So i was recently trying to exploit a kernel mode driver (HEVD) and I encountered a problem which doesn't make sense to me.
After i jumped to my shellcode (which resides in usermode and was made executable by VirtualProtect), the thread keeps generating a pagefault. Then the Thread keeps generating this pagefault in an endless loop (it doesn't throw an exception).
So i investegated what kind of page fault was triggered and i got the following output for the pagefault-handler:
Breakpoint 0 hit
nt!KiPageFault+0x8:
fffff806`7d201d08 488dac2480000000 lea rbp,[rsp+80h]
0: kd> dd %rsp + 0x158
ffff9307`717e9768 51dce820 ffffad82 00000011 00000000
ffff9307`717e9778 ee49cf8e 000001a4 00000010 00000000
ffff9307`717e9788 00010206 00000000 717e97a0 ffff9307
ffff9307`717e9798 00000018 00000000 00000003 00000000
ffff9307`717e97a8 c00000bb 00000000 0000004d 00000000
ffff9307`717e97b8 00000018 00000000 00000003 00000000
ffff9307`717e97c8 7a0e643b fffff806 00000000 00000000
ffff9307`717e97d8 55555555 55555555 7a0e8b10 fffff806
Here we see the stack after the page fault occured. The Breakpoint is triggered immediatly after i resume the execution (and yes, the breakpoint is conditional for only this thread). The frame is exactly the same every time the breakpoint was triggered.
So I tried to decode the stackframe. A pagefault pushes some infomation on the stack. I decoded it manually (I don't know a better way to do this) and I got that the following:
RFLAGS 10206
CS 10
RIP 000001a4ee49cf8e
Errcode 11 --> P, I bits set
CR2 000001a4ee49cf8e
This should be right decoded (maybe it's erroneous. The top of the stack was ffff9307`717e9770, so the 0x...11 was the last thing pushed onto the stack). So the errocodes says that the exception is thrown while the page was present and it was during an instruction fetch.
Now: I think this should mean a access violation because of the NX-Bit. But this cannot be because the address is executable, as this snippet shows:
VA 000001a4ee49cf8e
PXE at FFFFA8D46A351018 PPE at FFFFA8D46A203498 PDE at FFFFA8D440693B90 PTE at
FFFFA880D27724E0
contains 8A00000027704867 contains 0A0000010DA05867 contains 0A0000010F9BE867 contains
0100000119B6C825
pfn 27704 ---DA--UW-V pfn 10da05 ---DA--UWEV pfn 10f9be ---DA--UWEV pfn 119b6c ----A-
UREV
The executable bit is set for the pte. So I asked myself why this happens. It's a bit weird also that it doesn't throw a access-violation and create a bugcheck.
Also the OS runs on virtualbox. I looked up the settings and I noticed that SMEP and SMAP aren't supported by my virtual machine (but by my host system. VirtualBox catches cpuid-Instructions and pretends that this feature is disabled). So this shoudn't be the problem either. Also if you dump the register, the SMEP/SMAP bits in CR4 are not set.
I have really no clue what could be the cause of this problem. Maybe it's because of the virtual machine. It could also be because Windows gets confused when a thread runs userocde or something else. I really thought a long time about it but maybe I just overlookeda a simple reason.
Thanks in advance
Related
I must be missing something very basic here. Searched SO but could not find the answer to this particular question. Here's my NASM code:
%include "io64.inc"
section .text
myvar db "This is not working", 0
global CMAIN
CMAIN:
mov rbp, rsp; for correct debugging
;write your code here
xor rax, rax
mov [myvar], rax
ret
It crashes on the move [myvar], rax line with SIGSEGV. I am simply trying to store some zeroes at that address.
Thanks!
PS: Using SASM to build / run / debug with 64 bit option ticked (default settings otherwise), on Windows 10 64 bit.
section .text
myvar db "This is not working", 0
Section .text is an executable section without write permissions. This is done to prevent some kinds of vulnerabilities. You should either place your myvar into a writable section, e.g. .data (if the variable should live for the whole duration of program execution), have the variable on the stack (if it's not supposed to outlive the function where it's created), or change .text to be writable (not recommended for security reasons, but possible).
