Expert Mod 8TB | INTRODUCTION
The Secure Hash Algorithm 2 is a series of cryptographic hash algorithms designed by the US National Security Agency (NSA) and published by the National Institute of Standards and Technology (NIST) as a government standard.
There is currently a competition being held by NIST to find a new family of algorithms for what will be named SHA-3. These new functions may not necessarily be derived from the SHA-2 algorithms.
One of the most widely used hash algorithms is the MD5 algorithm but substantial weaknesses have been found with it and it has been strongly recommended that MD5 be discontinued. WHAT IS SHA2
SHA2 is a hash algorithm. A hash algorithm is basically a one-way encryption that outputs a fixed length. You can encrypt a text but you can not decrypt it. It is often used for password storage and message authentication.
Instead of storing a plaintext password, what you do is precalculate the hash of the password. When they enter in the password, you hash their input and compare it to the hash that you have stored. This way, even if they know what the result of the hash is, they do not know the original password that created the hash. It would be infeasible for them to calculate a password that would result in the same hash.
On a side note, this is how Linux based utilities reset Windows passwords. They overwrite the hash value that is stored with the user account. If security is an issue, what you should do is encrypt all the files of that user using the password. That way, even if they overwrite the hash and log in as that user, they can not view the files because they don't have the password to decrypt the data.
Hashes are often used in internet communication to authenticate messages. The sender will hash the original data and send it along with the encrypted data. The receiver will then decrypt the data and recalculate the hash from the unencrypted data. If the hashes match, then he is assured that the data was received as intended. If you were to send just the encrypted data, someone could conceivably intercept the message, flip a bit in the message, and pass it along. The receiver would never know the message was changed. GENERAL WEAKNESSES OF HASH ALGORITHMS
Because a hash algorithm is intended to return a fixed length regardless of the size of the input, there will invariably be collisions. Collisions are when two different texts produce the same hash. The longer the hash, the less likely the chance of a collision.
When using hashes to store passwords, it does not prevent brute force cracking of a password. Also, since the same text hashed using an algorithm always produces the same hash, it is strongly recommended that a nonce is used to defeat rainbow table attacks. A nonce, initialization vector, or salt are basically random bits that are used with the key so that even though you are using the same key, each message is different because the random bits in effect change the key that is used. A rainbow table is a precomputed table of hashes of popular passwords. All the hacker would have to do is compare the precomputed hashes with the stored hashes to see if there's a match. A nonce defeats this by changing the actual text that is being hashed. HOW SHA2 WORKS
SHA2 breaks messages into 64 byte chunks, does mathematical transformations on each chunk, and adds them to a hash value. It does this for every chunk and adds them to the same value. In the end, you get a fixed length output. SPECIFIC WEAKNESSES OF SHA2
There are currently no known successful attacks that will break all rounds of the SHA2 256-bit algorithm. SAMPLE IMPLEMENTATION
This is an implementation of the SHA2 256-bit algorithm. It even works in a Visual Basic Script and was, in fact, coded specifically for VBScript. But it should be directly portable to VBA. It takes a string, hashes it, and returns the result as a 32 item array containing the hashed value. I validated the output against official SHA2 hashes.
