path)
ql.hook_address(set_regs, base+0x1195)
if __name__=='__main__':
ql = Qiling([r"qilinglab-x86_64"], r'./qiling/examples/rootfs/x8664_linux', verbose=QL_VERBOSE.OFF)
#ql.verbose = 0
#ql.debugger = "gdb:0.0.0.0:9999"
#base_addr = ql.mem.get_lib_base(ql.path)
#ql.hook_address(win, base_addr + 0x12C5)
challenge1(ql)
challenge2(ql)
challenge3(ql)
challenge4(ql)
challenge5(ql)
challenge6(ql)
challenge7(ql)
challenge8(ql)
challenge9(ql)
challenge10(ql)
challenge11(ql)
ql.run()
s1nec-1o@s1nec1o:~/Qiling$ python3 solve.py
Welcome to QilingLab.
Here is the list of challenges:
Challenge 1: Store 1337 at pointer 0x1337.
Challenge 2: Make the 'uname' syscall return the correct values.
Challenge 3: Make '/dev/urandom' and 'getrandom' "collide".
Challenge 4: Enter inside the "forbidden" loop.
Challenge 5: Guess every call to rand().
Challenge 6: Avoid the infinite loop.
Challenge 7: Don't waste time waiting for 'sleep'.
Challenge 8: Unpack the struct and write at the target address.
Challenge 9: Fix some string operation to make the iMpOsSiBlE come true.
Challenge 10: Fake the 'cmdline' line file to return the right content.
Challenge 11: Bypass CPUID/MIDR_EL1 checks.Checking which challenge are solved...
Note: Some challenges will results in segfaults and infinite loops if they aren't solved.
Challenge 1: SOLVED
Challenge 2: SOLVED
Challenge 3: SOLVED
Challenge 4: SOLVED
Challenge 5: SOLVED
Challenge 6: SOLVED
[*] debug1 0x55555575a690
[*] debug2 0x3dfcd6ea00000539
[*] debug3 0x80000000dd54
Challenge 7: SOLVED
Challenge 8: SOLVED
Challenge 9: SOLVED
Challenge 10: SOLVED
Challenge 11: SOLVED
You solved 11/11 of the challenges
看雪ID：s1nec-1o
https://bbs.kanxue.com/user-home-989115.htm
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