Implementing AES-128 CBC by passing array as input data - c

Question

I am working on AES-128/192/256, basically I am getting data from a broker as string, I just need to encrypt that data and I need to verify that.
I already come across these https://github.com/empyreanx/tiny-AES128-C, https://github.com/kokke/tiny-AES-c links.
My code is:
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");
}
}
I also printed cipher text after encryption it is printing some undefined character.
I don't know but I am comparing "cipher" with "input", Its FAILURE finally!
Please anyone can tell me where I am doing wrong.
Thanks in advance.

Answer 1

This is logical to cipher to be different from input, isn't?
To check that encryption worked, you should decrypt the encoded message and check that they are equal:
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]);
}
}
}