Generating MAC by encrypting data

问题

I hope this question will not get closed as not constructive. I've been banging my head to the wall to solve this and still got nowhere. so, this is my last hope:
Question is basically about generating the MAC field of an ISO 8583-Rev 93 message.
I have some dumps of valid messages and I'm trying to generate the correct MAC.

Encryption Algorithm is DES, Mode=CBC, IV = new byte[] {0, 0, 0, 0, 0, 0, 0, 0}
Encryption key is: new byte[] { 0x11, 0x11, 0x11, 0x11, 0x11, 0x11, 0x11, 0x11 }
Data To encrypt is (the message) (208 bytes long):

new byte[] { 49, 50, 48, 48, 70, 54, 51, 52, 56, 52, 48, 49, 56, 56, 69, 49, 49, 48, 48, 48, 48, 49, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 22, 54, 50, 56, 48, 50, 51, 49, 49, 49, 48, 48, 48, 48, 54, 52, 50, 51, 49, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 48, 49, 49, 54, 48, 56, 48, 56, 52, 53, 51, 57, 55, 57, 56, 54, 49, 50, 48, 49, 49, 54, 49, 49, 51, 56, 52, 53, 48, 48, 48, 48, 48, 49, 49, 54, 54, 49, 48, 53, 48, 48, 54, 49, 51, 49, 55, 67, 6, 54, 50, 56, 48, 50, 51, 6, 54, 50, 56, 48, 50, 51, 53, 49, 52, 50, 52, 49, 52, 51, 50, 53, 52, 51, 48, 57, 57, 57, 57, 52, 48, 50, 48, 48, 48, 48, 48, 57, 57, 57, 57, 52, 48, 50, 32, 32, 32, 7, 84, 101, 108, 66, 97, 110, 107, 0, 0, 51, 55, 54, 69, 55, 49, 67, 68, 69, 49, 68, 56, 54, 54, 51, 54, 0, 0 };

I should encrypt the data and (probably after converting the result to hex) get to the valid BF327C0CED48F26B value which goes into the message's 128th field.

Here's what I have tried (along many more implementations):

var mac = Encrypt(dataToEncrypt);
// Convert to hex representation.
var hexMac = IsoUtils.ByteArrayToHex(mac);
// No BF327C0CED48F26B is in hexMac! something is probably wrong!


/// <summary>
/// Encrypts the input data. Tried many other ways with the same result.
/// </summary>    
public byte[] Encrypt(byte[] input)
{
    DESCryptoServiceProvider desProvider = new DESCryptoServiceProvider();
    desProvider.Padding = PaddingMode.None;
    desProvider.Mode = CipherMode.CBC;
    ICryptoTransform cryptoTransform = desProvider.CreateEncryptor(this._key, this._iv);
    MemoryStream encryptedStream = new MemoryStream();
    CryptoStream cryptStream = new CryptoStream(encryptedStream, cryptoTransform, CryptoStreamMode.Write);

    cryptStream.Write(input, 0, input.Length);
    cryptStream.FlushFinalBlock();
    encryptedStream.Position = 0;
    byte[] result = new byte[encryptedStream.Length];
    encryptedStream.Read(result, 0, (int)encryptedStream.Length);
    cryptStream.Close();
    return result;
}  


/// <summary>
/// Gets the hex representation of the byte array. The length is 2xdata.Length
/// </summary>    
public static string ByteArrayToHex(byte[] data)
{
    string result = string.Empty;
    foreach (byte ascii in data)
    {
        var n = (int)ascii;
        result += n.ToString("X").PadLeft(2, '0');
    }

    return result;
}

In short, How can I achieve BF327C0CED48F26B by encrypting dataToEncrypt variable?

I'd really appreciate any help.

