\documentclass{article}\usepackage{../style}\usepackage{../langs}\usepackage{../graphics}\usepackage{../data}\begin{document}\section*{Handout 3 (Buffer Overflow Attacks)}By far the most popular attack method on computers are bufferoverflow attacks or variations thereof. The first Internetworm (Morris) exploited exactly such an attack. The popularityis unfortunate because we nowadays have technology in place toprevent them effectively. But these kind of attacks are stillvery relevant even today since there are many legacy systemsout there and also many modern embedded systems often do nottake any precautions to prevent such attacks. The plot below showsthe percentage of buffer overflow attacks listed in the US National Vulnerability Database.\footnote{Search for ``Buffer errors'' at \url{http://web.nvd.nist.gov/view/vuln/statistics}.}\begin{center}\begin{tikzpicture}\begin{axis}[ xlabel={year}, ylabel={\% of total attacks}, ylabel style={yshift=-1em}, enlargelimits=false, xtick={1997,1998,2000,...,2014}, xmin=1996.5, xmax=2015, ymax=21, ytick={0,5,...,20}, scaled ticks=false, axis lines=left, width=12cm, height=5cm, ybar, nodes near coords= {\footnotesize $\pgfmathprintnumber[fixed,fixed zerofill,precision=1,use comma]{\pgfkeysvalueof{/data point/y}}$}, x tick label style={font=\footnotesize,/pgf/number format/1000 sep={}}]\addplot table [x=Year,y=Percentage] {bufferoverflows.data};\end{axis}\end{tikzpicture}\end{center}\noindent This statistics indicates that in the lastfive years or so the number of buffer overflow attacks isaround 10\% of all attacks (whereby the absolute numbers ofattacks grow each year).To understand how buffer overflow attacks work, we have to havea look at how computers work ``under the hood'' (on themachine level) and also understand some aspects of the C/C++programming language. This might not be everyday fare forcomputer science students, but who said that criminal hackersrestrict themselves to everyday fare? Not to mention thefree-riding script-kiddies who use this technology withouteven knowing what the underlying ideas are. If you want to bea 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 innocentdesign decisions, which are really benign on their own, needto conspire against you. All these decisions were taken at atime when there was no Internet: C was introduced around 1973;the Internet TCP/IP protocol was standardised in 1982 by whichtime there were maybe 500 servers connected (and all userswere well-behaved, mostly academics); Intel's first 8086 CPUsarrived around 1977. So nobody of the ``forefathers'' canreally be blamed, but as mentioned above we should already beway beyond the point that buffer overflow attacks are worth athought. Unfortunately, this is far from the truth. I let youponder why?One such ``benign'' design decision is how the memory is laidout 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 (usuallythis region is read-only). The heap stores all data theprogrammer explicitly allocates. For us the most interestingregion is the stack, which contains data mostly associatedwith the control flow of the program. Notice that the stackgrows from higher addresses to lower addresses (i.e.~from theback to the front). That means that older items on the stackwill be stored behind, or after, newer items. Let's look a bitcloser 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 thefunction \code{foo} with three arguments. \code{Foo} createstwo (local) buffers, but does not do anything interesting withthem. The only purpose of this program is to illustrate whathappens behind the scenes with the stack. The interestingquestion is what will the stack be after Line 3 has beenexecuted? 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} iscalled; on the right is the stack after \code{foo} finishes.The function call to \code{foo} in Line 7 pushes the argumentsonto the stack in reverse order---shown in the middle.Therefore first 3 then 2 and finally 1. Then it pushes thereturn address onto the stack where execution should resumeonce \code{foo} has finished. The last stack pointer(\code{sp}) is needed in order to clean up the stack to thelast level---in fact there is no cleaning involved, but justthe top of the stack will be set back to this address. So thelast stack pointer also needs to be stored. The two buffersinside \pcode{foo} are on the stack too, because they arelocal data within \code{foo}. Consequently the stack in themiddle is a snapshot after Line 3 has been executed. In caseyou are familiar with assembly instructions you can also readoff this behaviour from the machine code that the \code{gcc}compiler generates for the program above:\footnote{You canmake \pcode{gcc} generate assembly instructions if you call itwith the \pcode{-S} option, for example \pcode{gcc -S outin.