This essay consists of a CRC tutorial and a way of how to reverse it. Many
Coders/Reversers don't know exactly how CRC works and almost no one knows how to
reverse it, while this knowledge could be very usefull.

First the tutorial will
learn you how to calculate CRC in general, you can use it as data/code
protection.

Second, the reverse part will learn you (mainly) how to reverse
CRC-32, you can use this to break certain CRC protections in programs or over
programs (like anti-virus).

There seem to be utilities who can 'correct' CRCs
for you, but I doubt they also explain what they're doing.

I'd like to warn you, since there is quite some math used in this essay. This
wont harm anyone, and will be well understood by the avarage Reverser or Coder.

Why? Well. If you dont know why math is used in CRC, I suggest that you click
that button with a X at the top-right of this screen. So I assume the reader has knowledge of binair arithmetic.

**Part 1: CRC Tutorial, what it is and how to calculate it**

**Cyclic Redundancy Code or CRC**

We all know CRC. Even if you don't recall, you will when you think of those
annoying messages RAR, ZIP and other compressors give you when the file is
corrupted due to bad connections or those !@#$% floppies.

The CRC is a value computed over a piece of data, for example for each file at the
time of compression. When the archiver is unpacking that file, it will read the
CRC and check it with the newly computed CRC of the uncompressed file. When
they match, there is a good chance that the files are identical. With CRC-32,
there is a chance of 1/2^32 of the check failing to recognize a change in data.

A lot of people think CRC is short for Cyclic Redundancy Check. If indeed CRC
is short for Cyclic Redundancy Check then a lot of people use the term incorrect.

If it was you could not say 'the CRC of the program is 12345678'. People are also
always saying a certain program has a CRC check, not a Cyclic Redundancy Check
check.

Conclusion: CRC stands for Cyclic Redundancy Code and NOT for Cyclic
Redundancy Check.

How is the calculation done? Well, the main idea is to see the file as one
large string of bits divided by some number, which will leave you with a
remainder, the CRC! You always have a remainder (can also be zero) which is at
most one bit less then the divisor (else it still has a divisor in it).
(9/3=3 remainder=0 ; (9+2)/3=3 remainder=2)
Only here dividing with bits is done a little different.

Dividing is repeatedly
substracting (x times) a number (divisor) from a number you want to divide, which
will leave you with the remainder. If you want the original number back you
multiply with the divisor or (idem) add x times the divisor with itself and
afterwards adding the remainder.
CRC computation uses a special way of substracting and adding, i.e. a
new 'arithmetic'. While computing the carry for each bit calculation is
'forgotten'.

Lets look at 2 examples, number 1 is a normal substraction, 2&3 are special.

-+ (1) 1101 (2) 1010 1010 (3) 0+0=0 0-0=0 1010- 1111+ 1111- 0+1=1 *0-1=1 ---- ---- ---- 1+0=1 1-0=1 0011 0101 0101 *1+1=0 1-1=0In (1), the second column from the right would evaluate to 0-1=-1, therefore a bit is 'borrowed' from the bit next to it, which will give you this substraction (10+0)-1=1. (this is like normal 'by-paper' decimal substraction)

The special case (2&3) 1+1 would normally have as answer 10, where the '1' is the carry which 'transports' the value to the next bit computation.

This value is forgotten. The special case 0-1 would normally have as answer '-1', which would have impact on the bit next to it (see example 1).

This value is also forgotten. If you know something about programming this looks like, or better, it IS the XOR operation.

Now look at an example of a divide:

In normal arithmetic:

1001/11110001101 13 9/12013 1001 - 09 -| ---- -- | 1100 30 | 1001 - 27 - ---- -- 0110 3 -> the remainder 0000 - ---- 1100 1001 - ---- 011 -> 3, the remainderIn CRC arithmetic:

1001/11110001110 9/12014 remainder 6 1001 - ---- 1100 1001 - ---- 1010 1001 - ---- 0110 0000 - ---- 110 -> the remainder (example 3)The quotient of a division is not important, and not efficient to remember, because that would be only a couple of bits less than the bitstring where you wanted to calculate the CRC from. What IS important is the remainder!

That's the thing that says something important over about the original file. That's basicly the CRC!