For this to work, it has to use 32-bit unsigned integers but seeing as how that is not available as a datatype in VBA, we have to use doubles instead. However, the MOD, XOR, AND, and NOT operators will overflow when using doubles. In addition to that, there is no bit shift operators or functions in VBA. So I had to create those. - Function SHA(ByVal sMessage)
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Dim i, result(32), temp(8) As Double, fraccubeprimes, hashValues
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Dim done512, index512, words(64) As Double, index32, mask(4)
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Dim s0, s1, t1, t2, maj, ch, strLen
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mask(0) = 4294967296#
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mask(1) = 16777216
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mask(2) = 65536
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mask(3) = 256
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hashValues = Array( _
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1779033703, 3144134277#, 1013904242, 2773480762#, _
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1359893119, 2600822924#, 528734635, 1541459225)
-
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fraccubeprimes = Array( _
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1116352408, 1899447441, 3049323471#, 3921009573#, 961987163, 1508970993, 2453635748#, 2870763221#, _
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3624381080#, 310598401, 607225278, 1426881987, 1925078388, 2162078206#, 2614888103#, 3248222580#, _
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3835390401#, 4022224774#, 264347078, 604807628, 770255983, 1249150122, 1555081692, 1996064986, _
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2554220882#, 2821834349#, 2952996808#, 3210313671#, 3336571891#, 3584528711#, 113926993, 338241895, _
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666307205, 773529912, 1294757372, 1396182291, 1695183700, 1986661051, 2177026350#, 2456956037#, _
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2730485921#, 2820302411#, 3259730800#, 3345764771#, 3516065817#, 3600352804#, 4094571909#, 275423344, _
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430227734, 506948616, 659060556, 883997877, 958139571, 1322822218, 1537002063, 1747873779, _
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1955562222, 2024104815, 2227730452#, 2361852424#, 2428436474#, 2756734187#, 3204031479#, 3329325298#)
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sMessage = Nz(sMessage, "")
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strLen = Len(sMessage) * 8
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sMessage = sMessage & Chr(128)
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done512 = False
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index512 = 0
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If (Len(sMessage) Mod 64) < 56 Then
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sMessage = sMessage & String(56 - (Len(sMessage) Mod 64), Chr(0))
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ElseIf (Len(sMessage) Mod 64) > 56 Then
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sMessage = sMessage & String(120 - (Len(sMessage) Mod 64), Chr(0))
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End If
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sMessage = sMessage & Chr(0) & Chr(0) & Chr(0) & Chr(0)
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sMessage = sMessage & Chr(Int((strLen / mask(0) - Int(strLen / mask(0))) * 256))
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sMessage = sMessage & Chr(Int((strLen / mask(1) - Int(strLen / mask(1))) * 256))
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sMessage = sMessage & Chr(Int((strLen / mask(2) - Int(strLen / mask(2))) * 256))
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sMessage = sMessage & Chr(Int((strLen / mask(3) - Int(strLen / mask(3))) * 256))
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Do Until done512
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For i = 0 To 15
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words(i) = Asc(Mid(sMessage, index512 * 64 + i * 4 + 1, 1)) * mask(1) + Asc(Mid(sMessage, index512 * 64 + i * 4 + 2, 1)) * mask(2) + Asc(Mid(sMessage, index512 * 64 + i * 4 + 3, 1)) * mask(3) + Asc(Mid(sMessage, index512 * 64 + i * 4 + 4, 1))
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Next
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For i = 16 To 63
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s0 = largeXor(largeXor(rightRotate(words(i - 15), 7, 32), rightRotate(words(i - 15), 18, 32), 32), Int(words(i - 15) / 8), 32)
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s1 = largeXor(largeXor(rightRotate(words(i - 2), 17, 32), rightRotate(words(i - 2), 19, 32), 32), Int(words(i - 2) / 1024), 32)
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words(i) = Mod32Bit(words(i - 16) + s0 + words(i - 7) + s1)
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Next
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For i = 0 To 7
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temp(i) = hashValues(i)
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Next
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For i = 0 To 63
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s0 = largeXor(largeXor(rightRotate(temp(0), 2, 32), rightRotate(temp(0), 13, 32), 32), rightRotate(temp(0), 22, 32), 32)
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maj = largeXor(largeXor(largeAnd(temp(0), temp(1), 32), largeAnd(temp(0), temp(2), 32), 32), largeAnd(temp(1), temp(2), 32), 32)
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t2 = Mod32Bit(s0 + maj)
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s1 = largeXor(largeXor(rightRotate(temp(4), 6, 32), rightRotate(temp(4), 11, 32), 32), rightRotate(temp(4), 25, 32), 32)
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ch = largeXor(largeAnd(temp(4), temp(5), 32), largeAnd(largeNot(temp(4), 32), temp(6), 32), 32)
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t1 = Mod32Bit(temp(7) + s1 + ch + fraccubeprimes(i) + words(i))
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temp(7) = temp(6)
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temp(6) = temp(5)
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temp(5) = temp(4)
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temp(4) = Mod32Bit(temp(3) + t1)
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temp(3) = temp(2)
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temp(2) = temp(1)
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temp(1) = temp(0)
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temp(0) = Mod32Bit(t1 + t2)
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Next
