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我想用 openssl 计算给定明文的 CBC-MAC。我有以下明文（hexdump）：

hexdump -C example.txt
00000000  4d 41 43 73 20 61 72 65  20 76 65 72 79 20 75 73  |MACs are very us|
00000010  65 66 75 6c 20 69 6e 20  63 72 79 70 74 6f 67 72  |eful in cryptogr|
00000020  61 70 68 79 21 20 20 20  20 20 20 20 20 20 20 20  |aphy!           |

如果我使用 openssl 的命令行功能，我会得到以下解决方案：

openssl aes-256-cbc -in example.txt -K 8000000000000000000000000000000000000000000000000000000000000001 -e -iv 00 | hexdump -C
00000000  e8 e9 a4 ce 5d 20 c4 ad  f5 52 b2 c6 38 2e 12 4e  |....] ...R..8..N|
00000010  20 f5 63 65 b4 b3 96 9f  ad 8d ca e4 e8 34 2a e5  | .ce.........4*.|
00000020  0d 82 0e 3a 1e 10 5d 30  72 16 fc 00 c7 a5 b4 49  |...:..]0r......I|
00000030  f5 63 9f 85 ff e3 a4 a4  23 6e 6f 09 20 ed b1 ae  |.c......#no. ...|

到现在为止还挺好。我有一个额外的块，因为第一个块应该是加密的 IV。现在最后一行应该是我的 CBC-MAC，如果我理解正确的话。接下来我尝试在 C 中做同样的事情，这里是示例代码：

#include <stdio.h>
#include <string.h>
#include <openssl/sha.h>
#include <openssl/aes.h>

int main(int argc, char *argv[])
{
    unsigned char obuf[64] = {0};
    unsigned char decbuf[48] = {0};
    unsigned char msg1[] = {0x4d, 0x41, 0x43, 0x73, 0x20, 0x61, 0x72, 0x65, 0x20, 0x76, 0x65, 0x72, 0x79, 0x20, 0x75, 0x73,
                            0x65, 0x66, 0x75, 0x6c, 0x20, 0x69, 0x6e, 0x20, 0x63, 0x72, 0x79, 0x70, 0x74, 0x6f, 0x67, 0x72, 
                            0x61, 0x70, 0x68, 0x79, 0x21, 0x20, 0x20, 0x20, 0x20, 0x20, 0x20, 0x20, 0x20, 0x20, 0x20, 0x20};
    unsigned char key[] =  {0x80, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 
                            0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x01};
    unsigned char ivenc[] ={0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00};
    unsigned char ivdec[] ={0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00};
    int i=0;

    AES_KEY enc_key, dec_key;
    AES_set_encrypt_key(key, 256, &enc_key);
    AES_cbc_encrypt(msg1, obuf, 48, &enc_key, ivenc, AES_ENCRYPT);

    for (i = 0; i < 64; i++) {
        if (!(i%16))
            printf("\n");
        printf("%02x ", obuf[i]);
    }
    printf("\n");

    AES_set_decrypt_key(key, 256, &dec_key);
    AES_cbc_encrypt(obuf, decbuf, 64, &dec_key, ivdec, AES_DECRYPT);

    for (i = 0; i < 48; i++) {
        if (!(i%16))
            printf("\n");
        printf("%02x ", decbuf[i]);
    }
    printf("\n");

    return 0;
}

之后我解密加密的消息以验证我的代码。我的代码输出非常令人惊讶：

e8 e9 a4 ce 5d 20 c4 ad f5 52 b2 c6 38 2e 12 4e 
20 f5 63 65 b4 b3 96 9f ad 8d ca e4 e8 34 2a e5 
0d 82 0e 3a 1e 10 5d 30 72 16 fc 00 c7 a5 b4 49 
00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 

4d 41 43 73 20 61 72 65 20 76 65 72 79 20 75 73 
65 66 75 6c 20 69 6e 20 63 72 79 70 74 6f 67 72 
61 70 68 79 21 20 20 20 20 20 20 20 20 20 20 20

The encrypted message is completely identical to the command line output, except the last line is all 0. I thought, that the first line is the encrypted IV and the three following lines are the encrypted message, so with my interpretation the last line of the message wasn't encrypted. But to my surprise the decryption results exactly to the text I used as input, so it seems that there is no loss of information.

Questions:

  • How is it possible that I can decrypt the output of my encryption, even if I don't have the last line?
  • What is my CBC-MAC? is it the last line from my command line output or the last line of my C-code output?
  • Am I doing something wrong in my C code? I used this so question as help.
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1 回答 1

5

Your mistake is here:

I have one extra block because the first block should be the encrypted IV.

The extra block is because OpenSSL adds padding to the plain text, so that it is a multiple of the block size (16 bytes for AES). In this case the plain text is already a multiple of 16 bytes, but the padding scheme used (PKCS7) always adds padding, so here an entire block is added before encryption.

It’s common to add the IV to the front of the ciphertext, but that’s not what’s happening here.

In order to get the same result from your code you will need to add this padding yourself. In this case it is fairly simple, just add sixteen 0x10 bytes to the end of msg1 (so its toal length is 64), and change the 48 in the call to AES_cbc_encrypt to 64. The zeros you are seeing a just the value you initialize obuf to, since you are only writing 48 bytes into this buffer.

于 2017-07-06T12:52:13.363 回答