To perform a CRC calculation we need to choose a divisor, we call it the 'poly' from now on. The width W of a poly is the position of the highest bit, so the width of poly 1001 is 3, and not 4. Note that the highest bit is always one, when you have chosen the width of the poly you only have to choose a value for the lower W bits.

If we want to calculate the CRC over a bitstring, we want to make sure all the bits are processed. Therefore we need to add W zero bits to the end of the bitstring. In the case of example 3, we could say the bitstring was 1111.

Look at a little bigger example:

Poly = 10011, width W=4 Bitstring + W zeros = 110101101 + 0000 10011/1101011010000110000101 (we don't care about the quotient) 10011|||||||| - -----|||||||| 10011||||||| 10011||||||| - -----||||||| 00001|||||| 00000|||||| - -----|||||| 00010||||| 00000||||| - -----||||| 00101|||| 00000|||| - -----|||| 01010||| 00000||| - -----||| 10100|| 10011|| - -----|| 01110| 00000| - -----| 11100 10011 - ----- 1111 -> the remainder -> the CRC! (example 4)There are 2 important things to state here:

1.Only when the highest bit is one in the bitstring we XOR it with the poly, otherwise we only 'shift' the bitstring one bit to the left.

2.The effect of XORring is, that it's XORed with the lower W bits, because the highest bit always gives zero.

You all should understand that an algorithm based on bitwise calculation will be very slow and inefficient. It would be far more efficient if you could calculate it on a per-byte basis.

But then we can only accept poly's with a width of a multiple of 8 bits (that's a byte ;).

Lets visualize it in a example poly with a width of 32 (W=32):

3 2 1 0 byte +---+---+---+---+ Pop! this is the poly, 4*8 bits (figure 1)This is a register you use to store the temporary result of the CRC, I call it the CRC register or just register from now on. You are shifting bits from the bitstring in at the right side, and bits out at the left side.

When the bit just shifted out at the left side is one, the whole register is XORred by the lower W bits of the poly (in this case 32). In fact, we are doing exactly the same thing as the divisions above.

What if (as I said) we would shift in & out a whole group of bits at once.

Look at an example of 8 bit CRC with 4 bits at once shifted in & out:

The register just before the shift : 10110100 Then 4 bits (at the top) are shifted out at the left side while shifting 4 new bits in at the right side. In this example 1011 is shifted out and 1101 (new) is shifted in. Then the situation is this: 8 bits currently CRC/Register : 01001101 4 top bits just shifted out : 1011 We use this poly : 101011100, width W=8 Now we calculate just as usual the new value of the register. Top Register ---- -------- 1011 01001101 the topbits and the register 1010 11100 + (*1) Poly is XORred on position 3 of top bits (coz there is a one) ------------- 0001 10101101 result of XORring Now we still have a one on bit position 0 of topbits: 0001 10101101 previous result 1 01011100+ (*2) Poly is XORred on position 0 of top bits (coz there is a one) ------------- 0000 11110001 result of second XORring ^^^^ Now there are all zero's in the topbits, so we dont have to XOR with the poly anymore for this sequence of topbits. The same value in the register you get if you first XOR (*1) with (*2) and the result with the register. This is because of the standard XOR property: (a XOR b) XOR c = a XOR (b XOR c) 1010 11100 poly on position 3 of top bits 1 01011100+ poly XORred on position 0 of top bits ------------- 1011 10111100 (*3) result of XORring The result (*3) is XORred with the register 1011 10111100 1011 01001101+ the top bits and the register ------------- 0000 11110001You see? The same result! Now (*3) is important, because with the top bits 1010 is always the value (*3)=10111100 (only the lower W=8 bits) bound (under the stated conditions, of course)

This means you can precompute the XOR values for each combination of top bits. Note that top bits always become zero after one iteration, this must be because the combination of XORring leads to it.

Now we come back to figure 1. For each value of the top byte (8 bits) just shifted out, we can precompute a value. In this case it would be a table consisting of 256 (2^8) entries of double words (32bit). (the CRC-32 table is in the appendix)

In pseudo-language our algoritm now is this:

While (byte string is not exhausted) Begin Top = top_byte of register ; Register = Register shifted 8 bits left ORred with a new byte from string ; Register = Register XORred by value from precomputedTable at position Top ; End

The algorithm proposed above can be optimized. The bytes from the byte string don't need to travel through the whole register before they are used.

