RFC 3610-AES CCM-测试向量

时间:2019-02-05 14:18:18

标签: aes ccm

我想验证AES CCM算法的实现。

在RFC 3610的末尾是一些测试向量。首先包含:

=============== Packet Vector #1 ==================
AES Key =  C0 C1 C2 C3  C4 C5 C6 C7  C8 C9 CA CB  CC CD CE CF
Nonce =    00 00 00 03  02 01 00 A0  A1 A2 A3 A4  A5
Total packet length = 31. [Input with 8 cleartext header octets]
          00 01 02 03  04 05 06 07  08 09 0A 0B  0C 0D 0E 0F
          10 11 12 13  14 15 16 17  18 19 1A 1B  1C 1D 1E
CBC IV in: 59 00 00 00  03 02 01 00  A0 A1 A2 A3  A4 A5 00 17
CBC IV out:EB 9D 55 47  73 09 55 AB  23 1E 0A 2D  FE 4B 90 D6
After xor: EB 95 55 46  71 0A 51 AE  25 19 0A 2D  FE 4B 90 D6   [hdr]
After AES: CD B6 41 1E  3C DC 9B 4F  5D 92 58 B6  9E E7 F0 91
After xor: C5 BF 4B 15  30 D1 95 40  4D 83 4A A5  8A F2 E6 86   [msg]
After AES: 9C 38 40 5E  A0 3C 1B C9  04 B5 8B 40  C7 6C A2 EB
After xor: 84 21 5A 45  BC 21 05 C9  04 B5 8B 40  C7 6C A2 EB   [msg]
After AES: 2D C6 97 E4  11 CA 83 A8  60 C2 C4 06  CC AA 54 2F
CBC-MAC  : 2D C6 97 E4  11 CA 83 A8
CTR Start: 01 00 00 00  03 02 01 00  A0 A1 A2 A3  A4 A5 00 01
CTR[0001]: 50 85 9D 91  6D CB 6D DD  E0 77 C2 D1  D4 EC 9F 97
CTR[0002]: 75 46 71 7A  C6 DE 9A FF  64 0C 9C 06  DE 6D 0D 8F
CTR[MAC ]: 3A 2E 46 C8  EC 33 A5 48
Total packet length = 39. [Authenticated and Encrypted Output]
          00 01 02 03  04 05 06 07  58 8C 97 9A  61 C6 63 D2
          F0 66 D0 C2  C0 F9 89 80  6D 5F 6B 61  DA C3 84 17
          E8 D1 2C FD  F9 26 E0

如何从 CBC IV in 行中的 CBC IV out 行中获取值?

CBC-MAC算法是通过零初始化矢量初始化的吗?

是使用随机IV还是忽略了某些东西?

谢谢您的回答。

0 个答案:

没有答案