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 callingthe function \pcode{foo}. You can see that the numbers 3, 2, 1are stored on the stack (the register \code{$esp} refers tothe top of the stack; \pcode{$0x1}, \pcode{$0x2} \pcode{$0x3}are the encodings for \pcode{1} to \pcode{3}). On the rightyou can see how the function \pcode{foo} stores the two localbuffers onto the stack and initialises them with the givendata (Lines 2 to 9). Since there is no real computation goingon inside \pcode{foo}, the function then just restores thestack to its old state and crucially sets the return addresswhere the computation should resume (Line 9 in the code on theleft-hand side). The instruction \code{ret} then transferscontrol back to the function \pcode{main} to the theinstruction just after the call to \pcode{foo}, that is Line9.Another part of the ``conspiracy'' of buffer overflow attacksis that library functions in C look typically as follows:\begin{center}\lstinputlisting[language=C,numbers=none]{../progs/app5.c}\end{center} \noindent This function copies data from a source \pcode{src}to a destination \pcode{dst}. The important point is that itcopies the data until it reaches a zero-byte (\code{"\\0"}). This is a convention of the C language which assumes allstrings are terminated by such a zero-byte.The central idea of the buffer overflow attack is to overwritethe return address on the stack. This address decides wherethe control flow of the program should resume once thefunction at hand has finished its computation. So if we can control this address, then we can modify the controlflow of a program. To launch an attack we need somewhere in a function a local a buffer, say\begin{center}\code{char buf[8];}\end{center}\noindent which is filled by some user input. Thecorresponding stack of such a function will look as follows\begin{center} \begin{tikzpicture}[scale=0.65] %\draw[step=1cm] (-3,-1) grid (3,8); \draw[gray!20,fill=gray!20] (-1, 0) rectangle (1,-1); \draw[line width=1mm] (-1,-1.2) -- (-1,6.4); \draw[line width=1mm] ( 1,-1.2) -- ( 1,6.4); \draw (0,-1) node[anchor=south] {\tt main}; \draw[line width=1mm] (-1,0) -- (1,0); \draw (0,0) node[anchor=south] {\tt arg$_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 8characters so that it overwrites the stack pointer and thenalso overwrites the return address. If, for example, we wantto 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 therim; the next four \pcode{B}s overwrite the stack pointer(with what data we overwrite this part is usually notimportant); then comes the address we want to jump to. Noticethat we have to give the address in the reverse order. Alladdresses on Intel CPUs need to be given in this way. Sincethe string is enclosed in double quotes, the C convention isthat the string internally will automatically be terminated bya zero-byte. If the programmer uses functions like\pcode{strcpy} for filling the buffer \pcode{buf}, then we canbe sure it will overwrite the stack in this manner---since itwill copy everything up to the zero-byte. Notice that thisoverwriting of the buffer only works since the newer item, thebuffer, is stored on the stack before the older items, likereturn address and arguments. If it had be the other wayaround, then such an overwriting by overflowing a local bufferwould just not work. Had the designers of C had just been ableto foresee what headaches their way of arranging the stackcaused in the time where computers are accessible fromeverywhere?What the outcome of such an attack is can be illustrated withthe code shown in Figure~\ref{C2}. Under ``normal operation''this program ask for a login-name and a password. Both ofwhich are stored in \code{char} buffers of length 8. Thefunction \pcode{match} tests whether two such buffers containthe same content. If yes, then the function lets you ``in''(by printing \pcode{Welcome}). If not, it denies access (byprinting \pcode{Wrong identity}). The vulnerable function is\code{get_line} in Lines 11 to 19. This function does not takeany precautions about the buffer of 8 characters being filledbeyond its 8-character-limit. Let us suppose the login nameis \pcode{test}. Then the buffer overflow can be triggeredwith 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 thefunction \pcode{welcome()}. This means even with this input(where the login name and password clearly do not match) theprogram will still print out \pcode{Welcome}. The onlyinformation we need for this attack to work is to know wherethe function \pcode{welcome()} starts in memory. Thisinformation can be easily obtained by starting the programinside the debugger and disassembling this function. \begin{lstlisting}[numbers=none,language={[x86masm]Assembler}, morekeywords={movl,movw}]$ gdb C2GNU 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 besomething like this\begin{lstlisting}[numbers=none,language={[x86masm]Assembler}, morekeywords={movl,movw}]0x0804852c <+0>: push %ebp0x0804852d <+1>: mov %esp,%ebp0x0804852f <+3>: sub $0x4,%esp0x08048532 <+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 programsthat 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 usedbuffer overflow attacks as shown above to jump directly tothe part of the program that was intended to be only availableafter the correct key was typed in. \subsection*{Payloads}Unfortunately, much more harm can be caused by buffer overflowattacks. This is achieved by injecting code that will be runonce the return address is appropriately modified. Typicallythe code that will be injected starts a shell. This gives theattacker the ability to run programs on the target machine andto have a good look around, provided the attacked process was notalready running as root.\footnote{In that case the attackerwould already congratulate him or herself to anothercomputer under full control.} In order to be send as part ofthe string that is overflowing the buffer, we need the code tobe represented as a sequence of characters. For example\lstinputlisting[language=C,numbers=none]{../progs/o1.c}\noindent These characters represent the machine code foropening a shell. It seems obtaining such a string requires``higher-education'' in the architecture of the target system. Butit is actually relatively simple: First there are many suchstring ready-made---just a quick Google query away. Second,tools like the debugger can help us again. We can just writethe code we want in C, for example this would be the programfor 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 charactersequence. While easy, obtaining this string is not entirely trivialusing \pcode{gdb}. Remember the functions in C that copy orfill buffers work such that they copy everything until thezero byte is reached. Unfortunately the ``vanilla'' outputfrom the debugger for the shell-program above will containsuch zero bytes. So a post-processing phase is needed torewrite the machine code in a way that it does not contain anyzero bytes. This is like some works of literature that havebeen 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 alipogram is a 50,000 words novel titled Gadsby, see\url{https://archive.org/details/Gadsby}, which avoids the letter `e' throughout.} For rewriting themachine 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 whenencoded as string, but produces a zero-byte on the stack whenrun. Having removed the zero-bytes we can craft the string thatwill be send to the target computer. This of course requiresthat the buffer we are trying to attack can at least containthe shellcode we want to run. But as you can see this is only47 bytes, which is a very low bar to jump over. Moreformidable is the choice of finding the right address to jumpto. 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 addresswith which we will overwrite the buffer. It has to beprecisely the first byte of the shellcode. While this is easywith the help of a debugger (as seen before), we typicallycannot run anything, including a debugger, on the machine yetwe target. And the address is very specific to the setup ofthe target machine. One way of finding out what the rightaddress is is to try out one by one every possibleaddress until we get lucky. With the large memories availabletoday, however, the odds are long. And if we try out too manypossible candidates too quickly, we might be detected by thesystem 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 thestring, 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 itspurpose in a CPU is to do nothing apart from waiting a smallamount of time. If we now use an address that lets us jump toany address in the grey area we are done. The target machinewill execute these \pcode{NOP} operations until it reaches theshellcode. That is why this NOP-part is often called\emph{NOP-sledge}. A moment of thought should convince youthat this trick can hugely improve our odds of finding theright address---depending on the size of the buffer, it mightonly take a few tries to get the shellcode to run. And then weare in. The code for such an attack is shown inFigure~\ref{C3}. It is directly taken from the original paperabout ``Smashing the Stack for Fun and Profit'' (see pointergiven at the end).\begin{figure}[p]\lstinputlisting[language=C]{../progs/C3.c}\caption{Overwriting a buffer with a string containing apayload.\label{C3}}\end{figure}By the way you might have the question how do attackers findout about vulnerable systems? Well, the automated version uses\emph{fuzzers}, which throw randomly generated user input atapplications and observe the behaviour. If an applicationseg-faults (throws a segmentation error) then this is a goodindication that a buffer overflow vulnerability can beexploited.