How does the JPEG of Death vulnerability operate?
This vulnerability was definitely a heap overflow.
How can writing 0XFFFFFFFE bytes (4 GB!!!!) possibly not crash the program?
It probably will, but on some occasions you got time to exploit before the crash happens (sometimes, you can get the program back to its normal execution and avoid the crash).
When the memcpy() starts, the copy will overwrite either some other heap blocks or some parts of the heap management structure (e.g free list, busy list, etc.).
At some point the copy will encounter a non allocated page and trigger an AV (Access Violation) on write. GDI+ will then try to allocate a new block in the heap (see ntdll!RtlAllocateHeap) ... but the heap structures are now all messed up.
At that point, by carefully crafting your JPEG image you can overwrite the heap management structures with controlled data. When the system tries to allocate the new block, it will probably unlink a (free) block from the free list.
Block are managed with (notably) a flink (Forward link ; the next block in the list) and blink (Backward link; the previous block in the list) pointers. If you control both of the flink and blink, you might have a possible WRITE4 (write What/Where condition) where you control what you can write and where you can write.
At that point you can overwrite a function pointer (SEH [Structured Exception Handlers] pointers were a target of choice at that time back in 2004) and gain code execution.
See blog post Heap Corruption: A Case Study.
Note: although I wrote about the exploitation using the freelist, an attacker might choose another path using other heap metadata ("heap metadata" are structures used by the system to manage the heap ; flink and blink are part of the heap metadata), but the unlink exploitation is probably the "easiest" one. A google search for "heap exploitation" will return numerous studies about this.
Does this write beyond the heap area and into the space of other programs and the OS?
Never. Modern OS are based on the concept of virtual address space so each process on has its own virtual address space that enables addressing up to 4 gigabytes of memory on a 32-bit system (in practice you only got half of it in user-land, the rest is for the kernel).
In short, a process can't access the memory of another process (except if it asks the kernel for it through some service / API, but the kernel will check if the caller has the right to do so).
I decided to test this vulnerability this week-end, so we could get a good idea on what was going on rather than pure speculation. The vulnerability is now 10 years old, so I thought it was OK to write about it, although I haven't explained the exploitation part in this answer.
Planning
The most difficult task was to find a Windows XP with only SP1, as it was in 2004 :)
Then, I downloaded a JPEG image composed only of a single pixel, as shown below (cut for brevity):
File 1x1_pixel.JPG
Address Hex dump ASCII
00000000 FF D8 FF E0|00 10 4A 46|49 46 00 01|01 01 00 60| ÿØÿà JFIF `
00000010 00 60 00 00|FF E1 00 16|45 78 69 66|00 00 49 49| ` ÿá Exif II
00000020 2A 00 08 00|00 00 00 00|00 00 00 00|FF DB 00 43| * ÿÛ C
[...]
A JPEG picture is composed of binary markers (which intrduce segments). In the above image, FF D8 is the SOI (Start Of Image) marker, while FF E0, for example, is an application marker.
The first parameter in a marker segment (except some markers like SOI) is a two-byte length parameter which encodes the number of bytes in the marker segment, including the length parameter and excluding the two-byte marker.
I simply added a COM marker (0xFFFE) right after the SOI, since markers have no strict order.
File 1x1_pixel_comment_mod1.JPG
Address Hex dump ASCII
00000000 FF D8 FF FE|00 00 30 30|30 30 30 30|30 31 30 30| ÿØÿþ 0000000100
00000010 30 32 30 30|30 33 30 30|30 34 30 30|30 35 30 30| 0200030004000500
00000020 30 36 30 30|30 37 30 30|30 38 30 30|30 39 30 30| 0600070008000900
00000030 30 61 30 30|30 62 30 30|30 63 30 30|30 64 30 30| 0a000b000c000d00
[...]
The length of the COM segment is set to 00 00 to trigger the vulnerability. I also injected 0xFFFC bytes right after the COM marker with a recurring pattern, a 4 bytes number in hex, which will become handy when "exploiting" the vulnerability.