With this new algorithm we can directly XOR a byte from a byte string with the byte shifted out of the register. The result points to a value in the precomputed table which will be XORred with the register. I don't know exactly why this gives the same result (it has to do with a XOR property), but it has the Big advantage you don't have to append zero bytes/bits to your byte string. (if you know why, pleaz tell me :)

Lets visuallize this algorithm:

+-----: : : : : +---+---+---+---+ | | | | | +---+---+---+---+ (figure 2)

To make things more complicated there is a 'reflected' version of this algorithm. A Reflected value/register is that it's bits are swapped around it's centre. For example 0111011001 is the reflection of 1001101110.

They came up with this because of the UART (chip that performs serial IO), which sends each byte with the least significant bit (bit 0) first and the most significant bit (bit 7) last, this is the reverse of the normal situation.

Instead then of reflecting each byte before processing, every else is reflected. An advantage is that it gives more compact code in the implementation. So, in calculating the table, bits are shifted to the right and the poly is reflected. In calculating the CRC the register is shifted to the right and (of course) the reflected table is used.

byte string (or file) -->---+ | 1. In the table each entry is reflected byte 3 2 1 0 V 2. The initial register is reflected +---+---+---+---+ | 3. The bytes from the byte string aren't | | | | |>---XOR reflected, because all the rest is. +---+---+---+---+ | | | XOR V ^ | +---+---|---+---+ | | | | | | | Precomputed table +---+---+---+---+ | : : : : :

To get everything settled here's the complete CRC-32 standard:

Name : "CRC-32" Width : 32 Poly : 04C11DB7 Initial value : FFFFFFFF Reflected : True XOR out with : FFFFFFFFAs a bonus for you curious people, here's the CRC-16 standard: :)

Name : "CRC-16" Width : 16 Poly : 8005 Initial value : 0000 Reflected : True XOR out with : 0000

'XOR out with' is the value that is XORred with the final value of the register before getting (as answer) the final CRC. There are also 'reversed' CRC poly's but they are not relevant for this tutorial. Look at my references if you want to know more about that.

For the assembly implementation I use 32 bit code in 16 bit mode of DOS... so you will see some mixing of 32 bit and 16 bit code... it is easy to convert it to complete 32 bit code. Note that the assembly part is fully tested to be working correctly, the Java or C code is derived from that.

Ok. Here is the assembly implementation for computing the CRC-32 table:

xor ebx, ebx ;ebx=0, because it will be used whole as pointer InitTableLoop: xor eax, eax ;eax=0 for new entry mov al, bl ;lowest 8 bits of ebx are copied into lowest 8 bits of eax ;generate entry xor cx, cx entryLoop: test eax, 1 jz no_topbit shr eax, 1 xor eax, poly jmp entrygoon no_topbit: shr eax, 1 entrygoon: inc cx test cx, 8 jz entryLoop mov dword ptr[ebx*4 + crctable], eax inc bx test bx, 256 jz InitTableLoopNotes: - crctable is an array of 256 dwords

- eax is shifted to the right because the CRC-32 uses reflected Algorithm

- also therefore the lowest 8 bits are processed...

In Java or C (int is 32 bit):

for (int bx=0; bx>=1; eax^=poly; } else eax>>=1; } crctable[bx]=eax; }

The implementation for computing CRC-32 using the table:

computeLoop: xor ebx, ebx xor al, [si] mov bl, al shr eax, 8 xor eax, dword ptr[4*ebx+crctable] inc si loop computeLoop xor eax, 0FFFFFFFFh

Notes: - ds:si points to the buffer where the bytes to process are

- cx contains the number of bytes to process

- eax contains current CRC

- crctable is the table computed with the code above

- the initial value of the CRC is in the case of CRC-32: FFFFFFFF

- after complete calculation the CRC is XORred with: FFFFFFFF

which is the same as NOTting.

In Java or C it is like this:

for (int cx=0; cx>=8; eax^=crcTable[ebx]; } eax^=0xFFFFFFFF;So now we landed at the end of the first part: The CRC tutorial If you want to make a little deeper dive in CRC I suggest reading the document I did, you will find the URL at the end of this document. Ok. On to the most interresting part of this document: Reversing CRC!