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For i = 0 To 7
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hashValues(i) = Mod32Bit(hashValues(i) + temp(i))
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Next
-
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If (index512 + 1) * 64 >= Len(sMessage) Then done512 = True
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index512 = index512 + 1
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Loop
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For i = 0 To 31
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result(i) = Int((hashValues(i \ 4) / mask(i Mod 4) - Int(hashValues(i \ 4) / mask(i Mod 4))) * 256)
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Next
-
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SHA = result
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End Function
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Function Mod32Bit(value)
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Mod32Bit = Int((value / 4294967296# - Int(value / 4294967296#)) * 4294967296#)
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End Function
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Function rightRotate(value, amount, totalBits)
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'To leftRotate, make amount = totalBits - amount
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Dim i
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rightRotate = 0
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For i = 0 To (totalBits - 1)
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If i >= amount Then
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rightRotate = rightRotate + (Int((value / (2 ^ (i + 1)) - Int(value / (2 ^ (i + 1)))) * 2)) * 2 ^ (i - amount)
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Else
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rightRotate = rightRotate + (Int((value / (2 ^ (i + 1)) - Int(value / (2 ^ (i + 1)))) * 2)) * 2 ^ (totalBits - amount + i)
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End If
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Next
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End Function
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Function largeXor(value, xorValue, totalBits)
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Dim i, a, b
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largeXor = 0
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For i = 0 To (totalBits - 1)
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a = (Int((value / (2 ^ (i + 1)) - Int(value / (2 ^ (i + 1)))) * 2))
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b = (Int((xorValue / (2 ^ (i + 1)) - Int(xorValue / (2 ^ (i + 1)))) * 2))
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If a <> b Then
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largeXor = largeXor + 2 ^ i
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End If
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Next
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End Function
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Function largeNot(value, totalBits)
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Dim i, a
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largeNot = 0
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For i = 0 To (totalBits - 1)
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a = Int((value / (2 ^ (i + 1)) - Int(value / (2 ^ (i + 1)))) * 2)
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If a = 0 Then
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largeNot = largeNot + 2 ^ i
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End If
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Next
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End Function
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Function largeAnd(value, andValue, totalBits)
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Dim i, a, b
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largeAnd = 0
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For i = 0 To (totalBits - 1)
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a = Int((value / (2 ^ (i + 1)) - Int(value / (2 ^ (i + 1)))) * 2)
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b = (Int((andValue / (2 ^ (i + 1)) - Int(andValue / (2 ^ (i + 1)))) * 2))
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If a = 1 And b = 1 Then
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largeAnd = largeAnd + 2 ^ i
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End If
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Next
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End Function
| | Share: |
SHA2 Cryptographic Hash Algorithm for VBA and VBScript
Here's a bit that will save some CPU cycles and such - 31: If (Len(sMessage) Mod 60) <> 0 Then
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32: For i = (Len(sMessage) Mod 60) To 59
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33: sMessage = sMessage & Chr(0)
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34: Next
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35: End If
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replace with - 31: sMessage = sMessage & String(60 - Len(sMessage) Mod 60, Chr(0))
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confirmed VBA 14 (Office 2010)
| | |
Hmmm...not sure what I am missing, but I get the same hash result for the following:
SHA(string(64,"X"))
SHA(string(65,"X"))
SHA(string(66,"X"))
and so on
| | Expert Mod 8TB |
I will fix the code for messages longer than 64 characters and repost.
YEA!
I was just trying to figure that out once I confirmed Gaultz's discovery
-z
| | Expert Mod 8TB |
The code has been fixed. I also noticed a bug for messages between 61 and 64 characters that has also been fixed. Thanks for pointing this out for me.
I can make no guarantee on my fix though as I have no test values to use that are 61+ characters long.
| | Expert Mod 4TB |
I caught the changes on lines 31 thru 35; however, I don't see any other changes.
When I put these into the VBA version, the same issue with truncation past 64 is occuring.
| | Expert Mod 8TB |
I'm seeing different results, can you post your input and output?
| | Expert Mod 8TB |
@z, that wasn't the only change. There's a small and vital change on line 44 that's the crux of the matter.
| | Expert Mod 8TB |
Lines 31-34 was to fix the issue for lengths greater than 60. Line 44 is the fix for lengths greater than 64.
| | |
Great work i really like it
| | Expert Mod 8TB |
The return value is an array of numbers. You'll want to first put it in a string format so you can store it. Which format you choose is up to you. If it doesn't need to be human readable, ASCII will work and use the least amount of space. Otherwise, hexadecimal and base64 are popular formats.