Debugging
Double clicking the image will immediately trigger the bug in the Windows shell (aka "explorer.exe"), somewhere in gdiplus.dll, in a function named GpJpegDecoder::read_jpeg_marker().
This function is called for each marker in the picture, it simply: reads the marker segment size, allocates a buffer whose length is the segment size and copy the content of the segment into this newly allocated buffer.
Here the start of the function :
.text:70E199D5 mov ebx, [ebp+arg_0] ; ebx = *this (GpJpegDecoder instance)
.text:70E199D8 push esi
.text:70E199D9 mov esi, [ebx+18h]
.text:70E199DC mov eax, [esi] ; eax = pointer to segment size
.text:70E199DE push edi
.text:70E199DF mov edi, [esi+4] ; edi = bytes left to process in the image
eax register points to the segment size and edi is the number of bytes left in the image.
The code then proceeds to read the segment size, starting by the most significant byte (length is a 16-bits value):
.text:70E199F7 xor ecx, ecx ; segment_size = 0
.text:70E199F9 mov ch, [eax] ; get most significant byte from size --> CH == 00
.text:70E199FB dec edi ; bytes_to_process --
.text:70E199FC inc eax ; pointer++
.text:70E199FD test edi, edi
.text:70E199FF mov [ebp+arg_0], ecx ; save segment_size
And the least significant byte:
.text:70E19A15 movzx cx, byte ptr [eax] ; get least significant byte from size --> CX == 0
.text:70E19A19 add [ebp+arg_0], ecx ; save segment_size
.text:70E19A1C mov ecx, [ebp+lpMem]
.text:70E19A1F inc eax ; pointer ++
.text:70E19A20 mov [esi], eax
.text:70E19A22 mov eax, [ebp+arg_0] ; eax = segment_size
Once this is done, the segment size is used to allocate a buffer, following this calculation:
alloc_size = segment_size + 2
This is done by the code below:
.text:70E19A29 movzx esi, word ptr [ebp+arg_0] ; esi = segment size (cast from 16-bit to 32-bit)
.text:70E19A2D add eax, 2
.text:70E19A30 mov [ecx], ax
.text:70E19A33 lea eax, [esi+2] ; alloc_size = segment_size + 2
.text:70E19A36 push eax ; dwBytes
.text:70E19A37 call _GpMalloc@4 ; GpMalloc(x)
In our case, as the segment size is 0, the allocated size for the buffer is 2 bytes.
The vulnerability is right after the allocation:
.text:70E19A37 call _GpMalloc@4 ; GpMalloc(x)
.text:70E19A3C test eax, eax
.text:70E19A3E mov [ebp+lpMem], eax ; save pointer to allocation
.text:70E19A41 jz loc_70E19AF1
.text:70E19A47 mov cx, [ebp+arg_4] ; low marker byte (0xFE)
.text:70E19A4B mov [eax], cx ; save in alloc (offset 0)
;[...]
.text:70E19A52 lea edx, [esi-2] ; edx = segment_size - 2 = 0 - 2 = 0xFFFFFFFE!!!
;[...]
.text:70E19A61 mov [ebp+arg_0], edx
The code simply subtracts the segment_size size (segment length is a 2 bytes value) from the whole segment size (0 in our case) and ends up with an integer underflow: 0 - 2 = 0xFFFFFFFE
The code then checks is there are bytes left to parse in the image (which is true), and then jumps to the copy:
.text:70E19A69 mov ecx, [eax+4] ; ecx = bytes left to parse (0x133)
.text:70E19A6C cmp ecx, edx ; edx = 0xFFFFFFFE
.text:70E19A6E jg short loc_70E19AB4 ; take jump to copy
;[...]
.text:70E19AB4 mov eax, [ebx+18h]
.text:70E19AB7 mov esi, [eax] ; esi = source = points to segment content ("0000000100020003...")
.text:70E19AB9 mov edi, dword ptr [ebp+arg_4] ; edi = destination buffer
.text:70E19ABC mov ecx, edx ; ecx = copy size = segment content size = 0xFFFFFFFE
.text:70E19ABE mov eax, ecx
.text:70E19AC0 shr ecx, 2 ; size / 4
.text:70E19AC3 rep movsd ; copy segment content by 32-bit chunks
The above snippet shows that copy size is 0xFFFFFFFE 32-bits chunks. The source buffer is controlled (content of the picture) and the destination is a buffer on the heap.