When I was thinking of a way to reverse it... I got stuck several times. I tried to 'deactivate' the CRC by thinking of such an sequence of bytes that it then shouldn't matter anymore what bytes you would place behind it. I couldn't do it... Then I realized it could NEVER work that way, because CRC algorithm is build in such a way it wouldn't matter which _bit_ you would change, the complete CRC _always_ (well always... almost) changes drasticly. Try that yourself (with some simple CRC programs)... :) I realized I only could 'correct' the CRC _after_ the bytes I wanted to change. So I could make such a sequence of bytes, that would 'transform' the CRC into whatever I wanted!

Lets visualize the idea:

Bunch of bytes: 01234567890123456789012345678901234567890123456789012 You want to change from ^ this byte to ^ this one. Thats position 9 to 26. We also need 4 extra bytes (until position 30 ^) for the sequence of bytes which will change the CRC back to its original value after the patched bytes.When you are calculating the CRC-32 it goes fine until the byte on position 9, in the patched bunch of bytes the CRC radically changes from that point on.

Even when pass position 26, from where the bytes are not changed, you never get the original CRC back. NOT! When you read the rest of this essay you know how.

In short you have do this when patching a certain bunch of bytes while maintainting the CRC:

1. Calculate the CRC until position 9, and save this value.

2. Continue calculating until position 27 and 4 extra bytes, save the resulting value.

3. Use the value of 1 for calculating the CRC of the 'new' bytes and the extra 4 bytes (this should be 27-9+4=22 bytes) and save the resulting value.

4. Now we have the 'new' CRC value, but we want the CRC to be the 'old' CRC value. We use the reverse algorithm to compute the 4 extra bytes.

We can to point 1 to 3, below you learn to do point 4.

I thought, to make it more easy for you, first to calculate the reverse of CRC-16. Ok. We are on a certain point after the patched code where you want to change the CRC back to its original. We know the original CRC (calculated before patching the data) and the current CRC register. We want to calculate the 2-bytestring which changes the current CRC register to the original CRC. First we calculate 'normally' the CRC with the unknown 2 bytes naming them X and Y, for the register I take a1 a0 , the only non-variable is zero (00). :) Look again at our latest CRC algorithm, figure 3, to understand better what im doing.

Ok, here we go:

Take a 2-bytestring 'X Y'. Bytes are processed from the left side. Take for register a1 a0. For a XOR operation I write '+' (as in the CRC tutorial) Processing first byte, X: a0+X this is the calculated topbyte (1) b1 b0 sequence in table where the topbyte points at 00 a1 to right shifted register 00+b1 a1+b0 previous 2 lines XORred with eachother Now the new register is: (b1) (a1+b0) Processing second byte, Y: (a1+b0)+Y this is the calculated topbyte (2) c1 c0 sequence in table where the topbyte points at 00 b1 to right shifted register 00+c1 b1+c0 previous 2 lines XORred with eachother Now the final register is: (c1) (b1+c0) I'll show it a little different way: a0 + X =(1) points to b1 b0 in table a1 + b0 + Y =(2) points to c1 c0 in table b1 + c0=d0 new low byte of register c1=d1 new high byte of register (1) (2)

Wow! Let this info work out on you for a while... :) Don't be afraid, a real value example is coming soon. What if you wanted the register to be some d1 d0 (the original CRC) and you know the value of the register before the transformation (so a1 a0)... what 2 bytes or what X and Y would you have to fed through the CRC calculation?

Ok. We will begin working from the back to the front. d0 must be b1+c0 and d1 must be c1... But how-the-hell, I hear you say, can you know the value of byte b1 and c0??? ShallI remember you about the Table? You can just lookup the value of the word C0 C1 in the Table because you know C1. Therefore you need to make a 'lookup' routine. If you found the value, be sure to remember the index to the value because that's the way to find the unknown topbytes e.g. (1)&(2)!

So now you found c1 c0, how to get b1 b0? If b1+c0=d0 then b1=d0+c0! Now you use the lookup routine to lookup the b1 b0 value too. Now we know everything to calculate X & Y ! Cool huh? a1+b0+Y=(2) so Y=a1+b0+(2) a0+X=(1) so X=a0+(1)

Lets look at an example with real values:

-register before: (a1=)DE (a0=)AD

-wanted register: (d1=)12 (d0=)34

Look up the entry beginning with 12 in the CRC-16 table in the appendix.

-This is entry 38h with value 12C0. Try to find another entry beginning with 12.

