\documentclass{article}

\usepackage{../style}

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\section*{Handout 3 (Buffer Overflow Attacks)}

By far the most popular attack method on computers are buffer

overflow attacks or variations thereof. The first Internet

worm (Morris) exploited exactly such an attack. The popularity

is unfortunate because we nowadays have technology in place to

prevent them effectively. But these kind of attacks are still

very relevant even today since there are many legacy systems

out there and also many modern embedded systems often do not

take any precautions to prevent such attacks.

To understand how buffer overflow attacks work, we have to have

a look at how computers work ``under the hood'' (on the

machine level) and also understand some aspects of the C/C++

programming language. This might not be everyday fare for

computer science students, but who said that criminal hackers

restrict themselves to everyday fare? Not to mention the

free-riding script-kiddies who use this technology without

even knowing what the underlying ideas are. If you want to be

a good security engineer who needs to defend such attacks, 

then better you get to know the details.

For buffer overflow attacks to work, a number of innocent

design decisions, which are really benign on their own, need

to conspire against you. All these decisions were taken at a

time when there was no Internet: C was introduced around 1973;

the Internet TCP/IP protocol was standardised in 1982 by which

time there were maybe 500 servers connected (and all users

were well-behaved, mostly academics); Intel's first 8086 CPUs

arrived around 1977. So nobody of the ``forefathers'' can

really be blamed, but as mentioned above we should already be

way beyond the point that buffer overflow attacks are worth a

thought. Unfortunately, this is far from the truth. I let you

ponder why?

One such ``benign'' design decision is how the memory is laid

out into different regions for each process. 

\begin{center}

  \begin{tikzpicture}[scale=0.7]

  %\draw[step=1cm] (-3,-3) grid (3,3);

  \draw[line width=1mm] (-2, -3) rectangle (2,3);

  \draw[line width=1mm] (-2,1) -- (2,1);

  \draw[line width=1mm] (-2,-1) -- (2,-1);

  \draw (0,2) node {\large\tt text};

  \draw (0,0) node {\large\tt heap};

  \draw (0,-2) node {\large\tt stack};

  \draw (-2.7,3) node[anchor=north east] {\tt\begin{tabular}{@{}l@{}}lower\\ address\end{tabular}};

  \draw (-2.7,-3) node[anchor=south east] {\tt\begin{tabular}{@{}l@{}}higher\\ address\end{tabular}};

  \draw[->, line width=1mm] (-2.5,3) -- (-2.5,-3);

  \draw (2.7,-2) node[anchor=west] {\tt grows};

  \draw (2.7,-3) node[anchor=south west] {\tt\footnotesize older};

  \draw (2.7,-1) node[anchor=north west] {\tt\footnotesize newer};

  \draw[|->, line width=1mm] (2.5,-3) -- (2.5,-1);

  \end{tikzpicture}

\end{center}

\noindent The text region contains the program code (usually

this region is read-only). The heap stores all data the

programmer explicitly allocates. For us the most interesting

region is the stack, which contains data mostly associated

with the control flow of the program. Notice that the stack

grows from higher addresses to lower addresses (i.e.~from the

back to the front). That means that older items on the stack

will be stored behind, or after, newer items. Let's look a bit

closer what happens with the stack when a program is running.

Consider the following simple C program.

\lstinputlisting[language=C]{../progs/example1.c} 

\noindent The \code{main} function calls in Line 7 the

function \code{foo} with three arguments. \code{Foo} creates

two (local) buffers, but does not do anything interesting with

them. The only purpose of this program is to illustrate what

happens behind the scenes with the stack. The interesting

question is what will the stack be after Line 3 has been

executed? The answer can be illustrated as follows:

\begin{center} 

 \begin{tikzpicture}[scale=0.65]

  \draw[gray!20,fill=gray!20] (-5, 0) rectangle (-3,-1);

  \draw[line width=1mm] (-5,-1.2) -- (-5,0.2);

  \draw[line width=1mm] (-3,-1.2) -- (-3,0.2);

  \draw (-4,-1) node[anchor=south] {\tt main};

  \draw[line width=1mm] (-5,0) -- (-3,0);

  \draw[gray!20,fill=gray!20] (3, 0) rectangle (5,-1);