\subsubsection*{Format String Attacks}Another question might arise, where do we get all thisinformation about addresses necessary for mounting a bufferoverflow attack without having yet access to the system? Theanswer are \emph{format string attacks}. While technicallythey are programming mistakes (and they are pointed out aswarning by modern compilers), they can be easily made andtherefore an easy target. Let us look at the simplest versionof a vulnerable program.\lstinputlisting[language=C]{../progs/C4.c}\noindent The intention is to print out the first argumentgiven on the command line. The ``secret string'' is never tobe printed. The problem is that the C function \pcode{printf}normally expects a format string---a schema that directs how astring should be printed. This would be for example a properinvocation 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 stringhas 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 howthe 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 simplewhen the application reads data from the user and promptsresponses containing the user input. Consider the slightvariant of the program above\lstinputlisting[language=C]{../progs/C5.c}\noindent Here the programmer actually tried to take extracare to not fall pray to a buffer overflow attack, but in theprocess made the program susceptible to a format stringattack. Clearly the \pcode{printf} function in Line 7 containsnow an explicit format string, but because the commandlineinput is copied using the function \pcode{snprintf} the resultwill be the same---the string can be exploited by embeddingformat strings into the user input. Here the programmer reallycannot be blamed (much) because by using \pcode{snprintf} heor she tried to make sure only 10 characters get copied intothe local buffer---in this way avoiding the obvious bufferoverflow attack.\subsubsection*{Caveats and Defences}How can we defend against these attacks? Well, a reflex couldbe to blame programmers. Precautions should be taken by themso that buffers cannot been overfilled and format stringsshould not be forgotten. This might actually be slightlysimpler nowadays since safe versions of the library functionsexist, which always specify the precise number of bytes thatshould be copied. Compilers also nowadays provide warningswhen format strings are omitted. So proper education ofprogrammers is definitely a part of a defence against suchattacks. However, if we leave it at that, then we have themess we have today with new attacks discovered almost daily. There is actually a quite long record of publicationsproposing defences against buffer overflow attacks. One methodis to declare the stack data as not executable. In this way itis impossible to inject a payload as shown above which is thenexecuted once the stack is smashed. But this needs hardwaresupport which allows one to declare certain memory regions tobe not executable. Such a feature was not introduced beforethe Intel 386, for example. Also if you have a JIT(just-in-time) compiler it might be advantageous to havethe stack containing executable data. So it is definitely a trade-off.Anyway attackers have found ways around this defence: theydeveloped \emph{return-to-lib-C} attacks. The idea is to notinject code, but already use the code that is present at thetarget computer. The lib-C library, for example, alreadycontains the code for spawning a shell. With\emph{return-to-lib-C} one just has to find out where thiscode is located. But attackers can make good guesses. In myexamples I took a shortcut and always made the stackexecutable. Another defence is called \emph{stack canaries}. The advantage is that they can be automatically inserted into compiled codeand do not need any hardware support. Though they will makeyour program run slightly slower. The idea behind \emph{stackcanaries} is to push a random number onto the stack just before local data is stored. For our very first function thestack would with a \emph{stack canary} 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) -- (-1,7.4); \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 (0,5.1) node[anchor=south] {\tt\small\textcolor{red}{\textbf{random}}}; \draw[line width=1mm] (-1,6) -- (1,6); \draw (0,6) node[anchor=south] {\tt buf}; \draw[line width=1mm] (-1,7) -- (1,7); \end{tikzpicture}\end{center}\noindent The idea behind this random number is that when thefunction finishes, it is checked that this random number isstill intact on the stack. If not, then a buffer overflow hasoccurred. Although this is quite effective, but requires suitable support for