You can't find another because we calculated each entry for each possible value of the topbyte and that's 256 values, remember!

Now we know (2)= 38, c1= 12 and c0= C0, so b1= C0+34=F4, now look up the entry of B1 beginning with F4.

-This is entry 4Fh with value F441.

Now we know (1)= 4F, b1= F4 and b0= 41. Now all needed values are known, to compute X and Y we do:

Y=a1+b0+(2)=DE+41+38=A7

X=a0+(1) =AD+4F =E2

Conclusion: to change the CRC-16 register from DEAD to 1234 we need the bytes E2 A7 (in that order).

You see, to reverse CRC you have to 'calculate' your way back, and remember the values along the way. When you are programming the lookup table in assembly, remember that intel saves values backwards in Little-Endian format.

Now you probably understand how to reverse CRC-16.... now CRC-32

Now we had CRC-16, CRC-32 is just as easy (or as difficult). You now work with 4 bytes instead of 2. Keep looking and comparing this with the 16bit version from above.

Take a 4-bytestring X Y Z W , bytes are taken from the LEFT side Take for register a3 a2 a1 a0 Note that a3 is the most significant byte and a0 the least. Processing first byte, X: a0+X this is the calculated topbyte (1) b3 b2 b1 b0 sequence in table where the topbyte points at 00 a3 a2 a1 to right shifted register 00+b3 a3+b2 a2+b1 a1+b0 previous 2 lines XORred with eachother Now the new register is: (b3) (a3+b2) (a2+b1) (a1+b0) Processing second byte, Y: (a1+b0)+Y this is the calculated topbyte (2) c3 c2 c1 c0 sequence in table where the topbyte points at 00 b3 a3+b2 a2+b1 to right shifted register 00+c3 b3+c2 a3+b2+c1 a2+b1+c0 previous 2 lines XORred with eachother Now the new register is: (c3) (b3+c2) (a3+b2+c1) (a2+b1+c0) Processing third byte, Z: (a2+b1+c0)+Z this is the calculated topbyte (3) d3 d2 d1 d0 sequence in table where the topbyte points at 00 c3 b3+c2 a3+b2+c1 to right shifted register 00+d3 c3+d2 b3+c2+d1 a3+b2+c1+d0 previous 2 lines XORred with eachother Now the new register is: (d3) (c3+d2) (b3+c2+d1) (a3+b2+c1+d0) Processing fourth byte, W: (a3+b2+c1+d0)+W this is the calculated topbyte (4) e3 e2 e1 e0 sequence in table where the topbyte points at 00 d3 c3+d2 b3+c2+d1 to right shifted register 00+e3 d3+e2 c3+d2+e1 b3+c2+d1+e0 previous 2 lines XORred with eachother Now the final register is: (e3) (d3+e2) (c3+d2+e1) (b3+c2+d1+e0) I'll show it a little different way: a0 + X =(1) points to b3 b2 b1 b0 in table a1 + b0 + Y =(2) points to c3 c2 c1 c0 in table a2 + b1 + c0 + Z =(3) points to d3 d2 d1 d0 in table a3 + b2 + c1 + d0 + W =(4) points to e4 e3 e2 e1 in table b3 + c2 + d1 + e0 =f0 c3 + d2 + e1 =f1 d3 + e2 =f2 e3 =f3 (1) (2) (3) (4) (figure 4)This is reversed in the same way as the 16bit version. I shall give an example with real values. For the table values use the CRC-32 table in the appendix.

Take for CRC register before, a3 a2 a1 a0 -> AB CD EF 66

Take for CRC register after, f3 f2 f1 f0 -> 56 33 14 78 (wanted value)

Here we go:

First byte of entries entry value e3=f3 =56 -> 35h=(4) 56B3C423 for e3 e2 e1 e0 d3=f2+e2 =33+B3 =E6 -> 4Fh=(3) E6635C01 for d3 d2 d1 d0 c3=f1+e1+d2 =14+C4+63 =B3 -> F8h=(2) B3667A2E for c3 c2 c1 c0 b3=f0+e0+d1+c2=78+23+5C+66=61 -> DEh=(1) 616BFFD3 for b3 b2 b1 b0 Now we have all needed values, then X=(1)+ a0= DE+66=B8 Y=(2)+ b0+a1= F8+D3+EF=C4 Z=(3)+ c0+b1+a2= 4F+2E+FF+CD=53 W=(4)+d0+c1+b2+a3=35+01+7A+6B+AB=8E (final computation)Conclusion: to change the CRC-32 register from ABCDEF66 to 56331478 we need this sequence of bytes: B8 C4 53 8E

If you look at the by-hand computation of the sequence of bytes needed to change the CRC register from a3 a2 a1 a0 to f3 f2 f1 f0 its difficult to transform this into a nice compact algorithm.

Look at an extended version of the final computation:

Position X =(1) + a0 0 Y =(2) + b0 + a1 1 Z =(3) + c0 + b1 + a2 2 W =(4) + d0 + c1 + b2 + a3 3 f0= e0 + d1 + c2 + b3 4 f1= e1 + d2 + c3 5 f2= e2 + d3 6 f3= e3 7 (figure 5)It is just the same as figure 4, only some values/bytes exchanged. This view will help us to get a compact algorithm. What if we take a buffer of 8 bytes that is, for every line you see in figure 5 one byte is reserved. Bytes 0 to 3 are filled with a0 to a3, bytes 4 to 7 are filled with f0 to f3. As before, we take the last byte e3 which is equal to f3 and lookup the complete value in the CRC table. Then we XOR this value (e3 e2 e1 e0) on position 4 (as in figure 5). Then we automatically know what the value of d3 is, because we already XORred f3 f2 f1 f0 with e3 e2 e1 e0, and f2+e2=d3. Because we now already know what the value of (4) is (the entry number), we can directly XOR the value into position 3. Now we know d3 use this to lookup the value of d3 d2 d1 d0 and XOR this on one position earlier, that is position 3 (look at the figure!). XOR the found entry number (3) for the value on position 2. We now know c3 because we have the value f1+e1+d2=c3 on position 5. We go on doing this until we XORred b3 b2 b1 b0 on position 1. Et voila! Bytes 0 to 3 of the buffer now contains the needed bytes X to W!

Summarized is here the algorithm:

1. Of the 8 byte buffer, fill position 0 to 3 with a0 to a3 (the start value of the CRC register), and position 4 to 7 with f0 to f3 (wanted end value of CRC register).

2. Take the byte from position 7 and use it to lookup the complete value.

3. XOR this value (dword) on position 4

4. XOR the entry number (byte) on position 3

5. Repeat step 2 & 3 three more times while decreasing the positions each time by one.

Now its time for some code. Below are the implementation of the reverse algorithm for CRC-32 in Assembly (it is not difficult to do this for other languages and/or CRC standards). Note that in assembly (on PC's) dwords are written to and read from memory in reverse order.

crcBefore dd (?) wantedCrc dd (?) buffer db 8 dup (?) mov eax, dword ptr[crcBefore] ;/* mov dword ptr[buffer], eax mov eax, dword ptr[wantedCrc] ; Step 1 mov dword ptr[buffer+4], eax ;*/ mov di, 4 computeReverseLoop: mov al, byte ptr[buffer+di+3] ;/* call GetTableEntry ; Step 2 */ xor dword ptr[buffer+di], eax ; Step 3 xor byte ptr[buffer+di-1], bl ; Step 4 dec di ;/* jnz computeReverseLoop ; Step 5 */ Notes: -Registers eax, di bx are used Implementation of GetTableEntry crctable dd 256 dup (?) ;should be defined globally somewhere & initialized of course mov bx, offset crctable-1 getTableEntryLoop: add bx, 4 ;points to (crctable-1)+k*4 (k:1..256) cmp [bx], al ;must always find the value somewhere jne getTableEntryLoop sub bx, 3 mov eax, [bx] sub bx, offset crctable shr bx, 2 retOn return eax contains a table entry, bx contains the entry number.

Well... your reached the end of this essay. If you now think: wow, all those programs which are protected by CRC can say 'bye, bye'. Nope. It is very easy to make an anti-anti-CRC code. To make a succesfull CRCreverse you have to know exactly from what part of the code the CRC is calculated and what CRC algorithm is used. A simple countermeasure is using 2 different CRC algorithms, or combination with another dataprotection algorithm. Anywayz... I hope all this stuff was interesting and that you enjoyed reading it as I enjoyed writing it.

Fnx go out to the beta-testers Douby/DREAD and Knotty Dread for the good comments on my work which made it even better!

