Revision as of 19:45, 5 April 2012

Basic Assembly

Preface

Understanding of assembly and stack growth are important for the machine coder. In assembly, there are a limited amount of temporary variables, also called registers, with only the stack or heap to contain additional application data. These registers are of a fixed size and fulfil various functions; the general purpose registers work as small variables for use by the machine coder, whereas special registers such as the instruction pointer fulfil specific functions.

There is a limited number of instructions that can be issued to a processor. While assembly instructions can be compared to the functions of higher-level languages, it is important to keep in mind that they are processor-specific operations that are performed, often peforming alterations on the registers mentioned in the previous paragraph. It is possible to write cross-operating-system shellcoded applications, also known as flat binaries, or flat binary applications.

Because of the limitations of most shellcoding and overflow tutorials, the code examples and notations in this article will be given in both AT&T and Intel Syntax. Intel Syntax and AT&T System V syntax are syntax standards for coding assembly. The two syntaxes are both translated into the same machine code, however they are translated from machine code to human-readable code differently. On Windows XP SP2 systems, we will be demonstrating our code in Intel 32 Bit Standard Syntax. On Linux and UNIX systems, we will be demonstrating our code in AT&T System V syntax, of course for the intel 32 bit processor.

Again, there is no functional difference between the two syntaxes. They are simply methods of displaying assembly (as much as assembly itself is a method of displaying machine code); opinions differ on which of the two is more readable.

On the low level, a machine is actually very simple. A processor is able to understand absolute hexadecimal values, then perform mathematical operations(add, subtract, multiply, divide) upon them as it is asked to. Other than that, the functionality of a processor is mostly limited to memory operations and execution procedures, such as the management of functions and stack growth.

The C programming language is more or less assembly that is arranged so that it is more easily written by humans, including the C libraries that provide a framework with which to work.

Instructions & Concepts

So, now we have a good framework of the concepts of assembly language, but how can we see them be utilized to actually do something? Well, we'll start with explaining a bit about registers. General purpose registers on an intel 32 bit processor are 4 bytes in length, also referred to as a "double word" or "DWORD". A Byte consists of 8 bits. It takes four bits to compose a hexadecimal digit, or a nybble.

The general-purpose registers include eax, ecx, ebx, edi, and edx. Other, non general purpose registers are esi, esp, ebp, and eip. So lets test out a few small instructions to get the hang of this. In intel syntax, to put the value of 1 into eax, we will say

<syntaxhighlight lang="asm">mov eax, 1</syntaxhighlight>

In AT&T System V syntax, we would say

<syntaxhighlight lang="asm">movl $1, %eax</syntaxhighlight>

Intel syntax uses the format:

 [instruction] [destination],[source]

Whereas AT&T System V syntax uses the syntax format:

 [instruction] [source],[destination]

Now in our small amount of code, eax = 1. If we want to add 3 to eax, we would say:

Intel:

<syntaxhighlight lang="asm">add eax, 3</syntaxhighlight>

AT&T:

<syntaxhighlight lang="asm">addl $3, %eax</syntaxhighlight>

This is the simple explanation of some instructions. A full list of instructions and a table of their purpose is as follows :

The Stack

In order to understand how special instructions and special jumps are used, it is necessary to have a solid understanding of the nature of functions and stack-based operations. The execution stack and the data stack are two wholly seperate things and must be understood as such.

When an application loads into memory, it begins at what is referred to as the "entry point" of the binary. Program execution then occurs in increasing memory addresses. Memory addresses are stored in DWORD values. For example, suppose that the entry point to app.exe is at the hexadecimal address value 0x101c879d in memory. Suppose that the code there is 'mov eax, 1' (or mov 1, %eax), and the next segment of instruction is 'add eax, 3'. The program would look something like the following in memory:

m4 NOTE TO SELF: this sections needs some serious work, it need much more detail with regards to the stack, C calling conventions, functions etc

Binary & Hexadecimal

Counting

A nybble consists of 4 bits, a half-byte, or one hexadecimal digit - all three of these are the same thing. Hex works in conjunction with binary based on this fact. Hexadecimal is base 16, using the characters 0-9 and a-f, each nybble is able to represent values 0-15. Binary works by base 2, using only the values 0 or 1, corresponding to the on or off state.

