\documentclass{article}
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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 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*{Paylods}

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 gray 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 gray 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}

\subsubsection*{Format String Attacks}

A question might arise, where do we get all this information
about addresses necessary for mounting a buffer overflow
attack without having yet access to the system? The answer are
\emph{format string attacks}. While technically they are
programming mistakes (and they are pointed out as warning by
modern compilers), they can be easily made and therefore an
easy target. Let us look at the simplest version of a 
vulnerable program.

\lstinputlisting[language=C]{../progs/C4.c}

\noindent The intention is to print out the first argument
given on the command line. The ``secret string'' is never to
be printed. The problem is that the C function \pcode{printf}
normally expects a format string---a schema that directs how a
string should be printed. This would be for example a proper
invocation of this function:

\begin{lstlisting}[numbers=none,language=C]
long n = 123456789;
printf("This is a long %lu!", n);
\end{lstlisting}

\noindent In the program above, instead, the format string
has been forgotten and only \pcode{argv[1]} is printed.
Now if we give on the command line a string such as

\begin{center}
\code{"foo \%s"}
\end{center}

\noindent then \pcode{printf} expects a string to 
follow. But there is no string that follows, and how
the argument resolution works in C will in fact print out 
the secret string! This can be handily exploited by 
using the format string \code{"\%x"}, which reads out the 
stack. So \code{"\%x....\%x"} will give you as much 
information from the stack as you need and over the 
Internet.

While the program above contains clearly a programming 
mistake (forgotten format string), things are not as simple
when the application reads data from the user and prompts
responses containing the user input. 

\subsubsection*{Caveats}

\bigskip\bigskip
\subsubsection*{A Crash-Course for GDB}

\begin{itemize}
\item \texttt{(l)ist n} -- listing the source file from line 
\texttt{n}
\item \texttt{disassemble fun-name}
\item \texttt{run args} -- starts the program, potential 
arguments can be given
\item \texttt{(b)reak line-number} -- set break point
\item \texttt{(c)ontinue} -- continue execution until next 
breakpoint in a line number

\item \texttt{x/nxw addr} -- print out \texttt{n} words starting 
from address \pcode{addr}, the address could be \code{$esp} 
for looking at the content of the stack
\item \texttt{x/nxb addr} -- print out \texttt{n} bytes 
\end{itemize}

 
\bigskip\bigskip \noindent If you want to know more about
buffer overflow attacks, the original Phrack article
``Smashing The Stack For Fun And Profit'' by Elias Levy (also
known as Aleph One) is an engaging read:

\begin{center}
\url{http://phrack.org/issues/49/14.html}
\end{center} 

\noindent This is an article from 1996 and some parts are
not up-to-date anymore. The article called
``Smashing the Stack in 2010''

\begin{center}
\url{http://www.mgraziano.info/docs/stsi2010.pdf}
\end{center}

\noindent updates, as the name says, most information to 2010.
 
\end{document}

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