Write condition
The copy will trigger an access violation (AV) exception when it reaches the end of the memory page (this could be either from the source pointer or destination pointer). When the AV is triggered, the heap is already in a vulnerable state because the copy has already overwritten all following heap blocks until a non-mapped page was encountered.
What makes this bug exploitable is that 3 SEH (Structured Exception Handler; this is try / except at low level) are catching exceptions on this part of the code. More precisely, the 1st SEH will unwind the stack so it gets back to parse another JPEG marker, thus completely skipping the marker that triggered the exception.
Without an SEH the code would have just crashed the whole program. So the code skips the COM segment and parses another segment. So we get back to GpJpegDecoder::read_jpeg_marker() with a new segment and when the code allocates a new buffer:
.text:70E19A33 lea eax, [esi+2] ; alloc_size = semgent_size + 2
.text:70E19A36 push eax ; dwBytes
.text:70E19A37 call _GpMalloc@4 ; GpMalloc(x)
The system will unlink a block from the free list. It happens that metadata structures were overwritten by the content of the image; so we control the unlink with controlled metadata. The below code in somewhere in the system (ntdll) in the heap manager:
CPU Disasm
Address Command Comments
77F52CBF MOV ECX,DWORD PTR DS:[EAX] ; eax points to '0003' ; ecx = 0x33303030
77F52CC1 MOV DWORD PTR SS:[EBP-0B0],ECX ; save ecx
77F52CC7 MOV EAX,DWORD PTR DS:[EAX+4] ; [eax+4] points to '0004' ; eax = 0x34303030
77F52CCA MOV DWORD PTR SS:[EBP-0B4],EAX
77F52CD0 MOV DWORD PTR DS:[EAX],ECX ; write 0x33303030 to 0x34303030!!!
Now we can write what we want, where we want...
Since I don't know the code from GDI, what's below is just speculation.
Well, one thing that pops into mind is one behavior that I've noticed on some OSes (I don't know if Windows XP had this) was when allocating with new / malloc, you can actually allocate more than your RAM, as long as you don't write to that memory.
This is actually a behavior of the linux Kernel.
From www.kernel.org :
Pages in the process linear address space are not necessarily resident in memory. For example, allocations made on behalf of a process are not satisfied immediately as the space is just reserved within the vm_area_struct.
To get into resident memory a page fault must be triggered.
Basically you need to make the memory dirty before it actually gets allocated on the system:
unsigned int size=-1;
char* comment = new char[size];
Sometimes it won't actually make a real allocation in RAM (your program will still not use 4 GB). I know I've seen this behavior on a Linux, but I cannot however replicate it now on my Windows 7 installation.
Starting from this behavior the following scenario is possible.
In order to make that memory existing in RAM you need to make it dirty (basically memset or some other write to it):
memset(comment, 0, size);
However the vulnerability exploits a buffer overflow, not an allocation failure.
In other words, if I'd were to have this:
unsinged int size =- 1;
char* p = new char[size]; // Will not crash here
memcpy(p, some_buffer, size);
This will lead to a write after buffer, because there's no such thing as a 4 GB segment of continuous memory.
You didn't put anything in p to make the whole 4 GB of memory dirty, and I don't know if memcpy makes memory dirty all at once, or just page by page (I think it's page by page).
Eventually it will end up overwriting the stack frame (Stack Buffer Overflow).
Another more possible vulnerability was if the picture was kept in memory as a byte array (read whole file into buffer), and the sizeof comments was used just to skip ahead non-vital information.
For example
unsigned int commentsSize = -1;
char* wholePictureBytes; // Has size of file
...
// Time to start processing the output color
char* p = wholePictureButes;
offset = (short) p[COM_OFFSET];
char* dataP = p + offset;
dataP[0] = EvilHackerValue; // Vulnerability here
As you mentioned, if the GDI didn't allocates that size, the program will never crash.