c - 通过传递数组作为输入数据来实现 AES-128 CBC

Question

我正在研究 AES-128/192/256，基本上我是从代理处以字符串形式获取数据，我只需要加密该数据并且需要验证它。

我已经遇到过这些 https://github.com/empyreanx/tiny-AES128-C , https://github.com/kokke/tiny-AES-c链接。

我的代码是:

static void test_encrypt_cbc(void)
{
    unsigned char input[] = 
"So_letmeknowRuinterested/towork@thiscompany.comElsewilllookother";   
 //64bits

    unsigned char cipher[sizeof input];

    printf("size of in:%lu\n",strlen(input));

    unsigned char key[] = "Gns7AauH3dnaod==";    //16 bits
    unsigned char iv[] = "vhdNaleuTHenaOlL";     //16 bits

    AES128_CBC_encrypt_buffer(cipher, input, 64, key, iv);

    if(0 == memcmp((char*) cipher, (char*) input, 64))
    {
        printf("SUCCESS!\n");
    }
    else
    {
        printf("FAILURE!\n");
    }
}

我还在加密后打印了密文，它打印了一些未定义的字符。

我不知道，但我正在比较“密码”和“输入”，最后失败了!

请任何人都可以告诉我哪里做错了。

提前致谢。

Answer 1

这对于密码来说与输入不同是合乎逻辑的，不是吗？

要检查加密是否有效，您应该解密编码消息并检查它们是否相等:

static void test_encrypt_cbc(void)
{
    /* 64 bytes, or 512 bits */
    unsigned char input[] = 
"So_letmeknowRuinterested/towork@thiscompany.comElsewilllookother";   

    unsigned char cipher[sizeof input];
    unsigned char output[sizeof input];

    printf("size of in:%lu\n",strlen(input));

    /* 16 bytes or 128 bits */
    unsigned char key[] = "Gns7AauH3dnaod==";    
    unsigned char iv[] = "vhdNaleuTHenaOlL";     

    /* input --> cipher */
    AES128_CBC_encrypt_buffer(cipher, input, 64, key, iv);

    /* cipher --> output */
    AES128_CBC_decrypt_buffer(output, cipher, 64, key, iv);

    if(0 == memcmp((char*) output, (char*) input, 64))
    {
        printf("SUCCESS!\n");
    }
    else
    {
        int i;
        printf("FAILURE!\nInput and output are different:\n");
        for (i = 0; i < sizeof input; ++i) 
        {
            printf("%02x - %02x\n", input[i], output[i]);
        }
    }
}