  \draw[line width=1mm] (3,-1.2) -- (3,0.2);

  \draw[line width=1mm] (5,-1.2) -- (5,0.2);

  \draw (4,-1) node[anchor=south] {\tt main};

  \draw[line width=1mm] (3,0) -- (5,0);

   %\draw[step=1cm] (-3,-1) grid (3,8);

  \draw[gray!20,fill=gray!20] (-1, 0) rectangle (1,-1);

  \draw[line width=1mm] (-1,-1.2) -- (-1,7.4);

  \draw[line width=1mm] ( 1,-1.2) -- ( 1,7.4);

  \draw (0,-1) node[anchor=south] {\tt main};

  \draw[line width=1mm] (-1,0) -- (1,0);

  \draw (0,0) node[anchor=south] {\tt arg$_3$=3};

  \draw[line width=1mm] (-1,1) -- (1,1);

  \draw (0,1) node[anchor=south] {\tt arg$_2$=2};

  \draw[line width=1mm] (-1,2) -- (1,2);

  \draw (0,2) node[anchor=south] {\tt arg$_1$=1};

  \draw[line width=1mm] (-1,3) -- (1,3);

  \draw (0,3.1) node[anchor=south] {\tt ret};

  \draw[line width=1mm] (-1,4) -- (1,4);

  \draw (0,4) node[anchor=south] {\small\tt last sp};

  \draw[line width=1mm] (-1,5) -- (1,5);

  \draw (0,5) node[anchor=south] {\tt buf$_1$};

  \draw[line width=1mm] (-1,6) -- (1,6);

  \draw (0,6) node[anchor=south] {\tt buf$_2$};

  \draw[line width=1mm] (-1,7) -- (1,7);

  \draw[->,line width=0.5mm] (1,4.5) -- (1.8,4.5) -- (1.8, 0) -- (1.1,0); 

  \draw[->,line width=0.5mm] (1,3.5) -- (2.5,3.5);

  \draw (2.6,3.1) node[anchor=south west] {\tt back to main()};

\end{tikzpicture}

\end{center} 

\noindent On the left is the stack before \code{foo} is

called; on the right is the stack after \code{foo} finishes.

The function call to \code{foo} in Line 7 pushes the arguments

onto the stack in reverse order---shown in the middle.

Therefore first 3 then 2 and finally 1. Then it pushes the

return address onto the stack where execution should resume

once \code{foo} has finished. The last stack pointer

(\code{sp}) is needed in order to clean up the stack to the

last level---in fact there is no cleaning involved, but just

the top of the stack will be set back. So the last stack

pointer also needs to be stored. The two buffers inside

\pcode{foo} are on the stack too, because they are local data

within \code{foo}. Consequently the stack in the middle is a

snapshot after Line 3 has been executed. In case you are

familiar with assembly instructions you can also read off this

behaviour from the machine code that the \code{gcc} compiler

generates for the program above:\footnote{You can make

\pcode{gcc} generate assembly instructions if you call it with

the \pcode{-S} option, for example \pcode{gcc -S out in.c}\;.

Or you can look at this code by using the debugger. How to do

this will be explained later.}

\begin{center}\small

\begin{tabular}[t]{@{}c@{\hspace{8mm}}c@{}}

{\lstinputlisting[language={[x86masm]Assembler},

  morekeywords={movl},xleftmargin=5mm]

  {../progs/example1a.s}} &

{\lstinputlisting[language={[x86masm]Assembler},

  morekeywords={movl,movw},xleftmargin=5mm]

  {../progs/example1b.s}}  

\end{tabular}

\end{center}

\noindent On the left you can see how the function

\pcode{main} prepares in Lines 2 to 7 the stack before calling

the function \pcode{foo}. You can see that the numbers 3, 2, 1

are stored on the stack (the register \code{$esp} refers to

the top of the stack). On the right you can see how the

function \pcode{foo} stores the two local buffers onto the

stack and initialises them with the given data (Lines 2 to 9).

Since there is no real computation going on inside

\pcode{foo}, the function then just restores the stack to its

old state and crucially sets the return address where the

computation should resume (Line 9 in the code on the left-hand

side). The instruction \code{ret} then transfers control back

to the function \pcode{main} to the the instruction just after

the call to \pcode{foo}, that is Line 9.