For a sample CRC-32 correcting patcher program visit my webpages: http://surf.to/anarchriz -> Programming -> Projects (it's still a preview but will give you a proof of my idea)

For more info on DREAD visit http://dread99.cjb.net

If you still have questions you can mail me at anarchriz@hotmail.com, or try the channels #DreaD, #Win32asm, #C.I.A and #Cracking4Newbies (in that order) on EFnet (on IRC).

CYA ALL! - Anarchriz

"The system makes its morons, then despises them for their ineptitude, and rewards its 'gifted few' for their rarity." - Colin Ward

Appendix

00h 0000 C0C1 C181 0140 C301 03C0 0280 C241 08h C601 06C0 0780 C741 0500 C5C1 C481 0440 10h CC01 0CC0 0D80 CD41 0F00 CFC1 CE81 0E40 18h 0A00 CAC1 CB81 0B40 C901 09C0 0880 C841 20h D801 18C0 1980 D941 1B00 DBC1 DA81 1A40 28h 1E00 DEC1 DF81 1F40 DD01 1DC0 1C80 DC41 30h 1400 D4C1 D581 1540 D701 17C0 1680 D641 38h D201 12C0 1380 D341 1100 D1C1 D081 1040 40h F001 30C0 3180 F141 3300 F3C1 F281 3240 48h 3600 F6C1 F781 3740 F501 35C0 3480 F441 50h 3C00 FCC1 FD81 3D40 FF01 3FC0 3E80 FE41 58h FA01 3AC0 3B80 FB41 3900 F9C1 F881 3840 60h 2800 E8C1 E981 2940 EB01 2BC0 2A80 EA41 68h EE01 2EC0 2F80 EF41 2D00 EDC1 EC81 2C40 70h E401 24C0 2580 E541 2700 E7C1 E681 2640 78h 2200 E2C1 E381 2340 E101 21C0 2080 E041 80h A001 60C0 6180 A141 6300 A3C1 A281 6240 88h 6600 A6C1 A781 6740 A501 65C0 6480 A441 90h 6C00 ACC1 AD81 6D40 AF01 6FC0 6E80 AE41 98h AA01 6AC0 6B80 AB41 6900 A9C1 A881 6840 A0h 7800 B8C1 B981 7940 BB01 7BC0 7A80 BA41 A8h BE01 7EC0 7F80 BF41 7D00 BDC1 BC81 7C40 B0h B401 74C0 7580 B541 7700 B7C1 B681 7640 B8h 7200 B2C1 B381 7340 B101 71C0 7080 B041 C0h 5000 90C1 9181 5140 9301 53C0 5280 9241 C8h 9601 56C0 5780 9741 5500 95C1 9481 5440 D0h 9C01 5CC0 5D80 9D41 5F00 9FC1 9E81 5E40 D8h 5A00 9AC1 9B81 5B40 9901 59C0 5880 9841 E0h 8801 48C0 4980 8941 4B00 8BC1 8A81 4A40 E8h 4E00 8EC1 8F81 4F40 8D01 4DC0 4C80 8C41 F0h 4400 84C1 8581 4540 8701 47C0 4680 8641 F8h 8201 42C0 4380 8341 4100 81C1 8081 4040CRC-32 Table00h 00000000 77073096 EE0E612C 990951BA 04h 076DC419 706AF48F E963A535 9E6495A3 08h 0EDB8832 79DCB8A4 E0D5E91E 97D2D988 0Ch 09B64C2B 7EB17CBD E7B82D07 90BF1D91 10h 1DB71064 6AB020F2 F3B97148 84BE41DE 14h 1ADAD47D 6DDDE4EB F4D4B551 83D385C7 18h 136C9856 646BA8C0 FD62F97A 8A65C9EC 1Ch 14015C4F 63066CD9 FA0F3D63 8D080DF5 20h 3B6E20C8 4C69105E D56041E4 A2677172 24h 3C03E4D1 4B04D447 D20D85FD A50AB56B 28h 35B5A8FA 42B2986C DBBBC9D6 ACBCF940 2Ch 32D86CE3 45DF5C75 DCD60DCF ABD13D59 30h 26D930AC 51DE003A C8D75180 BFD06116 34h 