UPDATE: Here's the message details in case it could be of any help:

------------------------------------------------------------------------
                           Data For Encryption                           
------------------------------------------------------------------------
                     HEX                                  NORMAL
31 32 30 30 46 36 33 34 38 34 30 31 38 38 45 31     1200F634840188E1
31 30 30 30 30 31 30 30 30 30 30 30 30 30 30 30     1000010000000000
30 30 30 30 16 36 32 38 30 32 33 31 31 31 30 30     000062802311100
30 30 36 34 32 33 31 30 30 30 30 30 30 30 30 30     0064231000000000
30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30     0000000000000000
30 30 30 30 31 31 36 30 38 30 38 34 35 33 39 37     0000116080845397
39 38 36 31 32 30 31 31 36 31 31 33 38 34 35 30     9861201161138450
30 30 30 30 31 31 36 36 31 30 35 30 30 36 31 33     0000116610500613
31 37 43 60 36 32 38 30 32 33 60 36 32 38 30 32     17C62802362802
33 35 31 34 32 34 31 34 33 32 35 34 33 30 39 39     3514241432543099
39 39 34 30 32 30 30 30 30 30 39 39 39 39 34 30     9940200000999940
32 20 20 20 70 54 65 6C 42 61 6E 6B 00 00 33 37     2   TelBank  37
36 45 37 31 43 44 45 31 44 38 36 36 33 36 00 00     6E71CDE1D86636  
------------------------------------------------------------------------
                        Message ready to be sent                        
------------------------------------------------------------------------
                     HEX                                  NORMAL
30 32 32 34 31 32 30 30 46 36 33 34 38 34 30 31     02241200F6348401
38 38 45 31 31 30 30 30 30 31 30 30 30 30 30 30     88E1100001000000
30 30 30 30 30 30 30 31 16 36 32 38 30 32 33 31     000000016280231
31 31 30 30 30 30 36 34 32 33 31 30 30 30 30 30     1100006423100000
30 30 30 30 30 30 30 30 30 30 30 30 30 30 30 30     0000000000000000
30 30 30 30 30 30 30 30 31 31 36 30 38 30 38 34     0000000011608084
35 33 39 37 39 38 36 31 32 30 31 31 36 31 31 33     5397986120116113
38 34 35 30 30 30 30 30 31 31 36 36 31 30 35 30     8450000011661050
30 36 31 33 31 37 43 60 36 32 38 30 32 33 60 36     061317C6280236
32 38 30 32 33 35 31 34 32 34 31 34 33 32 35 34     2802351424143254
33 30 39 39 39 39 34 30 32 30 30 30 30 30 39 39     3099994020000099
39 39 34 30 32 20 20 20 70 54 65 6C 42 61 6E 6B     99402   TelBank
00 00 33 37 36 45 37 31 43 44 45 31 44 38 36 36       376E71CDE1D866
33 36 00 00 42 46 33 32 37 43 30 43 45 44 34 38     36  BF327C0CED48
46 32 36 42 00 00 00 00                         F26B    
------------------------------------------------------------------------
                                 Fields                                 
------------------------------------------------------------------------
[LLVar    n      ..19 0016] 002 [6280231110000642]
[Fixed    n         6 0006] 003 [310000]
[Fixed    n        12 0012] 004 [000000000000]
[Fixed    n        12 0012] 006 [000000000000]
[Fixed    n        10 0010] 007 [0116080845]
[Fixed    n         6 0006] 011 [397986]
[Fixed    n        12 0012] 012 [120116113845]
[Fixed    n         4 0004] 014 [0000]
[Fixed    n         4 0004] 017 [0116]
[Fixed    an       12 0012] 022 [61050061317C]
[LLVar    n      ..11 0006] 032 [628023]
[LLVar    n      ..11 0006] 033 [628023]
[Fixed    an       12 0012] 037 [514241432543]
[Fixed    an        8 0008] 041 [09999402]
[Fixed    ans      15 0015] 042 [000009999402   ]
[LLVar    ans    ..40 0007] 043 [TelBank]
[LLLVar   ans   ..999 0000] 048 []
[Fixed    an       16 0016] 052 [376E71CDE1D86636]
[LLLVar   ans   ..999 0000] 072 []
[Fixed    an       16 0016] 128 [BF327C0CED48F26B]

Update 2: Well, those idiots forgot to mention in their documents that field 128 should be filled with 0 and then be sent for encryption. Found this out by try and fault!

回答1:

The MAC is only the last 8 bytes of the CBC-encryption.
You should only calculate the MAC over part of the message. I do not believe ISO 8583 specifies which parts, so you will need to look at the specification that you should have been provided by the bank.

回答2:

CBC Mac, if that is what they are doing, is typically done over length || message where || is concatenation. So you simply need to call desEncrypt(length||message,key,cbcmode) or something analogous

来源：https://stackoverflow.com/questions/9261526/generating-mac-by-encrypting-data

标签

cryptography

iso

des

iso8583