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(Visual FoxPro) Tips on Matching Encryption with another System

This example provides tips on matching encryption results produced by another system.

Chilkat ActiveX Downloads

ActiveX for 32-bit and 64-bit Windows

LOCAL loCrypt
LOCAL lcIvHex1
LOCAL lcIvHex2
LOCAL lcKeyHex

* This example assumes the Chilkat API to have been previously unlocked.
* See Global Unlock Sample for sample code.

loCrypt = CreateObject('Chilkat_9_5_0.Crypt2')

* Let's examine 256-bit AES encryption in CBC mode.
* CBC mode is Cipher Block Chaining, and it uses an IV (initialization vector)
loCrypt.CryptAlgorithm = "aes"
loCrypt.CipherMode = "cbc"
loCrypt.KeyLength = 256
loCrypt.PaddingScheme = 0
lcIvHex1 = "000102030405060708090A0B0C0D0E0F"
lcIvHex2 = "FF0102030405060708090A0B0C0D0E0F"
loCrypt.SetEncodedIV(lcIvHex1,"hex")
lcKeyHex = "000102030405060708090A0B0C0D0E0F101112131415161718191A1B1C1D1E1F"
loCrypt.SetEncodedKey(lcKeyHex,"hex")

* Matching encryption requires all of the above settings to be matched exactly.
* Let's get our output in hex format so we can easily see the values of the encrypted bytes.
loCrypt.EncodingMode = "hex"

* Encrypt something small:
? loCrypt.EncryptStringENC("Hello")
* The result is 5B827AB3B4F9F2292C2B74C8A6C99A3D
* This 16 bytes -- exactly one AES encryption block.

* Let's change only the padding scheme.
loCrypt.PaddingScheme = 3

* Encrypt again:
? loCrypt.EncryptStringENC("Hello")
* The result is entirely different: 469C28CC576069F807891FEE2DE76D68

* The padding scheme only affects the very last block of output.  Therefore,
* if all settings match except for the padding scheme, we're unable to
* know if we encrypt a very small amount of data. However, if we encrypt
* a larger amount of data, the single difference becomes apparent:
? "-- Only the padding scheme differs --"
loCrypt.PaddingScheme = 0
? loCrypt.EncryptStringENC("HelloHelloHelloHelloHelloHelloHello")
loCrypt.PaddingScheme = 3
? loCrypt.EncryptStringENC("HelloHelloHelloHelloHelloHelloHello")

* Now examine the outputs:
* F6A201F8E0B6595FA20E4A212A2AD9A5046DAF29E8B35AD15CEE56A1A69F2A3A7B347A7C15E26E7A6760533C7A8E0D44
* F6A201F8E0B6595FA20E4A212A2AD9A5046DAF29E8B35AD15CEE56A1A69F2A3A292CA61D03A85E1AC39B50D4DA71691E
* We can see the output matches except for the last block, which is affected by the padding scheme.

* If we are able to easily use ECB mode w/ the other system
* we are trying to match, then eliminate the IV from the picture.
* If the encryption matches in ECB mode, but not in CBC mode,
* then we know all correct except for the IV.
* For example, you can see how the IV changes everything with CBC mode,
* but it's not used in ECB mode:
loCrypt.PaddingScheme = 0
loCrypt.CipherMode = "cbc"
? "-- Only the IV differs, CBC mode produces different output. --"
loCrypt.SetEncodedIV(lcIvHex1,"hex")
? loCrypt.EncryptStringENC("HelloHelloHelloHelloHelloHelloHello")
loCrypt.SetEncodedIV(lcIvHex2,"hex")
? loCrypt.EncryptStringENC("HelloHelloHelloHelloHelloHelloHello")

loCrypt.CipherMode = "ecb"
? "-- Only the IV differs, ECB does not use the IV.  The outputs are the same. --"
loCrypt.SetEncodedIV(lcIvHex1,"hex")
? loCrypt.EncryptStringENC("HelloHelloHelloHelloHelloHelloHello")
loCrypt.SetEncodedIV(lcIvHex2,"hex")
? loCrypt.EncryptStringENC("HelloHelloHelloHelloHelloHelloHello")

* If we can eliminate the padding scheme and IV from the degrees of freedom,
* then the only remaining likely differences are (1) the secret key,
* and (2) the input data itself.

* The secret key is composed of binary bytes of exactly KeyLength bits.
* For 256-bit AES encrytion, the key length is 256, and therefore the 
* secret key is exactly 32 bytes.  (32 * 8 bits/byte = 256 bits)
* If the secret key is derived from an arbitrary password string, then one must
* exactly duplicate the derivation scheme (such as PBKDF2, for example)
* The input bytes to the derivation scheme must also match.  For example,
* is it the utf-8 byte representation of the password string that is used
* as the starting point for the derivation, or perhaps utf-16, or ANSI (1 byte per char)?

* Likewise, if the data being encrypted is a string, what byte representation of
* the string is being encrypted?  If the bytes presented to the encryptor are different,
* then the output is different.

RELEASE loCrypt


 

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