在#include "CMAC/aes_cmac.h文件中找到下面这个例子，
但是这个例子只能做16整数的字节的算法CMAC！问题：
1，如果我要做任意字节的数据来算CMAC，怎么实现？
2，我们需要替换OPENSSL中的，CMAC_Init,CMAC_Update,CMAC_Final三个函数！
/*!
* \page Tutorial_AES_CMAC AES-CMAC Tutorial
*
* Please remember that before starting to call the Encryption function the context \b requires user
* initialization. The Key Size, Tag Size, Flags member must be initialized prior to calling the
* init function.
* Before the last call to AES_CMAC_Encrypt_Append the flag E_SK_FINAL_APPEND must be set in the context.
* Look at the each function documentation to see what is needed prior of calling.
*
* The API functions to be used are:
* - \ref AES_CMAC_Encrypt_Init initialize an \ref AESCMACctx_stt context with key, tag size, flags.
* - \ref AES_CMAC_Encrypt_Append process the input. It doesn't produce any output.
* It can be called multiple times but the input size must be multiple of 16.
* A last append call is allowed with any, positive, input size, but prior to that the flag
* E_SK_FINAL_APPEND must be set.
* - \ref AES_CMAC_Encrypt_Finish can be called only one time for the finalization process and it will generate
* the authentication TAG
*
* Here follows an example of AES-128-CMAC authentication, it will process a NIST test vector and generate the
* authentication TAG.
*
* \code
* #include <stdio.h>
* #include "crypto.h"
*
* int main()
* {
* const uint8_t KEY_CMAC[] ={0x2b,0x7e,0x15,0x16,0x28,0xae,0xd2,0xa6,0xab,0xf7,0x15,0x88,0x09,0xcf,0x4f,0x3c};
* const uint8_t PLAIN_TEXT_CMAC[] = {0x6b,0xc1,0xbe,0xe2,0x2e,0x40,0x9f,0x96,0xe9,0x3d,0x7e,0x11,0x73,0x93,0x17,0x2a,
* 0xae,0x2d,0x8a,0x57,0x1e,0x03,0xac,0x9c,0x9e,0xb7,0x6f,0xac,0x45,0xaf,0x8e,0x51,
* 0x30,0xc8,0x1c,0x46,0xa3,0x5c,0xe4,0x11,0xe5,0xfb,0xc1,0x19,0x1a,0x0a,0x52,0xef,
* 0xf6,0x9f,0x24,0x45,0xdf,0x4f,0x9b,0x17,0xad,0x2b,0x41,0x7b,0xe6,0x6c,0x37,0x10};
* uint8_t TagAesCmac[16] = {0,};
* int32_t TagSize=0;
* uint32_t error_status = 0;
*
* AESCMACctx_stt tst_AesCmac;
*
* // Set flag field to default value
* tst_AesCmac.mFlags = E_SK_DEFAULT;
*
* // Set key size to 16 (correspondng to AES-128)
* tst_AesCmac.mKeySize = 16;
*
* // Pointer to original Key buffer
* tst_AesCmac.pmKey = (uint8_t *) KEY_CMAC;
*
* // Size of returned authentication TAG
* tst_AesCmac.mTagSize = 16;
*
* // Initialize the operation, by passing the key and IV
* error_status = AES_CMAC_Encrypt_Init(&tst_AesCmac);
*
* // check for initialization errors
* if(error_status == AES_SUCCESS)
* {
* // Message is 4 blocks of 16 bytes each, do first append of 16 bytes
* error_status = AES_CMAC_Encrypt_Append(&tst_AesCmac, (uint8_t *) PLAIN_TEXT_CMAC, 16);
* if(error_status == AES_SUCCESS)
* {
* // Do second append of 32 bytes
* error_status = AES_CMAC_Encrypt_Append(&tst_AesCmac, (uint8_t *) PLAIN_TEXT_CMAC + 16, 32);
* }
* if(error_status == AES_SUCCESS)
* {
* // Do last append of the remaining 16 bytes, but first set the flag
* tst_AesCmac.mFlags |= E_SK_FINAL_APPEND;
* error_status = AES_CMAC_Encrypt_Append(&tst_AesCmac, (uint8_t *) PLAIN_TEXT_CMAC + 16 + 32, 16);
* if(error_status == AES_SUCCESS)
* {
* // Do the Finalization, write the TAG at the end of the encrypted message
* error_status = AES_CMAC_Encrypt_Finish(&tst_AesCmac, (uint8_t*) TagAesCmac, &TagSize);
* }
* }
* }
* printf("Returned: %d\n", error_status);
* print_buffer("Returned TAG:",TagAesCmac,TagSize);
* return(0);
* }
* \endcode
*/
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