I am in trouble with the definition 'memory location'. According to the 'Intel 64 and IA-32 Software Developer's Manual' many instruction can use a memory location as operand.
For example MOVBE (move data after swapping bytes):
Instruction: MOVBE m32, r32
The question is now how a memory location is defined;
I tried to use variables defined in the .bss section:
section .bss
memory: resb 4 ;reserve 4 byte
memorylen: equ $-memory
section .text
global _start
_start:
MOV R9D, 0x6162630A
MOV [memory], R9D
SHR [memory], 1
MOVBE [memory], R9D
EDIT:->
MOV EAX, 0x01
MOV EBX, 0x00
int 0x80
<-EDIT
If SHR is commented out yasm (yasm -f elf64 .asm) compiles without problems but when executing stdio shows: Illegal Instruction
And if MOVBE is commented out the following error occurs when compiling: error: invalid size for operand 1
How do I have to allocate memory for using the 'm' option shown by the instruction set reference?
[CPU=x64, Compiler=yasm]
If that is all your code, you are falling off at the end into uninitialized region, so you will get a fault. That has nothing to do with allocating memory, which you did right. You need to add code to terminate your program using an exit system call, or at least put an endless loop so you avoid the fault (kill your program using ctrl+c or equivalent).
Update: While the above is true, the illegal instruction here is more likely caused by the fact that your cpu simply does not support the MOVBE instruction, because not all do. If you look in the reference, you can see it says #UD If CPUID.01H:ECX.MOVBE[bit 22] = 0. That is trying to tell you that a particular flag bit in the ECX register returned by the 01 leaf of the CPUID instruction shows support of this instruction. If you are on linux, you can conveniently check in /proc/cpuinfo whether you have the movbe flag or not.
As for the invalid operand size: you should generally specify the operand size when it can not be deduced from the instruction. That said, SHR accepts all sizes (byte, word, dword, qword) so you should really not get that error at all, but you might get an operation of unexpected default size. You should use SHR dword [memory], 1 in this case, and that also makes yasm happy.
Oh, and +1 for reading the intel manual ;)
In my quest to learn more about the computer in general, I stumbled upon a book which has some chapters about disassembling, the x86 assembly language, and the relationship between C and x86 assembly. Now I have been reading this GDB command but I am unable to fully understand it.
The command, along with its results, follows:
(gdb) x/32xw $esp
0xbffff7e0: 0xb8000ce0 0x08048510 0xbffff848 0xb7eafebc
0xbffff7f0: 0x00000002 0xbffff874 0xbffff880 0xb8001898
0xbffff800: 0x00000000 0x00000001 0x00000001 0x00000000
0xbffff810: 0xb7fd6ff4 0xb8000ce0 0x00000000 0xbffff848
0xbffff820: 0x40f5f7f0 0x48e0fe81 0x00000000 0x00000000
0xbffff830: 0x00000000 0xb7ff9300 0xb7eafded 0xb8000ff4
0xbffff840: 0x00000002 0x08048350 0x00000000 0x08048371
0xbffff850: 0x08048474 0x00000002 0xbffff874 0x08048510
Now, from what I understand, the command that I issue tells the debugger to:
x (first one): examine the memory
32: get 32 of what follows
x: enable hexadecimal representation
w: show me Word size data.
**Note:** I know that I ask about the esp register, but I don't quite fully understand what $ is doing in front of it. When I try not to use it, I get a missing symbol error, so I get it has something to do with reference/de-reference?
What has been bugging me is how did I find all those bytes? Since I am examining a register, who's size is 32 bit, shouldn't I get only 32 bits, or 4 bytes (only 1 row of the above)? If I am correct with my assumption, then were did we find the rest of the data? Does it have to do something with the stack, and a particular stack frame, which I currently am unaware of?
I would appreciate your input so that I can clarify things in my mind.
(gdb) help x
Examine memory: x/FMT ADDRESS.
Giving $esp as the address will make gdb fetch whatever is in that register and use that as the memory address for the x command - and will show you the following 32 words in memory starting at that address.
variables within gdb itself are names prefixed with a $ , gdb sets up predefined variables for all the cpu registers.
If you want to inspect the esp register, use the command info registers esp, as you'll see with your example (x/32xw $esp), the esp register contains the first address shown, 0xbffff7e0
It's giving you 32 words of memory where the esp register is pointing (apparently that register contains the address 0xbffff7e0).
(gdb) x/32xw $esp
it means show me the 32 words field where esp points.
pieces:32,
format:hex ,
size:word (1 word= 32 bit on gdb)
show me the 32 words field where esp points
(hex)0xbffff7e0 - (hex)0xbffff7f0 = (dec)3221223392 - (dec)3221223408 = 16bytes=4words
Segments of memory - BSS, Stack, Heap, Data, Code/Text (Are there any more?).
Say I have a 128MB RAM, Can someone tell me:
How much memory is allocated for each of these memory segments?
Where do they start? Please specify the address range or something like that for better clarity.
What factors influence which should start where?
That question depends on the number of variables used. Since you did not specify what compiler or language or even operating system, that is a difficult one to pin down on! It all rests with the operating system who is responsible for the memory management of the applications. In short, there is no definite answer to this question, think about this, the compiler/linker at runtime, requests the operating system to allocate a block of memory, that allocation is dependent on how many variables there are, how big are they, the scope and usage of the variables. For instance, this simple C program, in a file called simpletest.c:
#include <stdio.h>
int main(int argc, char **argv){
int num = 42;
printf("The number is %d!\n", num);
return 0;
}
Supposing the environment was Unix/Linux based and was compiled like this:
gcc -o simpletest simpletest.c
If you were to issue a objdump or nm on the binary image simpletest, you will see the sections of the executable, in this instance, 'bss', 'text'. Make note of the sizes of these sections, now add a int var[100]; to the above code, recompile and reissue the objdump or nm, you will find that the data section has appeared - why? because we added a variable of an array type of int, with 100 elements.
This simple exercise will prove that the sections grows, and hence the binary gets bigger, and it will also prove that you cannot pre-determine how much memory will be allocated as the runtime implementation varies from compiler to compiler and from operating system to operating system.
In short, the OS calls the shot on the memory management!
you can get all this information compiling your program
# gcc -o hello hello.c // you might compile with -static for simplicity
and then readelf:
# readelf -l hello
Elf file type is EXEC (Executable file)
Entry point 0x80480e0
There are 3 program headers, starting at offset 52
Program Headers:
Type Offset VirtAddr PhysAddr FileSiz MemSiz Flg Align
LOAD 0x000000 0x08048000 0x08048000 0x55dac 0x55dac R E 0x1000
LOAD 0x055dc0 0x0809edc0 0x0809edc0 0x01df4 0x03240 RW 0x1000
NOTE 0x000094 0x08048094 0x08048094 0x00020 0x00020 R 0x4
Section to Segment mapping:
Segment Sections...
00 .init .text .fini .rodata __libc_atexit __libc_subfreeres .note.ABI-tag
01 .data .eh_frame .got .bss
02 .note.ABI-tag
The output shows the overall structure of hello. The first program header corresponds to the process' code segment, which will be loaded from file at offset 0x000000 into a memory region that will be mapped into the process' address space at address 0x08048000. The code segment will be 0x55dac bytes large and must be page-aligned (0x1000). This segment will comprise the .text and .rodata ELF segments discussed earlier, plus additional segments generated during the linking procedure. As expected, it's flagged read-only (R) and executable (X), but not writable (W).
The second program header corresponds to the process' data segment. Loading this segment follows the same steps mentioned above. However, note that the segment size is 0x01df4 on file and 0x03240 in memory. This is due to the .bss section, which is to be zeroed and therefore doesn't need to be present in the file. The data segment will also be page-aligned (0x1000) and will contain the .data and .bss ELF segments. It will be flagged readable and writable (RW). The third program header results from the linking procedure and is irrelevant for this discussion.
If you have a proc file system, you can check this, as long as you get "Hello World" to run long enough (hint: gdb), with the following command:
# cat /proc/`ps -C hello -o pid=`/maps
08048000-0809e000 r-xp 00000000 03:06 479202 .../hello
0809e000-080a1000 rw-p 00055000 03:06 479202 .../hello
080a1000-080a3000 rwxp 00000000 00:00 0
bffff000-c0000000 rwxp 00000000 00:00 0
The first mapped region is the process' code segment, the second and third build up the data segment (data + bss + heap), and the fourth, which has no correspondence in the ELF file, is the stack. Additional information about the running hello process can be obtained with GNU time, ps, and /proc/pid/stat.
example taken from:
http://www.lisha.ufsc.br/teaching/os/exercise/hello.html
memory depend on the global variable and local variable
When using copy-on-write semantics to share memory among processes, how can you test if a memory page is writable or if it is marked as read-only? Can this be done by calling a specific assembler code, or reading a certain spot in memory, or through the OS's API?
On Linux you can examine /proc/pid/maps:
$ cat /proc/self/maps
002b3000-002cc000 r-xp 00000000 68:01 143009 /lib/ld-2.5.so
002cc000-002cd000 r-xp 00018000 68:01 143009 /lib/ld-2.5.so
002cd000-002ce000 rwxp 00019000 68:01 143009 /lib/ld-2.5.so
002d0000-00407000 r-xp 00000000 68:01 143010 /lib/libc-2.5.so
00407000-00409000 r-xp 00137000 68:01 143010 /lib/libc-2.5.so
00409000-0040a000 rwxp 00139000 68:01 143010 /lib/libc-2.5.so
0040a000-0040d000 rwxp 0040a000 00:00 0
00c6f000-00c70000 r-xp 00c6f000 00:00 0 [vdso]
08048000-0804d000 r-xp 00000000 68:01 379298 /bin/cat
0804d000-0804e000 rw-p 00004000 68:01 379298 /bin/cat
08326000-08347000 rw-p 08326000 00:00 0
b7d1b000-b7f1b000 r--p 00000000 68:01 226705 /usr/lib/locale/locale-archive
b7f1b000-b7f1c000 rw-p b7f1b000 00:00 0
b7f28000-b7f29000 rw-p b7f28000 00:00 0
bfe37000-bfe4d000 rw-p bfe37000 00:00 0 [stack]
The first column is the virtual memory address range, the second column contains the permissions (read, write, execute, and private), columns 3-6 contain the offset, major and minor device numbers, the inode, and the name of memory mapped files.
On Win32, the best way is to use VirtualQuery. It returns a MEMORY_BASIC_INFORMATION for the page an address falls in. One of the members is Protect, which is some combination of these flags, which contain the possible protection modes. The function also tells you if the memory is free, committed, reserved, and whether it is private, part of an image or shared memory section.
The OS's API is the best way to detirmine the a page's protection. The CPU reads the protection mode from a page descriptor, which is only accessible from kernel mode.
Are you talking abou the variety of shared memory allocated via shmget (on Unix)? I.e.
int shmget(key_t, size_t, int);
If so, you can query that memory using
int shmctl(int, int, struct shmid_ds *);
For example:
key_t key = /* your choice of memory api */
int flag = /* set of flags for your app */
int shmid = shmget(key, 4096, flag);
struct shmid_ds buf;
int result = shmctl(shmid, IPC_STAT, &buf);
/* buf.ipc_perm.mode contains the permissions for the memory segment */
If you're using Win32, there are the calls IsBadReadPtr and IsBadWritePtr. However, their use is discouraged:
"The general consensus is that the IsBad family of functions (IsBadReadPtr, IsBadWritePtr, and so forth) is broken and should not be used to validate pointers."
The title of Raymond Chen's take on this says it all: "IsBadXxxPtr should really be called CrashProgramRandomly"
Chen has some helpful advice about how to deal with this issue here.
The upshot is, you shouldn't be testing this kind of thing at run-time. Code so that you know what you're being handed, and if it's not what's expected, treat it as a bug. If you really have no choice, look into SEH for handling the exception.