AES/CFB8 IV size

Question

AFAIK, CFB8 mode has block size of 1byte. So I can induce that IV is also 1byte length. However, when I do a test passing same iv of just 1 byte into common crypto create function for encrypt and decrypt function, encrypted and decrypted message mismatch.

So I think that the API should have taken more than 1 byte to use as IV. I would like to know why? Any thing wrong with my understanding?

CCCryptorStatus result = CCCryptorCreateWithMode(operation,
                                                 kCCModeCFB8,
                                                 kCCAlgorithmAES128,
                                                 ccNoPadding,
                                                 iv.bytes,
                                                 key.bytes,
                                                 key.length,
                                                 NULL,
                                                 0,
                                                 0,
                                                 0,
                                                 &_cryptor);

if (result == kCCSuccess)
    result = CCCryptorUpdate(_cryptor,
                             data.bytes,
                             data.length,
                             cipherData.mutableBytes,
                             cipherData.length,
                             &outLength);

if (result == kCCSuccess)
    result = CCCryptorFinal(_cryptor,
                            cipherData.mutableBytes,
                            cipherData.length,
                            &outLength);

if (result == kCCSuccess)
    result = CCCryptorRelease(_cryptor);

Answer 1

It doesn't have block size of 1 byte, it is just resynchronized each 1 byte. The IV is actually 16 bytes (for AES).

Answer 2

The IV size must match the symmetric algorithm block size. Hence, for AES you should have an IV of 16 bytes.

CFB-8 has a shift size of a byte. It bears no relation to the block size of the cipher.