智能设备-初探Qiling framework_游戏逆向|游戏安全|yxfzedu.com

前言

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！同时接下来每周都有着很多考试，希望一切顺利~

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

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

1	`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);

1	`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

1	`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

1	`ql.mem.read(address, size)`

从内存中读取

1	`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 0

def 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 Qiling

if __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 QlFsMappedObject
 
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
 
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

1	`ql.arch.regs.read("EAX")`

从 Unicorn Engine const 读取

1	`ql.arch.regs.read(UC_X86_REG_EAX)`

读eax

1	`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

1	`ql.arch.regs.write(UC_X86_REG_EAX,` `0xFF)`

将 0xFF 写入 eax

1	`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

1	`ql.arch.regs.register_mapping()`

在 64 位环境中，这将返回 64

1	`ql.arch.reg_bits("rax")`

在 64 位环境中，这将返回 32

1	`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_VERBOSE
 
def 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

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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 struct
 
def 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

智能设备-初探Qiling framework

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