generating random numbers. This is alwayshard to get right and attackers are happy to exploit the resulting weaknesses.Another defence is \emph{address space randomisation}. Thisdefence tries to make it harder for an attacker to guess addresses where code is stored. It turns out that addresseswhere code is stored is rather predictable. Randomising theplace where programs are stored mitigates this problem somewhat.As mentioned before, modern operating systems have thesedefences enabled by default and make buffer overflow attacksharder, but not impossible. Indeed, I as an amateur attackerhad to explicitly switch off these defences. I run my exampleunder an Ubuntu version ``Maverick Meerkat'' from October 2010 and the gcc 4.4.5. I have not tried whether newer versionswould work as well. I tested all examples inside a virtual box\footnote{\url{https://www.virtualbox.org}} insulating my main system from any harm. When compiling the programs I called the compiler with the following options:\begin{center}\begin{tabular}{l@{\hspace{1mm}}l}\pcode{/usr/bin/gcc} & \pcode{-ggdb -O0}\\ & \pcode{-fno-stack-protector}\\ & \pcode{-mpreferred-stack-boundary=2}\\ & \pcode{-z execstack} \end{tabular}\end{center}\noindent The first two are innocent as they instruct thecompiler to include debugging information and also producenon-optimised code (the latter makes the output of the code abit more predictable). The third is important as it switchesof defences like the stack canaries. The fourth again makes ita bit easier to read the code. The final option makes thestack executable, thus the the example in Figure~\ref{C3}works as intended. While this might be consideredcheating....since I explicitly switched off all defences, Ihope I was able convey that this is actually not too far fromrealistic scenarios. I have shown you the classic version ofthe buffer overflow attacks. Updated variants do exist. Alsoone might argue buffer-overflow attacks have been solved oncomputers (desktops or servers) but the computing landscape ofnowadays is wider than ever. The main problem nowadays areembedded systems against which attacker can equally cause alot of harm and which are much less defended. Anthony Bonkoskimakes a similar argument in his security blog:\begin{center}\url{http://jabsoft.io/2013/09/25/are-buffer-overflows-solved-yet-a-historical-tale/}\end{center}There is one more rather effective defence against buffer overflow attacks: Why not using a safe language? Java at its inception was touted as a safe language because it hidesall explicit memory management from the user. This definitelyincurs a runtime penalty, but for bog-standard user-inputprocessing applications, speed is not of such an essence anymore. There are of course also many other programming languages that are safe, i.e.~immune to buffer overflowattacks.\bigskip\subsubsection*{Further Reading}If you want to know more about buffer overflow attacks, theoriginal Phrack article ``Smashing The Stack For Fun AndProfit'' by Elias Levy (also known as Aleph One) is anengaging read:\begin{center}\url{http://phrack.org/issues/49/14.html}\end{center} \noindent This is an article from 1996 and some parts arenot 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.There is another Phrack article about return-into-lib(c) exploits from 2012:\begin{center}\url{http://phrack.org/issues/58/4.html}\end{center}\noindentThe main topic is about getting around the non-executability of stackdata (in case it is protected). This article gives some furtherpointers into the recent literature about buffer overflow attacks.Buffer overflow attacks are not just restricted to Linux and ``normal'' computers. There is a book\begin{quote}\rm ``iOS Hacker's Handbook'' by Miller et al, Wiley, 2012\end{quote}\noindentwhich seem to describe buffer overflow attacks on iOS. A book from thesame publisher exists also for Android (from 2014) which seem to alsofeature buffer overflow attacks. Alas I do not own copies of thesebooks.\subsubsection*{A Crash-Course for GDB}If you want to try out the examples from KEATS it might behelpful to know about the following commands of the GNU Debugger:\begin{itemize}\item \texttt{(l)ist n} -- lists the source file from line \texttt{n}, the number can be omitted \item \texttt{disassemble fun-name} -- show the assembly code of a function\item \texttt{run args} -- starts the program, potential arguments can be given\item \texttt{(b)reak line-number} -- sets break point\item \texttt{(c)ontinue} -- continue execution until next breakpoint\item \texttt{x/nxw addr} -- prints 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} -- prints out \texttt{n} bytes \end{itemize}\bigskip\bigskip \noindent \end{document}%%% Local Variables: %%% mode: latex%%% TeX-master: t%%% End: