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The Cracking Manual
ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
Written By The Cyborg - April 3, 1992
Disclaimer
The author of this text shall hold no liability for special,
incidental, or consequential damages arising out of or
resulting from the use/misuse of the information in this
file.
The Cracking Manual
INTRODUCTION
Introduction
------------
Welcome to the wonderful world of cracking. What is
cracking? If you don't know and you're reading this, ask
yourself why? Anyway, cracking is the art of removing copy
protected coding from programs. Why do this? In recent
years, software companies have been fighting to keep copy
protection in their software to avoid their work to be
illegally copied. Users feel that such copy protection is
ridiculous in that it violate their own rights to make
backups of their sometimes expensive investments.
Whichever side you may favor, this manual will go into
some detail on removing copy protection from programs. If
you feel offended by this, then I would suggest you stop
here. Please note, I do not endorse cracking for the illegal
copying of software. Please take into consideration the hard
work and effort of many programmers to make the software.
Illegal copying would only increase prices on software for
all people. Use this manual with discretion as I place into
your trust and judgement with the following knowledge.
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The Cracking Manual
WHAT YOU WILL NEED
What You Will Need
------------------
Like all programming, cracking is the debugging stage of
software development. It is the most tedious and hectic part
of programming as you shall see. However, unlike software
development, you are given no source code, only the machine
level code commonly called machine language. Cracking
demands patience. No patience, no cracking.
Before we begin, you will need certain tools. These
include:
- A decent computer. By this, I mean at minimum a 286
computer with 2 or more megs of RAM. A 386 is the
ideal since it can load a debugger into usable memory.
- A source level debugger (eg. Turbo Debugger)
- A low level debugger (eg. DEBUG)
- An assembler system (eg. MASM, LINK, EXE2BIN)
- A hex dumping program (eg. Norton Utilities)
The source level debugger is what you will try to be using
most of the time. It provides many features that are a
convenience to the cracker, such as interrupt redirection.
Become comfortable with its features. However, in some
instances, the source level debugger may not be suitable for
cracking huge games since the debugger itself may take up too
much memory. In such a case, a low level debugger must be
used since their memory usage may be considered negligible.
This manual will focus on its use.
The assembler package will be used in the creation of
the famed loaders, which provide the cracker with dynamic
memory alterations without changing the original program.
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CRASH COURSE IN ASSEMBLY LANGUAGE
Crash Course in Assembly Language
---------------------------------
If you are already well familiar with the assembly
language, you may wish to skip this section. Cracking
demands the knowledge of assembly language. If you wish to
become a "serious" cracker, you might like to read up more
about this fascinating language. This section will only give
you enough info for intermediate level cracking.
At this point, you should familiarize yourself with
DEBUG and its commands as we will be using them shortly.
Registers
---------
One of the neato things that you will be fooling around
most often with are called the registers. Registers are like
variables (such as in BASIC) that are located within the CPU
itself. These registers may hold a positive integer from 0
to 255 or from 0 to 65535. They can also hold negative
integers from -128 to 127 or from -32768 to 32767. The
registers are given names as follows:
AX => accumulator - this register is most commonly used
for mathematical or I/O operations
BX => base - this register is used commonly as a base or
a pointer register (we'll talk more about this
later)
CX => count - used commonly for counting instructions
such as loops
DX => displacement - much like the base register
The registers stated above are considered general purpose
registers, since they can basically be used to store whatever
the user wants. Let's try putting some number in these
registers. Type in "R {enter}". You should see a bunch of
info, of which are four of the above mentioned registers.
Now, type in "RAX {enter}". Then type in a number like
8FABh. Type in "R" again and noticed how the accumulator
(AX) has change its number.
These general purpose registers can also be "split" in
half into its higher and lower order components. Instead of
having one register AX, you can have two registers, AH and
AL. Note however that while you have a range of 0 to FFFFh
for AX, you will now have a range of 0 to FF for AH and AL.
You cannot change these directly in debug, but be aware that
programs will use it. If AX contains 0A4Ch, then AH will
contain 0Ah and AL will contain 4Ch.
The following are called the segment registers:
CS => code segment - the block of memory where the code
(instructions are located)
DS => data segment - the block of memory where data can
be accessed. In block move operations in which
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huge blocks of memory are moved, this is commonly
the segment in which the CPU reads from.
ES => extra segment - also another data segment. In
block move operations in which huge blocks of
memory are moved, this is commonly the segment in
which the CPU writes to.
SS => stack segment - this is the block of memory in
which the CPU uses to store return addresses from
subroutines. (more on this later)
In introductory level of cracking, we don't mess around with
these registers. Later, we will see how we can use these to
trick a program into thinking other things, but that's later.
You can also change these registers in debug. Type in "RCS
{enter}". Then enter "0 {enter}" and notice how the CS
register changed.
There are other registers that we use to see what the
program is doing. These registers can also be change in
debug. Included are the following:
SI => source index - this register is used in
conjunction with block move instructions. This is
a pointer within a segment (usually DS) that is
read from by the CPU.
DI => destination index - this register is also used in
conjunction with block move instructions. This is
a pointer within a segment (usually ES) that is
written to by the CPU.
BP => base pointer - a pointer used commonly with the
stack segment
SP => stack pointer - another pointer used commonly with
the stack segment (this one, you don't touch)
By now, you may probably be confused about this
segment/pointer bit. Here is an analogy that my straighten
things out.
Pretend you are in kindergarden learning to read. There
are four black boards surrounding the room. These black
boards are like SEGMENTS. Let's pretend the front blackboard
is the code segment (CS). The teacher has written some
instructions on pronunciation rules. This is what the
students refer to when they try to pronounce words. In a
program, this is what the CPU refers to when it follows
directions.
Okay, now the teacher has gone to the blackboard on the
left of the classroom. We will call this board the data
segment (DS). The teacher has also written a set of words on
the board. Then she uses a wooden stick or a POINTER to
point to a word. Let's pretend this stick is the source
index (SI). She points to the word "their". Now, the
students look at the front blackboard (CS) to see how to
pronounce the word and they say "their".
Now, the instructor wants the students to learn how to
write. She points the stick to the word "apple". The
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students pronounce the word. Then she goes to the blackboard
on the right. We shall call this one the extra segment (ES).
She then uses her finger as a different POINTER and points to
a location on the board where Mary Jane will write "apple".
That's basically what segments and pointers are.
Segments are the blackboards and pointers are the teacher's
stick (we're not talking sexually here) or finger.
One last important register is the flags register.
These registers control how certain instruction work, such as
the conditional jumps (in BASIC, they are like IF-THEN's).
They are stored as bits (0's or 1's) in the flags register.
We will most often use:
zero => ZR/NZ (zero/not zero) - tells you whether an
instruction (such as subtraction) yielded a zero
as an answer
sign => NG/PL (negative/positive) - tells you whether an
instruction yielded a positive or negative
number
carry => CY/NC (carry/no carry) - tells you whether an
instruction needed to carry a bit (like in
addition, you carry a number over to the next
digit). Various system (BIOS) functions use
this flag to denote an error.
direction => DN/UP (decrement/increment) - tells a block
instruction to either move forward or backwards
in reads and writes
Try changing some of these bits. Type in "RF {enter}". Then
type in "DN {enter}" to change the direction flag to its
decrement position.
The Instructions
----------------
MOV - move
----------
Now we get to the actual instructions or commands that
the CPU will use. The first instruction you will see most
often is the move instruction. Its form is
MOV {destination},{source}. Let's try programming now. Exit
(q) and reenter debug again. Now, type in "A {enter}". You
will see a bunch of number to the left. You can think of
these as line numbers. Now type in "MOV AX,7A7A {enter}".
Then type "MOV DX,AX" and so on until your program looks
similar to the one below: (type "U 100" to see)
xxxx:0100 B8A77A MOV AX,7AA7
xxxx:0103 89C2 MOV DX,AX
xxxx:0105 B90000 MOV CX,0000
xxxx:0108 88D1 MOV CL,DL
xxxx:010A 890E0005 MOV [0500],CX
xxxx:010E 8B160005 MOV DX,[0500]
xxxx:0112 BB0200 MOV BX,0002
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xxxx:0115 26A30005 MOV ES:[0500],AX
Press enter again until you see the "-" prompt again. You
are ready to run your first program. Type "R {enter}" and
note the values of the general purpose registers. Then type
in "T {enter}". Debug will automatically display the
registers after the execution of the instruction. What is in
the AX register? It should be 7AA7h. Now, "T" again. What
is in the DX register? It should also be 7AA7h. Trace again
using "T" and note that CX should be 0 if it was not already.
Trace again and note what is in the CX register. It should
be 00A7h. Now trace another step. What is this instruction
doing? It is now moving the contents of CX into memory
location 500h in the data segment (DS). Dump the memory by
typing in "D 500". The first two two-digit numbers should be
the same as in the CX register. But wait a minute you say.
They are not the same. They are backwards. Instead of
00A7h, it is A700h. This is important. The CPU stores 16
bit numbers in memory backwards to allow for faster access.
For 8 bit numbers, it is the same. Now, continue tracing.
This instruction is moving the memory contents of address
500h into the DX register. DX should be 00A7h, the same as
CX regardless of how it looked in memory. The next trace
should be nothing new. The next trace again moves the
contents of a register into memory. But notice it is using
the BX register as a displacement. That means it adds the
contents of BX and 500h to get the address, which turns out
to be 502h. But also not the "ES:" in front of the address.
This additional statement tells the CPU to use the extra
segment (ES) rather than the data segment (DS which is the
default). Now dump address 502h by entering "D ES:502" and
you should see A77Ah, which is backwards from 7AA7h.
CMP/J? - compare/conditional jump
---------------------------------
Another instruction you will see quite often is the CMP
or compare instruction. This instruction compares the two
"variables" and changes the flags register accordingly. The
source and destination operands are the same as those for the
move instruction.
Let's consider an example in which the AX register holds
21 and the BX register holds 22. Then "CMP AX,BX" is
performed. The compare instruction is like a subtraction
instruction, but it doesn't change the contents of the AX
register. So, when 22 is subtracted from 21, the answer will
be -1, but we will never see the answer, only the flags which
have resulted from the operation. Number 21 is less than 22,
so the carry flag and the sign flag should be set. Just
remember that when the carry flag is set, the first number is
less than the second number. The same is true for the sign
flag. Why have two flags if they tell us the same thing?
This is more complicated and you should not concern yourself
with it. It requires knowledge of hexadecimal arithmetic,
the denotation of signed and unsigned integers.
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So, now that we have done the compare instruction, there
will most likely be a conditional jump instruction after. If
we wanted to jump if AX is less than BX (which it is), then
there would be an instruction like "JB 200". This
instruction says Jump if Below to instruction 200h. What
about if we wanted to jump if AX is greater than BX. Then we
might have "JA 200". This is read Jump if Above to
instruction 200. What about AX equal to BX. We would then
have "JZ 200" or "JE 200". (Please note that the previous
instructions are synonymous.) This is read Jump if Equal to
instruction 200h. Here are the jumps you will most likely
encounter:
Mnemonic Flag(s) Checked Description
-------------------------------------------------------------
JB/JNAE CF=1 Jump if below/not above or
equal (unsigned)
JAE/JNB CF=0 Jump if above or equal/not
above (unsigned)
JBE/JNA CF=1 or ZF=1 Jump if below or equal/not
above (unsigned)
JE/JZ ZF=1 Jump if equal/zero
JNE/JNZ ZF=0 Jump if not equal/not zero
JL/JNGE SF not equal Jump if less/not greater or
to OF equal (signed)
JGE/JNL SF=OF Jump if greater or equal/not
less (signed)
JLE/JNG ZF=1 or SF Jump is less or equal/not
not equal OF greater (signed)
JG/JNLE ZF=0 or SF=OF Jump if greater/not less or
equal (signed)
JS SF=1 Jump if sign
JNS SF=0 Jump if no sign
JC CF=1 Jump if carry
JNC CF=0 Jump if no carry
JO OF=1 Jump if overflow
JNO OF=0 Jump if not overflow
JP/JPE PF=1 Jump if parity/parity even
JNP/JPO PF=0 Jump if no parity/parity odd
There are all the possible combinations of conditional jumps
that you will encounter. I realize that we have not
discussed some of the flags such as overflow or parity, but
be aware that they exist and programs sometimes use them.
JMP - jump
----------
This instruction does what it suggests. It jumps too
different sections of code. Several forms of the jump
instruction include:
2E0B:0208 EBF6 JMP 0200
2E0B:020A 3EFF24 JMP DWORD PTR DS:[SI]
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The first instruction jumps to an address within the segment.
The latter instruction jumps to an address pointed to by ds:
si. The DWORD says that this will be a far jump, a jump to a
different segment (a different blackboard). So, if the
double word that is pointed to by ds:si contains 1000:0040h,
then, the instruction will jump to 1000:0040h whereas the
previous jump instruction will jump within the current
segment (or blackboard).
CALL - procedural transfer
--------------------------
This instruction is the baby that you will be carefully
watching out for most often. This instruction calls another
procedure and upon it's completion, will return to calling
address. For example, consider the following block of code:
2E0B:1002 E8BB46 CALL 56C0
2E0B:1005 7209 JB 1010
2E0B:1007 0C00 OR AL,00
The first line calls another procedure at "line number"
56C0h. Upon its completion, the instruction pointer will
point to the second line. Note that there is a "JC"
instruction. Remember that programs often use the carry flag
to signal errors. If the call instruction called a copy
protection instruction and you entered a wrong code or
something, it may return with the carry flag set. The next
instruction would then jump if there was an error to an
exiting procedure.
Note, this is a near call. A program can also have far
calls just like jumps.
INT - generate an interrupt
---------------------------
This instruction is much like the call instruction. It
also transfers control to another procedure. However, the
number after the INT instruction does not point to an
address. Instead, it is a number pointing to an address that
is located in something called an interrupt vector. You will
commonly see "INT 10", "INT 21", "INT 13". Just know (for
now) that they are like calls to procedures.
LODSB/LODSW/STOSB/STOSW - load/store a byte/word
------------------------------------------------
These instructions either load in or store a byte or a
word to or from memory. The DS:SI register pair points to
the source data. These are the registers the CPU will use
when reading from memory using the LODS instruction. The
AX/AL register will hold the number to either read from or
write to the memory. So, if DS:SI points to a byte which is
maybe 60, then a "LODSB" instruction will load in the number
60 into the AL register. A LODSB or STOSB will use the AL
register while the LODSW or STOSW will use the AX register.
The STOS writes whatever is in the AX/AL register to the
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memory pointed to by ES:DI. So, if ES:DI points to 100:102h
and if AL held 50, then the byte at 100:102h will hold 50.
After the instruction is finished, the CPU will either
increment or decrement SI or DI according to the status of
the direction flag. So, if SI was 100h and a "LODSW"
instruction was performed with a cleared direction flag
(forward), the SI will now point to 102h.
MOVSB/MOVSW - copies a byte/word from source to destination
-----------------------------------------------------------
This instruction gets a byte or a word from the data
pointed to by DS:SI and copies it to the data pointed to by
the ES:DI address. When the instruction is finished, SI and
DI will be incremented or decremented accordingly with the
status of the direction flag. So, if DS:SI pointed to a byte
with the number 30, a "MOVSB" instruction would copy into the
byte pointed to by ES:DI the number 30.
REP - repeat
------------
The REP instruction in front of a MOVS/LODS/STOS would
cause the MOVS/LODS/STOS instruction to be repeated for a
number of times specified in the CX register. So, if CX
contained 5, then "REP STOSB" would store whatever was in the
AL register into the byte pointed to by ES:DI five times,
increasing DI each time.
LOOP - looping
--------------
The LOOP instruction repeats a block of instructions for
a certain number of times. This number will be held in the
CX register. Each time we reach this instruction, the CPU
will decrement the CX register and jump to a specified
instruction until CX becomes zero. This instruction looks
like "LOOP 1A00" where the number indicates the instruction
address to loop to.
Arithmetic Operators
--------------------
Arithmetic instructions allow you to perform various
arithmetic function of data. "ADD" and "SUB" work the same
way as "MOV" instructions do in that it subtracts whatever is
in the source register from the destination register and
stores it in the destination register.
The "MUL" and "DIV" instructions are a bit more
complicated and they are not used as intensively as the "ADD"
or "SUB" since they are slow, so we will not talk about them.
There are also a multitude of other instructions that
you should familiarize yourself with if you are thinking of
becoming a serious cracker. The instructions given above are
only the BARE minimum that you need. There is no way around
learning assembly for better cracking.
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THE CRACKING
The Cracking
------------
Now the fun stuff begins. First, we must discuss the
different forms of copy protection schemes. They are
basically divided into the disk based and manual based copy
protection schemes.
With disk based schemes, the software often reads from
specific sectors on a disk to determine the disk's validity.
How can this be done? When you perform a disk format, the
disk is formatted with specific sector sizes. Once the
sector size changes, DOS cannot recognize it, thinking that
it is a bad sector. Since this looks like a bad sector, a
simple DISKCOPY will not work in copying such disks.
Interrupt 13h (the assembly mnemonic is INT 13) was commonly
used to handle such copy protections. It is now very rare to
encounter the once famed INT 13h copy protection method
nowadays since it was quite easy to defeat. Any professional
commercial software will often use their own custom based
disk I/O routines. This involves intimate access to I/O
ports using IN and OUT instructions. This is beyond the
scope of the first release of this manual. However, if you
are lucky, the I/O functions might be called from a "CALL"
instruction in which case you may defeat the protection
without much difficulty. Another disk based scheme used to
denote legality of software is used during the installation
process of the software. With certain programs, when you
install it, it copies the files into the hard drive. But it
also sets a specific sector in the hard drive so that the
program can recognize it. This is also similar to diskette
copy protections, but can be defeated in much the same way.
Thank goodness that disk based copy protections are
almost completely out of the software industry. However, a
sometimes more difficult copy protection scheme has arisen
that may sometimes prove to be even more difficult to crack.
These schemes are commonly known as the doc checks in which
the user must have a copy of the manual to bypass the
protection. With programs compiled as true assembly (you can
call then "normal" programs), these protections are not too
bad to trace through and crack. With programs that run
scripts (such as Sierra games), this can he a real chore
however. Why? It is because it is like running a program
within a program. You just have to be very very patient in
this case, carefully tracing through the instructions.
As if these copy protection schemes weren't enough,
software companies have also added trace inhibition schemes
to their code. What does this mean? This means that you
will have a hell of a time trying to trace through code.
However, if you know how these things work, it should not be
too much of a problem.
Run-time compression/decompression and
encryption/decryption of files also make changes to the
program difficult. In this case, the loader sure comes in
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handy. Also, when the data within the file changes due to
overlays, loaders are also good to use.
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DISK BASED COPY PROTECTIONS
Disk Based Copy Protection
--------------------------
Since disk based copy protection schemes are rarely
used, we will not go into great depth in its discussion.
INT 13h
-------
I have previously mentioned that INT 13h copy protection
schemes are hardly ever used anymore. Nevertheless, it would
be good practice for the beginner to learn how to defeat the
code. You will most likely see INT 13h used with function 2,
read sector. This means that:
AH => will contain the number 2 (function 2)
AL => the number of sectors to read in. This is
commonly only 1 since you just want to check a few
sectors for disk validity.
CH => will contain the cylinder number
CL => will contain the sector number
DH => will contain the head number
DL => will contain the drive number
00h - 7Fh for floppies
80h - FFh for fixed disks
ES:BX => will point to the address into which the data
read from the disk will be written to
Upon the return for this interrupt, if the carry flag is
set, that means that the program could not read the sector,
and therefore the disk is valid. If the carry flag is clear,
that meant that INT 13h could read the sector properly and so
the disk would be bad in the eyes of the program, thinking it
was a copied disk.
Okay, now that we know to look for INT 13h in the
program code, we can begin tracing. First, we must know the
difference between debug's "T" and "P". "T" is the trace
instruction, which tells it to follow instructions step by
step. That also means that in LOOP or REP instruction, the
trace will patiently go through the loop until finished.
Also, during CALL instructions, trace will go into the call
and execute the instructions pointed to by the call
instruction. The "P" command is similar to the "T" but with
the difference in that it traces over instructions. That
means that if it encounter a LOOP or REP, it will quickly
finish up the loop and point to the next instruction. With a
CALL, the "P" (proceed) will not go into the subroutine.
Instead, it will just execute the procedure, then point to
the next instruction.
Okay, before you start tracing for hours through a
program, you must first notice when and where the copy
protection appears. Run the program in DOS first and make
careful note of when things happen. You might see an intro
screen, then the music pops up, then the menu comes out.
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Notice this so you will know where you are in the program.
Once you have done that, you can begin debugging the
program. Whenever you start out with a program, you use "P"
to trace through the program. Be patient as this might take
a while. While you are tracing, watch out for CALLs and
INTerrupts. When you are just about to execute the step, try
to remember the segment and offset of the instruction. The
segment is the number to the left of the colon while the
offset is the number to the right. As you continue tracing
through the program, you will find that the screen might
blank and display the intro screen or something like that.
This is a good sign and it tells you that you are headed in
the right direction. Start slowing down when you feel that
you are near to the copy protection.
Situation 1 - Exit from copy protected CALL
-------------------------------------------
Oops, you have traced over a call that accessed drive A.
Unfortunately, you also exited the program. That's good.
You have just narrowed down the location of the copy
protection code. Now I hope you remembered the address of
that CALL. If not, you gotta start all over to find it.
Anyway, restart the program now. Now Go to that instruction
by "G {segment:address}".
Did something go wrong? Did the computer freeze or
something? It is most likely that this is an overlay or
encrypted code or something that caused the code at that
location to change. In this case, you will have to remember
the addresses of various instructions along the way.
Instructions that you want to take note of are far calls (if
you remember, calls with a segment:offset address as their
operand). You don't have to do this for every call. As you
crack more and more, you will get the hang of which
instructions to keep track of.
Okay, let's assume you have gotten back into the
location of the code again. It is a CALL instruction that
will access the disk drive. At this point, try skipping the
CALL instruction. To do this, type in "RIP {enter}". Then
type in the address of the next instruction. Then execute
the do or die instruction, "G". If the program runs fine
without asking for the copy protection, congratulations! You
have cracked the program.
If the program freezes or does something weird, restart
the program and trace back to the suspected copy protected
location. Now use the "T" command once and start using "P"
again. Remember to write down the address of that CALL
instruction you just traced into so you can come back to it
quickly. As you keep tracing, using the above procedures,
pretend you eventually come up to an INT 13h instruction.
See what it does by tracing over it. Make sure you have a
disk in drive A too. If there was no error, force an error
by turning on the carry flag and proceeding. With INT 13h
copy protections, this should be sufficient to crack the
program.
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Situation 2 - Return from copy protected CALL
---------------------------------------------
Okay, the CALL that you just traced over accessed the
disk drive, but it didn't kick you out. Keep on proceeding
and this point. If there is an instruction that causes you
to jump because of a carry flag, try fooling around with this
carry flag and see how the program reacts. INT 13h copy
protections are usually simple enough for you to just change
the carry flag to allow the program to bypass the copy
protection.
Access to the Hard Drive
------------------------
The cracking for installation software is also the same
as cracking for the INT 13h. You just keep tracing until you
see some disk activity. At that point, you try messing
around with some of the conditional jumps to see what
happens. If you have the original program, you should run it
also to see the differences between the valid and invalid
copies.
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DOC CHECK COPY PROTECTIONS
Doc Check Copy Protections
--------------------------
Okay, we have just quickly scanned over disk based copy
protections because they are rarely used nowadays. Doc
checks will be discussed in greater detail for the rest of
this manual.
Unlike the disk based protections, which are based on
hardware identification, doc checks are based on software
identification. Therefore, the only information that will
indicate that a copy protection is happening is the screen,
unlike the whirr of the disk drive. The moral, watch the
screen. Because this copy protection is software based, it
will be more of a challenge to trace, but of course, that is
the "fun" part of cracking.
The Basics
----------
Make sure you have the COMPLETE version of the program
you are about to crack. When you do, run the program in DOS.
While the program is loading, take note of exactly what goes
on with the screens, sounds, etc. Here is what you might
want to note:
1) What comes up first? Is it a standard text output
that asks you for the type of graphics adaptor you
have, the number of joysticks, the sound card?
2) When does the intro screen come up? Is it after the
music starts? After the copyright notice? After
the text prompt for the graphics mode you will be
operating in?
3) What happens now? An animated sequence that brings
you through the beginning plot of a game? If so,
can you press a key and escape from it?
4) Now what? Is there a main menu? When you start the
game by selecting the "START GAME" option from the
menu, does the copy protection come up immediately?
5) If it doesn't come up immediately, when does it come
up?
6) Does the copy protection only appear when you are
playing the game, or does it come up also when you
select "CHANGE OPTIONS" from the main menu?
Obviously, these questions are merely prompts for you to
follow. Use your own mind in discovering what to take note
of. There are no set rules for cracking. It is a puzzle
that you must use your mind on.
Okay, once you have run the program, go into your
debugger (in our case, DEBUG) and load up the program. One
tip to use when you first start out programming is to use the
"P" command to trace through code. As you become a more
advanced cracker, you might start seeing patterns in coding.
These patterns are characteristic of high level programming
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languages (Pascal, C, etc.) and are usually the
initialization code for the rest of the program. Use "P" for
each instruction, one at a time. Be patient as this might
take a while.
Okay, you have been tracing for some time now and
finally, you notice something happen. The screen might have
blanked or maybe a message prompting you to enter the
graphics mode may have popped up. Was this what you have
noted before? It should be and you can assure yourself that
you are headed in the right direction. As you keep tracing
programs, you notice that CALLs usually do something
significant. A CALL might clear the screen or sound some
music. When it does something rad like this, write down its
address as the segment:offset pair. The segment is the
number to the left of the colon while the offset is the
number to the right of the colon. Don't be a dork and set a
breakpoint there. Write it down on paper or something. We
will see later on why breakpoints fail miserably in the cool
wares.
Why take note of these instructions? As you trace
deeper and deeper into programs, the coding often loads up
overlays or maybe decompresses code to the memory location
that you have just traced over. Therefore, if you set a
breakpoint there, or execute a "G" instruction to that
address, you will fuck up the program and cause your computer
to freeze. We will see why when we examine how breakpoints
and single stepping works.
Also, while you are tracing using "P", mentally remember
the addresses of the CALLs. That way, if you trace over a
call that brought you immediately to the copy protection, you
won't have to retrace the code again. You don't have to
write down all of the addresses, of course, just remember one
at a time and write them down if they do anything
significant.
Code Guards Through Keyword Entry
---------------------------------
Okay, you know that the copy protection is one in which
the program waits for you to type in a keyword that you have
to look up in the manual or something. Here are then
following steps you should take.
Situation 1 - Return from a copy protected CALL
-----------------------------------------------
When a copy protection coding reveals itself on the
screen, you can have a situation in which you are returned to
the debugger, waiting for the next instruction to be
executed. Now, suppose that the CALL asked you to enter a
code. You entered an incorrect code and were returned to the
debugger, but you have not exited the program. Make sure
that you have previously recorded the address of this CALL.
Now, you can do two things, (1) you can try skipping over the
CALL, (2) you can trace on further. As you become more
experienced, you will be able to better decide. As one with
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experience, however, I can say that 90% of the time, you will
have to trace further on, but hey, you might get lucky.
For now, let's say you are lazy and decide that you want
to skip over the call to see what happens. To do this, you
must restart the program. Then trace your way back to the
CALL where the copy protection was located. Use "G
{segment:offset}" to do this. If, for some reason, the
computer freezes when you do this, you will have to use "G"
followed by the addresses of the CALLs that you have noted
down to be significant. If that doesn't work, resort to
retracing the code over again. As you become more
experienced, you will find that you rarely have to retrace
the entire code since you can "feel" what is going on. Okay,
now that you are at the location of the CALL, this is the
time to skip over the instruction. To do this, enter "RIP"
and then the address of the next instruction's address. Now
enter the "G" command and see what happens. If the program
runs just fine, you've cracked the program. If the program
kicks you out or crashes, you have to do some more tracing.
Okay, so you've decided to continue tracing from the
point of the copy protection. There are usually a bunch of
CMP and J? CMPS? instructions after the call. This point on
is the difficulty of cracking for a beginner since you don't
know what the fuck is going on. All those compares and jumps
don't mean shit to you are you are about to pass out in
frustration. Don't distress, here are a few tips I can give
you. If these don't work, you gotta find out your own
solutions to the problem.
Okay, in all probability, the CALL that you just traced
over was acting as a read string procedure (like BASIC's
INPUT). That means somewhere in the computer's memory, there
lies the code that you typed in and the code that you were
supposed to have typed in. What this would mean is that the
code after the CALL will do some sort of string comparison.
Look out for these. It might be hidden inside another CALL
if you're lucky. In such a case, does the program kick you
out? If it does, you have to trace into the call using "T"
to see what is going on. Okay, the string comparison will
most likely take the form of some kind of loop. Maybe "REP
CMPSB" or "LOOP". In the case of the REP CMPSB, there might
be a JZ/JNZ or JCXZ/JECXZ that follows it. When strings
match, the CX register will be zero. If CX is not zero, the
strings are not the same and the conditional jump will
probably jump to an exit routine. All you have to do is to
change the status of the zero flag. Then, try out the "G"
instruction. If it still didn't work, start over and do some
more tracing. If the string compare is not of the REP form,
there will be some kind of loop that will check between two
memory locations. In such a case, you will just have to
become accustomed to realizing that the code is a string
compare. There is no standard code for this. If you know
you have entered a wrong code, trace through the loop and see
where in the loop you are thrown out of the loop. At this
point, you can go back to it, change some flags to make sure
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you stay in the loop. When you exit through a different
location, you have probably bypassed the code and now, you
can enter "G" to see what happens.
Situation 2 - Exit from a copy protected CALL
---------------------------------------------
When a copy protection coding reveals itself on the
screen, you can have a situation in which you are not
returned to the debugger, instead, causing you to exit the
program. In this case, you have to restart the program and
trace into the CALL using "T". After that, you can start
using "P" again to uncover the location of the code. You
will most likely encounter a condition that will resemble
situation 1. Follow its instructions.
Shortcuts For Keyword Entry Protections
---------------------------------------
With keyword entry systems, you might be lucky to have
the codes stuck somewhere into file in its
uncompressed/unencrypted form. This means that you can "see"
the keywords in its ASCII format. This case is cool because
you won't have to do any tracing to crack the program. All
you have to do is to dump the contents of the files to find
something that looks like a keyword. (Always backup the file
that you are about to alter.) When you have found such a
file and the location of the codes, all you have to do now is
to change the codes to values that you know. For example,
one code might call for you to enter "PIRATE". It's a bitch
if you don't know the code. But if you change the code to
your name or something else you will never forget ("CYBORG"),
then you'd be set.
However, in most instances, you can't simple just type
over the old code with your new code. In high level
languages, these codes are stored as strings. In 'C',
strings are stored in their ASCII equivalent. They are then
terminated with a NULL character (this is a 0). In Pascal,
the lengths of the strings are first stored in the first
position. Then, the ASCII is stored.
NULL Terminated Strings
-----------------------
So, if you see zeros after the codes, this is a NULL
terminated string. Now, start at the beginning of the string
and enter your code. Then, enter the '0'. Make sure your
string is less than the original string since 'C' refers to
these strings also with pointers.
Pre-Length Indentifier
----------------------
If you see numbers before strings, enter your own code.
Then change the length of the code appropriately. Make sure
you do not exceed the length of the original string.
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Code Guards Through Pointed Icons
---------------------------------
We have a case where we do not type in keywords.
Rather, we must use a pointer device such as the cursor keys
on the keyboard, the mouse, or joystick. These protections
are a bit more complicated since there are no strings to
compare against. Rather, the input will be a number stored
in memory or a register. This is what makes this copy
protection more difficult to crack. We have to hunt through
code to find out which compare instruction is the key.
What you have to do is to find the general location of
the copy protection code as before. Then, instead of typing
in the keyword, you select the icon. Like before, you must
step slowly through the code and go until the program JUST
STOPS asking you for the code. For example:
2E0B:0000 E8740E CALL 0E77
2E0B:0003 38D0 CMP AL,DL
2E0B:0005 7569 JNZ 0070
2E0B:0007 CB RETF
You might decide to trace over the call at address xxxx:0000.
But then, you see that the screen displayed the icons and you
got to select the code. Then, the procedure does some disk
activity and you return to address xxxx:0003. If you see
something happen after you have just finished entering the
code or if it is slow in returning you to debug, then,
some code must have been performed before you returned. In
this case, you must trace into the CALL to see what has
happened. If not, there is still a small probability that
there were some instructions that formatted the code you
entered and saved it to a memory location. (We'll talk about
multiple doc checks later.)
Realize that most of the programs that you will be
cracking have been programed by C or some other high level
language. These languages often use the stack (SS:SP) to
pass parameters (variables) or to create local variables for
a procedure's use. Most likely, you will see compares to
data contained within the stack such as "CMP AX,WORD PTR
[BP+10]" or "MOV DX,WORD PTR [BP+10]". This is what you hope
to find, although not always the case. If you do see some
access via the stack using the BP register as a pointer, you
may have something there. Then, all you would have to do is
to mess around the flags register (most likely, JZ/JE will be
used) at the compare instruction.
Multiple Doc Checks
-------------------
There are some wares that invoke multiple doc checks,
doc checks that pop up either systematically or randomly. In
addition, there could also be two types of this protection.
The doc check could be a similar type (eg. typing the code
found on page...) or they could be different (eg. typing in
the code on page... then select the correct icon), although
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the latter is more rarely used due to its extensive memory
usage.
Situation 1 - Similar doc checks
--------------------------------
Cracking multiple doc checks that are similar is just
like cracking with just one doc check. The procedure to
trace is still the same. Keep Proceeding until you come up
to the CALL that contains the copy protection. Just use the
sequences mentioned above. When you are absolutely positive
that the call contains the copy protection (skip the CALL and
see what happens; if the protection has been bypassed but
appears at other times, you got something), here is what you
do.
1) Note what type of CALL it was. Near if the operand
(number after the CALL) was a four digit number or
far if the operand contained the segment:offset
pair.
2) Trace INTO the call.
3) At the first instruction, note the address inside
the CALL.
4) Then, type in "A" then the address of that very
first instruction.
5) If there was a near call performed, now type in
"RETN", otherwise, type in "RETF".
6) Now run the program ("G") and see what happens.
If this call was definitely the copy protection, you should
have bypassed the copy protection completely. Otherwise, you
might have a case like situation 2.
Situation 2 - Different doc check types
---------------------------------------
Again, cracking multiple doc checks are like cracking
single doc checks. You follow the same procedures until you
come up to a copy protected location. Then, you would trace
into the code as explained in situation 1 just to make sure
that the code is not called up again. Different doc checks
are a bitch to do because you have to manually keep tracing
until you find each one to effectively rid yourself of the
copy protection. There is not sure way of getting rid of all
the doc checks any other way. But luckily, there are very
few wares out there like this. Remember, the more the
company shoves into the program's memory, the more money it's
gonna cost them.
Of course, I cannot cover every single type of doc check
since there are too many of them. You'd just have to use
your own imagination to solve some of them.
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SPECIAL SITUATIONS
Special Situations
------------------
What all crackers are faced with at one time or another
are situations that call for intuitive thinking to overcome
the barrier. Remember, there is no one sure way of cracking.
INT 3 - Problems During Tracing
-------------------------------
Sometimes, when you start cracking, you just find your
instruction pointer messing up. You keep tracing and
tracing, then your computer freezes. But then, when you type
"G" at the beginning of the program, it works just fine.
What is happening here? There are several things that the
program could do to impede tracing. Unless you have a
hardware debugger, you have to settle in for more primitive,
intuitive methods. First, we have to find out how a software
debugger works.
I now introduce you to INT 3 and INT 1. They are the
breakpoint and single stepping interrupts respectively. We
will be looking at INT 3 the most.
What happens when you set breakpoints? Well, here is
what the debugger does. At the address you have specified,
the debugger will read in the byte at that address and store
it somewhere else in its own memory. This byte is part of
the whole instruction located at that address. For example,
if there was an "INT 13" at that location, the machine
language equivalent will be CD13h. Debug will read in the
first byte, CDh, and save it in memory. The CDh will then be
replaced by INT 3 (CCh). So, the code will now look like
CC13h in machine language. When you unassemble this at the
address, you will see "INT 3" (the instruction only takes up
one byte) and some gibberish after that. So, when the CPU
comes up to this address, it will encounter INT 3 and will
return control to the debugger. The debugger then replaces
the INT 3 with the CDh byte used before.
With single stepping, the same thing occurs. Debug will
also insert the INT 3 instruction at the instruction after
the one you are about to execute. Then, internally, a "G"
instruction is performed until it reaches the INT 3, at which
point, the byte will be replaced and everything will be cool.
Use of INT 3 to Call Up Other Interrupts
----------------------------------------
This INT 3 deal seems to be cool, working in many
situations. But what if the software vendor reprograms INT 3
to point to an INT 21? Many programs use INT 21 to access
DOS functions like reading a file, etc. There would be a
conflict now as the program uses INT 3 to call up DOS while
debug wants to use INT 3 for its breakpoints. There is also
another problem. INT 21 uses two bytes (CD21h) while INT 3
uses only one byte (CCh). Therefore, you cannot replace INT
3 with the INT 21.
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Also, INT 3 could be reprogrammed so that everytime it
is used, the program will just exit to its higher process.
So everytime you single step, you will be kicked out of the
program.
Parity Errors with INT 3
------------------------
The tough copy protections use the change of memory to
obstruct tracing. Examine the code below:
2E0B:0500 FC CLD
2E0B:0501 B80000 MOV AX,0000
2E0B:0504 BB0000 MOV BX,0000
2E0B:0507 BE0005 MOV SI,0500
2E0B:050A BF0010 MOV DI,1000
2E0B:050D B90005 MOV CX,0500
2E0B:0510 AC LODSB
2E0B:0511 345A XOR AL,5A ;'Z'
2E0B:0513 01C3 ADD BX,AX
2E0B:0515 AA STOSB
2E0B:0516 E2F8 LOOP 0510
2E0B:0518 3B1E0043 CMP BX,[4300]
2E0B:051C 7403 JZ 0521
2E0B:051E E9EF2A JMP 3010
2E0B:0521 D1E0 SHL AX,1
Notice what the program is doing. It is performing a simple
decryption of a block of code from address 500h and putting
it in address 1000h. In addition, there is a checksum being
performed at address . The program is adding all those bytes
up, then comparing the number with some other number (a
checksum value) in memory at address 4300h. So what you may
say. When the program is run without any set breakpoints,
the program will run fine. But when you start tracing
through the code, or putting a breakpoint somewhere after the
loop, the program will cause you to exit. If you decide to
change the program so that it will let you pass regardless of
the checksum value, somewhere along the line, the program
will fuck up.
This goes back to the idea of INT 3. Right before debug
executes an instruction, it places an INT 3 at the next
instruction. In this program, when debug places this
interrupt and executes an instruction, the program is reading
in this INT 3 at the address and copies it to a different
address. INT 3 is obviously a different number than the
other instructions, so the checksum value will be different.
So, now that INT 3 is copied to another location in memory,
debug also cannot replace that with it's original byte value.
Therefore, if you try to force the checksum to match and
continue running the program, the program will crash because
the INT 3 is causing the instructions after itself to be
interpreted incorrectly by the CPU.
To bypass this, you have to make sure not to get your
INT 3 placed in the wrong place at the wrong time. Looking
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at the program, you can keep tracing normally until the SI
register points to any byte past the CMP instruction at
address 519h. Then, you can do a "G 518" to finish off the
loop quicker. Debug will place a temporary INT 3 at address
518h, but it doesn't matter now since SI will be past 518h.
This is obviously a simple example, but it gets the point
across that you have to watch where you trace.
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OVERLAYS/LOADERS
Overlays/Loaders
----------------
Sometimes, programs will have an initialization code and
upon its completion, call up another program or overlay.
These programs present unique situations in which it is
sometimes difficult, after finding the copy protection code,
to write the changes to disk. Let's see what these programs
do before we go on to the next topic of making changes
permanent.
Loaders are usually small programs that might first ask
you for the graphics mode or what sound card you have. When
finished, it will load up another program. Sometimes, this
is done with DOS' interrupt 21h, function 4B00h (load and
execute). This is the same interrupt DOS uses to load up
programs when you type them in at the DOS prompt. You can
tell what file is going to be executed by tracing up to the
INT 21 instruction and dumping the address pointed to by
DS:DX (type in "D DS:DX"). Also, internal procedures could
be used to call up the program. Use what you've learned to
trace through them.
Code decryptions or dynamic heap allocation where data
is to be loaded presents problems as well. Code that changes
as the program progresses makes code changes difficult in the
file itself. And when you want to alter sometime in the data
area, something called a heap is often used to store the
data. The thing with the heap is that it can be allocated at
anytime and depending on what is currently in memory, you
can't tell where the memory is going to be located. In these
cases, you might choose to go with run-time memory overlays
(discussed later).
Writing the Changes Out to the File
-----------------------------------
Okay, so you've found the copy protection. You also
know how to bypass it. Now, the next problem you will most
likely encounter is writing it out to a file. But first,
let's assume a simple case.
Using a Hex Dump Program
------------------------
Included is this package is one of the files from Norton
Utilities which does a decent job of finding and changing the
contents of files. Before we exit that debugger, we must
know what to look for.
1) At the location of the instruction, copy down the
machine language equivalent of the instruction. At
instructions after that, also take down their
machine level equivalents. This is what you will
use to search for the code in the file.
a) If there is a near call or a near jump or a near
memory access, you can just write down all the
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hex numbers.
b) If there is a far call (CALL DS:[5C10+BX]) or a
far jump (JMP DWORD PTR ES:[5080+BX]) or a far
memory access (MOV AX,WORD PTR ES:[10+SI]), then
do not write these instructions down. In .EXE
files, anything that is located in different
segments will have different displacement
values. This is a value in the file. At the
beginning of the file is a table that tells DOS
where these instructions are located. When the
program is loaded into memory, the pointers are
changed appropriately to match the memory
location. So, write down other near
instructions like CLD, JZ 100, INC AX, etc.
2) After you know what to search for, you must now know
what you will have to be changing. Very often,
NOP's are used to "delete" code. For example, if
there is a CALL 3140 and we want to skip this call,
we can NOP it out. The near call takes up three
bytes. The NOP takes up one byte. So, type in "A"
at the address of the call and enter "NOP" three
times. Then unassemble the code to make sure that
the code still looks okay. Take down the machine
level equivalents of the NOP's (90h). Same thing
with conditional jumps. Suppose you have a JZ 90
and you want it to jump to address 90 everytime,
then type in "A" at the jump instruction and enter
"JMP 90". Then, just write down the machine code as
before. One thing, however. You cannot do what I
have just said above with far calls. Remember, the
numbers will be different in the file as compared to
memory. So what do you do? No problemo. At the
call instruction, trace into the call and place a
"RETF" instruction at the address of the callee.
This will be the location that you will search for
(write down the bytes here) and where you will be
writing to (RETF is CBh in machine language).
3) Finally, after all this is through, you can enter
your file editor and search for the numbers you
wrote down. Then, you can change the numbers. Now
run the program and it should be cracked. But
remember, always backup the file you are about to
change.
Using a Memory Overlay
----------------------
When do you use these things? You would use memory
overlays when step 3 (stated above) has failed in some way.
Maybe you couldn't find the code, or when you change it, the
program freezes up. Don't fret, the memory overlay is here.
What is a memory overlay? It is an external program (TSR)
that when it reaches a certain point during program
execution, it will change the location in memory you have
specified. It overlays the code during run time.
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Here is what you will need to do to make the overlay
work. First, you must find some way for the program to call
up the overlay code. This can most easily be done by
reprogramming interrupts. So, the first thing you have to do
is look for an interrupt usage near the copy protection code
(usually an INT 21h or INT 10h). When you find this
interrupt (it must be fairly close to the code), write down
the address of the NEXT instruction. You must get down the
segment and the offset. Also, get down the current status of
the registers. For interrupts like INT 21h and INT 10h,
write down the functions numbers (eg. AX,AL,BX,DX,etc.).
Then, keep tracing until the copy protection code. Get the
address of the instruction that you want to change (the
segment and the offset). Also get down the machine language
equivalent of the changed code. This should be all you need
for the overlay program. Here is the overlay program:
INT_SEG equ 1DA5h ;SEG:OFF of instruction after the
INT_OFF equ 05D1h ; calling interrupt
CHANGE_SEG equ 2DA5h ;SEG:OFF of instruction to change
CHANGE_OFF equ 0432h
OVERLAY segment para 'code'
assume cs:OVERLAY,ds:OVERLAY
org 100h ;This will be a .COM program
START: jmp INITCODE ;Initialization code
;**************************************************************************
OLDINT dw 0,0 ;Storage for old interrupt address
ADDR_OFF equ <word ptr [bp+2]>
ADDR_SEG equ <word ptr [bp+4]>
CR equ 0Dh ;Carriage return
LF equ 0Ah ;Line feed
BEEP equ 07h ;Beep
EOS equ '$' ;End of DOS string
DISPLACEMENT equ CHANGE_SEG - INTSEG
;**************************************************************************
NEWINT proc far
push bp ;Establish stack frame
mov bp,sp
push ax ;Save necessary registers
push bx
push cx
push dx
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push si
push di
push ds
push es
mov bx,ADDR_OFF ;Get offset
cmp bx,INT_OFF
jnz EXIT
cmp ax,0201h ;Check for AX=0201h <=(1)
jnz EXIT
cmp bx,0001h ;Check for BX=0001h <=(2)
jnz EXIT
mov bx,ADDR_SEG ;Get segment
add bx,DISPLACEMENT
mov ds,bx ;This will be the segment of change
;change the number at the next line to point to the offset of
; the address to be changed
mov bx,1C12h ;This is the offset of the change
mov al,0EBh ;This is the byte to be changed
mov [bx],al
;change the number at the next line to point to the offset of
; the address to be changed
mov bx,1C20h ;This is the new offset of the change
mov ax,0B8h ;This is the byte to be changed
mov [bx],ax
mov al,0 ;This is the next byte to be changed
mov [bx+2],al
pop es ;Restore necessary registers
pop ds
pop di
pop si
pop dx
pop cx
pop bx
pop ax
pop bp
iret ;Interrupt return
EXIT: pop es ;Restore necessary registers
pop ds
pop di
pop si
pop dx
pop cx
pop bx
pop ax
pop bp
jmp dword ptr cs:OLDINT ;Jump to old interrupt
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NEWINT endp
;**************************************************************************
FINISH equ $
MESSAGE db "This is an overlay loader.",CR,LF
db "Written by The Cyborg.",CR,LF,BEEP,EOS
INITCODE:
mov ax,cs
mov ds,ax ;DS point to CS
mov ah,9 ;Print string
mov dx,offset MESSAGE ;The address of the message
int 21h
mov ax,3510h ;Get old interrupt address
int 21h
mov OLDINT[0],bx ;Save in memory for later use
mov OLDINT[2],es
mov ax,2510h ;Set new interrupt address
mov dx,offset NEWINT ;Point to new procedure
int 21h
lea dx,FINISH ;CS:DX of last byte of code to remain
int 27h ; in memory. Terminate and stay
; resident.
OVERLAY ends
end START
All you have to do is set the first four values in the first
four lines of the file. They are the segment:offset pairs of the
interrupt address and the address of the bytes to be changed.
Also, change the functions to check for at (1) and (2) to
appropriately check for proper code entry. Then, specify which
bytes you will be changing at the specified lines. Then compile
this crack ("ASM OVL {enter}").
The next program demonstrates a simple loader. It also
demonstrates what you can do if you have a program that utilizes
scripts or dynamically allocated data areas in heap spaces. This
program scans for a known segment in memory for a "keyword". When
it finds this, it can then begin writing new code to overlay the
old data. Note, KEYWORD specifies the keyword to look for. Then,
CRK (0's) is the list of bytes to replace the data areas pointed
to by addresses listed in LIST. The addresses in LIST are
displacement addresses. This means that at the address the
keyword was found in, the appropriate number listed in LIST is
added to that address. There are thirteen addresses whose data
are to be changed in this case.
Also interesting to note is that this program is using two
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interrupt vectors, INT F1h and INT 21h. INT 21h is used in the
same way as the above overlay program uses it. It replaces two
bytes at offset 1FE5h with CDF1h. This is the machine language
equivalent of INT F1h. Now, let's examine what INT F1h actually
does. First, it changes the return address in the stack so that
instead of returning to the address right after the INT F1h
instruction, it will return to another instruction, located at
offset 1FE5. This is the location of the INT F1h instruction.
This interrupt, upon its completion, will replace the INT F1h
instruction with the original instruction and run the program
normally.
The loader itself is simple. It reallocates the memory
located to itself to accommodate a "daughter" program, the program
that it is going to load. If it can't find the program or if an
error has occurred trying to execute the program, the loader will
load itself up as a TSR. Then, you can run the program via DOS.
This loader also checks if INT F1h has been occupied and returns
an error if it is.
LOADER segment para 'code'
assume cs:LOADER,ss:LOADER
org 100h
BEGIN: jmp INIT
CR equ 0Dh
LF equ 0Ah
BEEP equ 07h
EOLN equ '$'
OPTION db 1 ;Options
CRC dw 0 ;Cyclic Redundency Checking data
START equ $
OLDINT1 dw 0,0
OLDINT2 dw 0,0
KEYWORD db "weat"
CRK db 0,0,0,0
LIST dw 0h,014h,019h,02Dh,041h,046h,05Ah,05Fh,073h,087h,08Ch,0A0h,0B4h
;********** New Interrupt 1 **********;
NEWINT1 proc far
push bp ;Establish stack frame
mov bp,sp
push ax ;Save registers
push bx
push cx
push dx
push di
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push si
push ds
mov ax,cs
mov ds,ax
mov ax,word ptr [bp+2] ;Get offset
cmp ax,1FE7h
jnz EXIT1
NEXT1: mov ax,1FE5h ;Where to return next
mov word ptr [bp+2],ax
mov ax,word ptr [bp+4] ;Get segment
mov ds,ax ;Put in data segment
mov bx,1FE5h ;Offset to change
mov ax,0D803h ;The new code to put in
mov [bx],ax ;Store changes
mov ax,cs ;Get current data segment
mov ds,ax
mov di,0 ;Where to start search
mov dx,0FF00h ;Search the entire segment
mov bx,0
COMP: mov di,bx ;Where to begin
mov si,offset KEYWORD ;Get keyword
mov cx,4 ;Lenght of keyword
repe cmpsb ;Compare until done
jz MATCH
inc bx
dec dx ;Done?
jz EXIT1 ;If no match, exit
jmp COMP
MATCH: mov dx,bx
mov ax,0E07h
int 10h
mov bx,offset LIST ;Get list of codes to change
mov cx,13 ;Number of locations to change
NEXT2: push cx
mov cx,4 ;Lenght of string
mov di,[bx] ;Get destination
add di,dx
mov si,offset CRK ;Get string to copy from
rep movsb ;Copy String
inc bx ;Next location
inc bx
pop cx
loop NEXT2
EXIT1: pop ds ;Restore registers
pop si
pop di
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pop dx
pop cx
pop bx
pop ax
pop bp
iret ;Interrupt return
NEWINT1 endp
;********** New Interrupt 2 **********;
NEWINT2 proc far
push bp ;Establish stack frame
mov bp,sp
push ax ;Save registers
push bx
push ds
mov bx,word ptr [bp+2] ;Get offset
cmp bx,0Ch ;See if called from the proper offset
jnz EXIT2 ;If not, exit
cmp ah,30h ;See if want this function call
jnz EXIT2 ;If not, exit
mov bx,word ptr [bp+4] ;Get segment
add bx,0F8Dh ;New segment
mov ds,bx
mov bx,1FE5h ;New offset
mov ax,0F1CDh ;The new instruction
mov [bx],ax ;Save changes in memory
EXIT2: pop ds ;Restore registers
pop bx
pop ax
mov sp,bp
pop bp
jmp dword ptr cs:OLDINT2 ;Call old interrupt
NEWINT2 endp
FINISH equ $
;********** Initialization Code **********;
PARAM dw 0
db 80h,0
PARAM1 dw 5 dup(0)
PROG db 8 dup('1234567890')
MESS db 'Savage Empire áeta Crack v1.0 July 15,1991',CR,LF
db 'Loader needed only after creating a character.',CR,LF
db "Press {ENTER} at the copy protection.",CR,LF,BEEP,EOLN
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ERR1 db 'ERROR: Not enough memory. '
db 'Activating TSR sequence.',CR,LF,BEEP,EOLN
ERR2 db 'ERROR: Could not load program. '
db 'Activating TSR sequence.',CR,LF,BEEP,EOLN
ERR3 db 'ERROR: Interrupt vector (0xF1) already occupied.',CR,LF
db ' Release memory before restarting.',CR,LF,LF,BEEP,EOLN
INIT: mov ah,9 ;Print string
mov dx,offset MESS
int 21h
mov ax,35F1h ;Get interrupt vector
int 21h
mov OLDINT1[0],bx ;Save in memory
mov OLDINT1[2],es
cmp word ptr es:[bx],8B55h ;Check for vector occupation
jnz CONT1
mov ah,9 ;Write string
mov dx,offset ERR3
int 21h
mov ax,4C03h ;Exit with error 3
int 21h
CONT1: mov ax,25F1h ;Set interrupt vector
mov dx,offset NEWINT1
int 21h
mov ax,3521h ;Get interrupt vector
int 21h
mov OLDINT2[0],bx ;Save in memory
mov OLDINT2[2],es
mov ax,2521h ;Change interrupt vector
mov dx,offset NEWINT2
int 21h
cmp OPTION,0 ;See if wants to run program
jz EXIT3
mov ax,cs
mov ds,ax
mov es,ax
mov bx,offset ENDCODE ;Get end of memory
shr bx,1 ;Convert to paragraphs
shr bx,1
shr bx,1
shr bx,1
inc bx
mov ah,4Ah ;Reallocate memory
int 21h
jnc OKAY1 ;If no error, continue
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mov ah,9h ;Write string
mov dx,offset ERR1
int 21h
jmp EXIT3
OKAY1: mov ax,cs
mov PARAM,ax
mov PARAM1,ax
mov bx,offset PARAM
mov dx,offset PROG
mov ax,4B00h ;Load and execute child
int 21h
jnc OKAY2 ;If no error, continue
mov ah,9h ;Write string
mov dx,offset ERR2
int 21h
jmp EXIT3
OKAY2: mov ax,25F1h ;Restore interrupt vector
lds dx,dword ptr OLDINT1
int 21h
mov ax,2521h ;Restore interrupt vector
lds dx,dword ptr OLDINT2
int 21h
mov ax,4C00h ;Exit with error code 0
int 21h
EXIT3: lea dx,FINISH ;Offset of booster
int 27h ;Exit with ejection of booster
LOADER ends
end BEGIN
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The Cracking Manual
CONCLUSION
Conclusion
----------
Okay, so we've seen the processes of cracking. If you are
just a beginner and don't know much about programming, you
probably got lost somewhere right after the introduction. I would
suggest that you spend some time learning assembly before doing
anything else. Actually, you don't have to start out with
assembly. I started programming using BASIC. When I got really
good at it, I jumped into Assembly, regardless of how difficult
people said it was. Assembly is not at all difficult if you have
had some previous knowledge of another language. It is only
difficult if you make it hard. And after you've learned assembly,
you get a "feel" for the other languages and can learn them in a
matter of days. Pascal, Modula-2, C, C++, ..., they're are based
on assembly language programming.
Cracking is like the debugging process of programming. To
become experienced with debugging is to become adept at cracking.
You just need lots o' practice as practice makes perfect.
One final note. I got this manual out kinda quickly so there
are bound to be errors, inconsistencies in what I've said, unclear
passages, etc. Well, too bad. If you really want a good manual,
tell me or something and I'll consider it. I got really bored
towards the last parts of the manual so it went pretty fast,
skipping over some stuff. If a lot (and I mean A LOT) of people
want a better manual, tell me and give me suggestions. I'll find
the time to do it somehow.
Anyways, have fun!
- The Cyborg
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