In order to construct binary numbers easily, it is useful to remember that the value of a place in a binary number is twice that in the place to the right. In other words, because binary assumes that 1 is on and 0 is off, we can only represent values with static place holders. A nybble takes the following layout :

 ? ? ? ?
 8 4 2 1

That is to say, 0001 would have a value of 1, whereas 0100 would have a value of 4. 0101, on the other hand would have a value of 4 + 1; 5. Binary is the same as our own numerary system, the key difference being that there are only 2 'numbers' (0-1 instead of 0-9) that a place value can hold.

In the above diagram, a "?" represents a 1 or 0. In hexadecimal, however, these values are represented differently.

Analyzing Code

This gives us a basic understanding of how hex values work, so lets revisit the previous shellcode of mov eax, 1 and add eax, 3.

Time to go to expanded view and show you WHY the addresses work the way they do, notice that the machine code in the middle contains a certain amount of bytes. Now we'll take a look at the expanded view of the execution stack or what is refered to as the .text segment. Instead of viewing it by one line of assembly instructions at a time, we'll view it one Byte at a time.

Now if we look at the expanded view, we can tell how 0x101c87a2 is the beginning of the next line of code, and even get a feel for certain raw instructions. How can we do this? Well, we can see the code for each instruction:

<syntaxhighlight lang="asm"> mov eax, 1 ; \xb8\x01\x00\x00\x00 add eax, 3 ; \x83\xc0\x03 </syntaxhighlight>

So lets rip this apart a little bit. We know that the eax register can contain a DWORD value, so lets extrapolate on the mov eax, 1 instruction. Because eax can hold a DWORD value, we can safely assume that the machine code instruction for moving or setting a value to the eax register must be done in proper DWORD format, that is, it must take 4 bytes. So we notice that the value \x01 is followed by \x00\x00\x00. These are three "null bytes". Seeing these null bytes and knowing the way that assembly operates, we can safely say that the value of eax is assigned backwards in DWORD format. Following all of these logical trains of thought, we can draw the following conclusions:

The machine code instruction "\xb8" is equivilent to mov eax, [dword].

<syntaxhighlight lang="asm"> mov eax, 1  ; \xb8 \x01\x00\x00\x00 mov eax, 0x00000001 ; \xb8 \x01\x00\x00\x00 </syntaxhighlight>

All of these instructions are synonymous. Now lets analyze the add eax, 3 instruction:

<syntaxhighlight lang="asm"> add eax, 3 \x83\xc0 \x03

add eax, \x03 \x83\xc0, 3 </syntaxhighlight>

Overflows

Can also be seen to be all synonymous. Therefore we can draw the conclusion that the machine code instructions \x83\xc0 is equivilent to the assembly instructions add eax, [byte]. Now lets get to the stack. The stack is a linear region of memory with a particularly allocated size, which cannot exceed 16 megabytes, for various reasons. Think of the stack like a stack of paper. While the location of each piece of paper is stored in a hexadecimal DWORD address, the stack grows backwards. This is to say that a stack is assigned a certain memory range and that range of memory is written to from the highest value to the lowest, not the lowest value to the highest. The stack is considered a part of the .data segment of an application, or the .bss segment. This is important because of something called DEP, or Data Execution Prevention. The stack region of an application is the region that can be "overflowed" to allow us to execute arbitrary machine code, however because it is a data region of the program certain safeguards have been put in place to prevent code located within the stack from executing.

Lets make a fake stack, for the sake of example and explanation so that the reader may better grasp this concept. lets say that our stack exists between addresses 0x100010ff and 0x10001000. The stack would look like this, in memory :

Now we can see that the top of the stack has a lower memory address than the bottom of the stack. Now we're going to need a bit more information about our non-general purpose registers and a bit more information about our stack operation special instructions.

Push & Pop

Obviously push applies to both registers as well as static data. For example, we can push direct data in hexadecimal format like push 0x1badc0de or we can push a byte like push 0x1b or we can push a register's dword value like push eax. Pushing a nybble is not supported. Lets go over the push and pop concept a bit more with our "fake stack" that we're creating for the sake of education. My fake machine code is as follows :

Alright, so say we execute that code starting with the top line first. The data stack will now look something like the following :

A detailed view of the bottom of the stack:

So we can sort of understand how the stack grows backwards now, as 1badcode was the first thing to get pushed onto the stack and is at the bottom of the stack. Think of the stack like a stack of papers. When you push something onto it, its like putting a piece of paper on it. When you push something else on it, you're putting another piece of paper on the first piece of paper. This is where these "special registers" come into play:

Special Registers

ESP

Stack Pointer - This register remembers the current index of the stack. At the beginning of the small code with the pushes, esp is 0x100010ff, and then it becomes 0x100010f8, f0, so on and so forth as DWORD values are pushed onto the stack. The program uses this to keep track of where it is pushing and popping data from inside of the stack.

EIP

Instruction Pointer - This register represents the location of the actual code executing, for example when add eax, 1 is executed, eip is the memory address of the \x83 byte.

EBP

Base Pointer - This points to the very bottom of the stack, however it can be used to hold addresses of other bases.

ESI

Function call register - when a function is called using the call instruction, the location of the beginning of the function is placed into esi.

Needless to say, our goal when attempting a buffer overflow is going to be manipulating the value of the eip register to make the computer execute our code as opposed to the original code. There are several ways to do this, but first we need to look at a few of our "special instructions" in assembly:

Special Instructions

JMP

Jumps to a location - this location can be defined by a byte offset (jump this many bytes forward or backward), but can also be defined with a DWORD memory address. The jmp instruction effectively modifies the value of the eip register to match that of the location to jump into.

CALL

Calls a function - this is a somewhat difficult concept in assembly, however a function will be demonstrated before the end of this section. When a call instruction is executed, the location of the function that is being called is placed in the esi register, while the location of the instruction immediately after the call is pushed to a stack for return purposes

RET

Exits a function - keep in mind that the call instruction pushes the location of the instruction after it to the stack, so at the end of the function, assuming that the esp register has the same value it contained when the function was first called, ret will jump back to the value that call pushed onto the stack. In other words, call pushes a pointer to the stack for the ret instruction to use when the function is finished executing so that the program can return to "normal" execution flow. This is what is being exploited during a buffer overflow attack.

NOP

NO OPERATION - this instruction literally does nothing. It is the equivilent of saying do nothing.

Call and Return

Lets make another fake flat binary. Reason being, we have to demonstrate the call and return sequence of assembly so that the reader can understand exactly what's going on here. Lets make our function that we are calling near pointless. There's not really much reason to have some uber-awesome code in this part of the article, we are still learning basic concepts. Our function will simply add 3 to the current value of eax. So here goes with another fake program.

Lets say our entry point is at 0x001010a1. This would put the first instruction at mov eax, 2. When the call instruction is executed, it pushes 0x001010ab onto the stack. When the ret instruction at 0x001010a0 is executed, it pops the value 0x001010ab off of the stack and jumps to that location, jumping to the line after the call function, where no operation is preformed.

Lets take a closer look at all of these registers for a moment so we can try to go ahead and get a better idea of where all our application's data is stored in the computer's memory. The eax register is a standard 32 bit register (4 bytes), however it has several sub-registers, like the other standard 32 bit registers. The registers break down as follows :

That's all for now, have fun with bytecode-asm comparison. I reccomend OllyDBG or GDB to test and expirement. Both are free and available from our friend google :)