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What size of initialization vector (IV) is needed for AES encryption? I am using either CBC or CFB modes.
Knowing that AES is a sysmmetrical block-cipher algorithm with a 128-bit block size, I think the answer for IV is still 16 bytes or 128 bits for AES 128, 192 and 256.
Reading this stackoverflow Q&A it seems as if the size of the IV is the size of the block which is always 128 bit (= 16 bytes) in AES, even if the keysize is larger than the block size. Is that correct?
For both CBC mode and CFB mode, the initialization vector is the size of a block, which for AES is 16 bytes = 128 bits. For CFB mode, the IV must never be reused for different messages under the same key; for CBC mode, the IV must never be reused for different messages under the same key, and must be unpredictable in advance by an attacker. Using successive integers as the IV is fine for CFB but not fine for CBC.
Beware: The Wikipedia articles are currently (2017-08-11) full of archaic drivel about self-synchronizing ciphers and error propagation and other fortunately forgotten relics of the dark ages of crypto engineering from a bygone century. I cite them only for the easily accessible statements of equations relating plaintext and ciphertext and their associated diagrams.
All that said, could I interest you in an authenticated encryption scheme instead, such as NaCl crypto_secretbox_xsalsa20poly1305? Or, if you find yourself reaching for the letters ‘AES’ and ‘CBC’, you're probably lost in a vat of acronym soup and you may need help navigating crypto protocols more than you need help picking the right parameter sizes for a confusing crypto API that asks you to choose mode and then specify IV.