21B4F4B5 56B3C423 CFBA9599 B8BDA50F 38h 2802B89E 5F058808 C60CD9B2 B10BE924 3Ch 2F6F7C87 58684C11 C1611DAB B6662D3D 40h 76DC4190 01DB7106 98D220BC EFD5102A 44h 71B18589 06B6B51F 9FBFE4A5 E8B8D433 48h 7807C9A2 0F00F934 9609A88E E10E9818 4Ch 7F6A0DBB 086D3D2D 91646C97 E6635C01 50h 6B6B51F4 1C6C6162 856530D8 F262004E 54h 6C0695ED 1B01A57B 8208F4C1 F50FC457 58h 65B0D9C6 12B7E950 8BBEB8EA FCB9887C 5Ch 62DD1DDF 15DA2D49 8CD37CF3 FBD44C65 60h 4DB26158 3AB551CE A3BC0074 D4BB30E2 64h 4ADFA541 3DD895D7 A4D1C46D D3D6F4FB 68h 4369E96A 346ED9FC AD678846 DA60B8D0 6Ch 44042D73 33031DE5 AA0A4C5F DD0D7CC9 70h 5005713C 270241AA BE0B1010 C90C2086 74h 5768B525 206F85B3 B966D409 CE61E49F 78h 5EDEF90E 29D9C998 B0D09822 C7D7A8B4 7Ch 59B33D17 2EB40D81 B7BD5C3B C0BA6CAD 80h EDB88320 9ABFB3B6 03B6E20C 74B1D29A 84h EAD54739 9DD277AF 04DB2615 73DC1683 88h E3630B12 94643B84 0D6D6A3E 7A6A5AA8 8Ch E40ECF0B 9309FF9D 0A00AE27 7D079EB1 90h F00F9344 8708A3D2 1E01F268 6906C2FE 94h F762575D 806567CB 196C3671 6E6B06E7 98h FED41B76 89D32BE0 10DA7A5A 67DD4ACC 9Ch F9B9DF6F 8EBEEFF9 17B7BE43 60B08ED5 A0h D6D6A3E8 A1D1937E 38D8C2C4 4FDFF252 A4h D1BB67F1 A6BC5767 3FB506DD 48B2364B A8h D80D2BDA AF0A1B4C 36034AF6 41047A60 ACh DF60EFC3 A867DF55 316E8EEF 4669BE79 B0h CB61B38C BC66831A 256FD2A0 5268E236 B4h CC0C7795 BB0B4703 220216B9 5505262F B8h C5BA3BBE B2BD0B28 2BB45A92 5CB36A04 BCh C2D7FFA7 B5D0CF31 2CD99E8B 5BDEAE1D C0h 9B64C2B0 EC63F226 756AA39C 026D930A C4h 9C0906A9 EB0E363F 72076785 05005713 C8h 95BF4A82 E2B87A14 7BB12BAE 0CB61B38 CCh 92D28E9B E5D5BE0D 7CDCEFB7 0BDBDF21 D0h 86D3D2D4 F1D4E242 68DDB3F8 1FDA836E D4h 81BE16CD F6B9265B 6FB077E1 18B74777 D8h 88085AE6 FF0F6A70 66063BCA 11010B5C DCh 8F659EFF F862AE69 616BFFD3 166CCF45 E0h A00AE278 D70DD2EE 4E048354 3903B3C2 E4h A7672661 D06016F7 4969474D 3E6E77DB E8h AED16A4A D9D65ADC 40DF0B66 37D83BF0 ECh A9BCAE53 DEBB9EC5 47B2CF7F 30B5FFE9 F0h BDBDF21C CABAC28A 53B39330 24B4A3A6 F4h BAD03605 CDD70693 54DE5729 23D967BF F8h B3667A2E C4614AB8 5D681B02 2A6F2B94 FCh B40BBE37 C30C8EA1 5A05DF1B 2D02EF8D

> A painless guide to CRC error detection algorithm url: ftp://ftp.adelaide.edu.au/pub/rocksoft/crc_v3.txt (I bet this 'painless guide' is more painfull then my 'short' one ;)

> I also used a random source of a CRC-32 algorithm to understand the algorithm better.

> Link to crc calculation progs... hmmm search for 'CRC.ZIP' or 'CRC.EXE' or something alike at ftpsearch (http://ftpsearch.lycos.com?form=advanced)

Copyright (c) 1998,1999 by Anarchriz

(this is REALLY the last line :)