After that, you can call it the same way you would any other function.
| | |
"You'll want to first put it in a string format so you can store it." - You mean convert all array fields into strings?
| | Expert Mod 8TB |
I mean append each array element into one long string in the format of your choice.
| | |
I've noticed that hashes for 56-to-59-character-long messages don't match those calculates by other programs (e.g. http://www.xorbin.com/tools/sha256-hash-calculator). I believe this can be fixed if lines 31-36 are changed to the following: - 31: If (Len(sMessage) Mod 64) < 56 Then
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32: sMessage = sMessage & String(56 - (Len(sMessage) Mod 64), Chr(0))
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33: ElseIf (Len(sMessage) Mod 64) > 56 Then
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34: sMessage = sMessage & String(120 - (Len(sMessage) Mod 64), Chr(0))
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35: End If
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36: sMessage = sMessage & Chr(0) & Chr(0) & Chr(0) & Chr(0)
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| | Expert Mod 8TB |
Thanks for pointing out the bug, I'll take a closer look when I get into the office tomorrow
| | Expert Mod 8TB |
Sorry, forgot I'm in training most of this week, will have to get to this later.
| | Expert Mod 8TB |
Thanks for finding the bug. I have fixed my code above. I guess at the time I thought the length of the message was 32 bits instead of 64 bits.
| | Expert Mod 4TB |
I am a tad confused here...
+ As noted in the opening paragraph, SHA2 refers to a class of algorithms: SHA-224, SHA-256, SHA-384, SHA-512, SHA-512/224, SHA-512/256 (as much as I hate to reference anything Wiki: SHA-2 Secure Hash Algorithm 2 it is nonetheless a good cliff from which to jump).
++ Which implementation are we using in this thread? I thought that Rabbit's version was the SHA256 implementation as stated in section 5 of the OP.
It was my understanding that SHA-256 uses 32 bit words; whereas, SHA-512 uses 64 bit words.
( Verification of a Cryptographic Primitive: SHA-256 - )Section 3 paragraph 4 (PDF - CS.Princton.EDU))
>>> caveat emptor
>>> I am an Analytical Chemist with a more than average background in mathematics and statistics - not a cryptologist :); thus, I wont even pretend to understand everything here... what is it that I am missing - I don't have the original code handy to compare with the revised code. :-(
| | Expert Mod 8TB |
The size of the word refers to the size of the data that is operated upon. The 256 bit version of SHA2 XORs, MODs, and SHIFTs operands of 32 bits.
The piece that was wrong in the code was the piece that defined the total length of the message being hashed. Part of the hash algorithm is to append the length of the message to the end of the message before hashing.
| | Expert Mod 4TB |
Thank you Rabbit,
Re-read the reference material this evening and between this and that think I understand. (@_@)
| | |
Excellent code. I have used this in Excel, but in order to avoid having to reference the Microsoft Access model (not all users of the spreadsheet will have Access), I changed the following line from: - 25: sMessage = Nz(sMessage, "")
to: - 25: If IsNull(smessage) Then smessage = ""
| | |
To return a string of hex digits (as I needed to), declare result as string rather than an array or variants and replace from line 84: - For i = 0 To 31
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result(i) = Int((hashValues(i \ 4) / mask(i Mod 4) - Int(hashValues(i \ 4) / mask(i Mod 4))) * 256)
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Next
with: - result = ""
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For i = 0 To 31
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result = result & Hex(Int((hashValues(i \ 4) / mask(i Mod 4) - Int(hashValues(i \ 4) / mask(i Mod 4))) * 256))
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Next
The Hex() function is built into VBA
Thanks to Rabbit for helping me avoid dependencies on external libraries.
| | |
Sorry, you are right - the Hex function returns only one character for 1-digit values but it needs to be zero filled to 2 characters: - Result = Result & Right("00" & Hex(Int((hashValues(i \ 4) / mask(i Mod 4) - Int(hashValues(i \ 4) / mask(i Mod 4))) * 256)), 2)
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