Another part of the ``conspiracy'' of buffer overflow attacks

is that library functions in C look typically as follows:

\begin{center}

\lstinputlisting[language=C,numbers=none]{../progs/app5.c}

\end{center} 

\noindent This function copies data from a source \pcode{src}

to a destination \pcode{dst}. The important point is that it

copies the data until it reaches a zero-byte (\code{"\\0"}). 

This is a convention of the C language which assumes all

strings are terminated by such a zero-byte.

The central idea of the buffer overflow attack is to overwrite

the return address on the stack. This address decides where

the control flow of the program should resume once the

function at hand has finished its computation. So if we 

can control this address, then we can modify the control

flow of a program. To launch an attack we need 

somewhere in a function a local a buffer, say

\begin{center}

\code{char buf[8];}

\end{center}

\noindent which is filled by some user input. The

corresponding stack of such a function will look as follows

\begin{center}

 \begin{tikzpicture}[scale=0.65]

  %\draw[step=1cm] (-3,-1) grid (3,8);

  \draw[gray!20,fill=gray!20] (-1, 0) rectangle (1,-1);

  \draw[line width=1mm] (-1,-1.2) -- (-1,6.4);

  \draw[line width=1mm] ( 1,-1.2) -- ( 1,6.4);

  \draw (0,-1) node[anchor=south] {\tt main};

  \draw[line width=1mm] (-1,0) -- (1,0);

  \draw (0,0) node[anchor=south] {\tt arg$_3$=3};

  \draw[line width=1mm] (-1,1) -- (1,1);

  \draw (0,1) node[anchor=south] {\tt arg$_2$=2};

  \draw[line width=1mm] (-1,2) -- (1,2);

  \draw (0,2) node[anchor=south] {\tt arg$_1$=1};

  \draw[line width=1mm] (-1,3) -- (1,3);

  \draw (0,3.1) node[anchor=south] {\tt ret};

  \draw[line width=1mm] (-1,4) -- (1,4);

  \draw (0,4) node[anchor=south] {\small\tt last sp};

  \draw[line width=1mm] (-1,5) -- (1,5);

  \draw (0,5.1) node[anchor=south] {\tt buf};

  \draw[line width=1mm] (-1,6) -- (1,6);

  \draw (2,5.1) node[anchor=south] {\code{$esp}};

  \draw[<-,line width=0.5mm] (1.1,6) -- (2.5,6);

  \draw[->,line width=0.5mm] (1,4.5) -- (2.5,4.5);

  \draw (2.6,4.1) node[anchor=south west] {\code{??}};

  \draw[->,line width=0.5mm] (1,3.5) -- (2.5,3.5);

  \draw (2.6,3.1) node[anchor=south west] {\tt jump to \code{\\x080483f4}};

\end{tikzpicture}

\end{center}

\noindent We need to fill this buffer over its limit of 8

characters so that it overwrites the stack pointer and then

also overwrites the return address. If, for example, we want

to jump to a specific address in memory, say,

\pcode{\\x080483f4} then we can fill the buffer with the data

\begin{center}

\code{char buf[8] = "AAAAAAAABBBB\\xf4\\x83\\x04\\x08";}

\end{center}

\noindent The first eight \pcode{A}s fill the buffer to the

rim; the next four \pcode{B}s overwrite the stack pointer

(with what data we overwrite this part is usually not

important); then comes the address we want to jump to. Notice

that we have to give the address in the reverse order. All

addresses on Intel CPUs need to be given in this way. Since

the string is enclosed in double quotes, the C convention is

that the string internally will automatically be terminated by

a zero-byte. If the programmer uses functions like

\pcode{strcpy} for filling the buffer \pcode{buf}, then we can

be sure it will overwrite the stack in this manner---since it

will copy everything up to the zero-byte. Notice that this

overwriting of the buffer only works since the newer item, the

buffer, is stored on the stack before the older items, like

return address and arguments. If it had be the other way

around, then such an overwriting by overflowing a local buffer

would just not work. If the designers of C had just been able

to foresee what headaches their way of arranging the stack

caused in the time where computers are accessible from

everywhere. 

What the outcome of such an attack is can be illustrated with

the code shown in Figure~\ref{C2}. Under ``normal operation''

this program ask for a login-name and a password. Both of

which are stored in \code{char} buffers of length 8. The

function \pcode{match} tests whether two such buffers contain

the same content. If yes, then the function lets you ``in''

(by printing \pcode{Welcome}). If not, it denies access (by

printing \pcode{Wrong identity}). The vulnerable function is

\code{get_line} in Lines 11 to 19. This function does not take

any precautions about the buffer of 8 characters being filled

beyond its 8-character-limit. Let us suppose the login name

is \pcode{test}. Then the buffer overflow can be triggered

with a specially crafted string as password:

\begin{center}

\code{AAAAAAAABBBB\\x2c\\x85\\x04\\x08\\n}

\end{center}

\noindent The address at the end happens to be the one for the

function \pcode{welcome()}. This means even with this input

(where the login name and password clearly do not match) the

program will still print out \pcode{Welcome}. The only

information we need for this attack to work is to know where

the function \pcode{welcome()} starts in memory. This

information can be easily obtained by starting the program

inside the debugger and disassembling this function. 

\begin{lstlisting}[numbers=none,language={[x86masm]Assembler},

  morekeywords={movl,movw}]

$ gdb C2

GNU gdb (GDB) 7.2-ubuntu

(gdb) disassemble welcome

\end{lstlisting}

\noindent \pcode{C2} is the name of the program and

\pcode{gdb} is the name of the debugger. The output will be

something like this

\begin{lstlisting}[numbers=none,language={[x86masm]Assembler},

  morekeywords={movl,movw}]

0x0804852c <+0>:     push   %ebp

0x0804852d <+1>:     mov    %esp,%ebp

0x0804852f <+3>:     sub    $0x4,%esp

0x08048532 <+6>:     movl   $0x8048690,(%esp)

0x08048539 <+13>:    call   0x80483a4 <puts@plt>

0x0804853e <+18>:    movl   $0x0,(%esp)

0x08048545 <+25>:    call   0x80483b4 <exit@plt>

\end{lstlisting}

\noindent indicating that the function \pcode{welcome()}

starts at address \pcode{0x0804852c} (top address in the 

left column).

\begin{figure}[p]

\lstinputlisting[language=C]{../progs/C2.c}

\caption{A vulnerable login implementation.\label{C2}}

\end{figure}

This kind of attack was very popular with commercial programs

that needed a key to be unlocked. Historically, hackers first 

broke the rather weak encryption of these locking mechanisms.

After the encryption had been made stronger, hackers used

buffer overflow attacks as shown above to jump directly to

the part of the program that was intended to be only available

after the correct key was typed in. 

\subsection*{Payloads}

Unfortunately, much more harm can be caused by buffer overflow

attacks. This is achieved by injecting code that will be run

once the return address is appropriately modified. Typically

the code that will be injected starts a shell. This gives the

attacker the ability to run programs on the target machine and

to have a good look around, provided the attacked process was not

already running as root.\footnote{In that case the attacker

would already congratulate him or herself to another

computer under full control.} In order to be send as part of

the string that is overflowing the buffer, we need the code to

be represented as a sequence of characters. For example

\lstinputlisting[language=C,numbers=none]{../progs/o1.c}

\noindent These characters represent the machine code for

opening a shell. It seems obtaining such a string requires

higher-education in the architecture of the target system. But

it is actually relatively simple: First there are many such

string ready-made---just a quick Google query away. Second,

tools like the debugger can help us again. We can just write

the code we want in C, for example this would be the program

for starting a shell:

\lstinputlisting[language=C,numbers=none]{../progs/shell.c} 

\noindent Once compiled, we can use the debugger to obtain 

the machine code, or even the ready-made encoding as character

sequence. 

While easy, obtaining this string is not entirely trivial.

Remember the functions in C that copy or fill buffers work

such that they copy everything until the zero byte is reached.

Unfortunately the ``vanilla'' output from the debugger for the

shell-program above will contain such zero bytes. So a

post-processing phase is needed to rewrite the machine code in

a way that it does not contain any zero bytes. This is like

some works of literature that have been written so that the

letter e, for example, is avoided. The technical term for such

a literature work is \emph{lipogram}.\footnote{The most

famous example of a lipogram is a 50,000 words novel titled

Gadsby, see \url{https://archive.org/details/Gadsby}.} For

rewriting the machine code, you might need to use clever

tricks like

\begin{lstlisting}[numbers=none,language={[x86masm]Assembler}]

xor %eax, %eax

\end{lstlisting}

\noindent This instruction does not contain any zero-byte when

encoded as string, but produces a zero-byte on the stack when

run. 

Having removed the zero-bytes we can craft the string that

will be send to the target computer. This of course requires

that the buffer we are trying to attack can at least contain

the shellcode we want to run. But as you can see this is only

47 bytes, which is a very low bar to jump over. More

formidable is the choice of finding the right address to jump

to. The string is typically of the form

\begin{center}

  \begin{tikzpicture}[scale=0.6]

  \draw[line width=1mm] (-2, -1) rectangle (2,3);

  \draw[line width=1mm] (-2,1.9) -- (2,1.9);

  \draw (0,2.5) node {\large\tt shell code};

  \draw[line width=1mm,fill=black] (0.3, -1) rectangle (2,-0.7);

  \draw[->,line width=0.3mm] (1.05, -1) -- (1.05,-1.7) --

  (-3,-1.7) -- (-3, 3.7) -- (-1.9, 3.7) -- (-1.9, 3.1);

  \draw (-2, 3) node[anchor=north east] {\LARGE \color{codegreen}{``}};

  \draw ( 2,-0.9) node[anchor=west] {\LARGE\color{codegreen}{''}};

  \end{tikzpicture}

\end{center}

\noindent where we need to be very precise with the address

with which we will overwrite the buffer. It has to be

precisely the first byte of the shellcode. While this is easy

with the help of a debugger (as seen before), we typically

cannot run anything, including a debugger, on the machine yet

we target. And the address is very specific to the setup of

the target machine. One way of finding out what the right

address is is to try out one by one every possible

address until we get lucky. With the large memories available

today, however, the odds are long. And if we try out too many

possible candidates too quickly, we might be detected by the

system administrator of the target system.

We can improve our odds considerably by following a clever 

trick. Instead of adding the shellcode at the beginning of the

string, we should add it at the end, just before we overflow 

the buffer, for example

\begin{center}

  \begin{tikzpicture}[scale=0.6]

  \draw[gray!50,fill=gray!50] (-2,0.3) rectangle (2,3);

  \draw[line width=1mm] (-2, -1) rectangle (2,3);

  \draw[line width=1mm] (-2,0.3) -- (2,0.3);

  \draw[line width=1mm] (-2,-0.7) -- (2,-0.7);

  \draw (0,-0.2) node {\large\tt shell code};

  \draw[line width=1mm,fill=black] (0.3, -1) rectangle (2,-0.7);

  \draw [line width=0.5,decoration={brace,amplitude=2mm},decorate] 

    (2.3,3) -- (2.3,0.3);

  \draw[line width=0.3mm] (1.05, -1) -- (1.05,-1.7) --

  (3,-1.7) -- (3,1.65) -- (2.6, 1.65);

  \draw (-2, 3) node[anchor=north east] {\LARGE \color{codegreen}{``}};

  \draw ( 2,-0.9) node[anchor=west] {\LARGE\color{codegreen}{''}};

  \end{tikzpicture}

\end{center}

\noindent Then we can fill up the grey part of the string with

\pcode{NOP} operations. The code for this operation is

\code{\\0x90}. It is available on every architecture and its

purpose in a CPU is to do nothing apart from waiting a small

amount of time. If we now use an address that lets us jump to

any address in the grey area we are done. The target machine

will execute these \pcode{NOP} operations until it reaches the

shellcode. A moment of thought can convince you that this

trick can hugely improve our odds of finding the right

address---depending on the size of the buffer, it might only

take a few tries to get the shellcode to run. And then we are

in. The code for such an attack is shown in Figure~\ref{C3}.

It is directly taken from the original paper about ``Smashing

the Stack for Fun and Profit'' (see pointer given at the end).

\begin{figure}[p]

\lstinputlisting[language=C]{../progs/C3.c}

\caption{Overwriting a buffer with a string containing a

payload.\label{C3}}

\end{figure}

By the way you might have the question how do attackers find

out about vulnerable systems? Well, the automated version uses

\emph{fuzzers}, which throw randomly generated user input at