On 03/11/2022 14:21, Wiktor Kwapisiewicz via openssl-users wrote:
I'd like to clarify one aspect of the API regarding EVP_EncryptUpdate
 that is the length of the output buffer that should be passed to
that function ("out" parameter). (Actually I'm using EVP_CipherUpdate
but the docs are more comprehensive for EVP_EncryptUpdate).
For the record I'm using AES-128 cipher in ECB mode and the docs say:
For most ciphers and modes, the amount of data written can be
anything from zero bytes to (inl + cipher_block_size - 1) bytes. For
wrap cipher modes, the amount of data written can be anything from
zero bytes to (inl + cipher_block_size) bytes. For stream ciphers,
the amount of data written can be anything from zero bytes to inl
AES-128-ECB doesn't appear to be a stream cipher (since the "block size"
returns 16 not the magical value of 1) and I'm unable to find any
mentions of "wrap cipher modes" in search engines. Apparently ECB is a
block cipher mode.
Does that mean that "wrap cipher modes" == "block cipher modes"?
The term "block cipher" is a feature of the underlying primitive - so
AES-128 is a block cipher. It encrypts in blocks of 16 bytes. ECB is a
particular mode for using a block cipher. "Wrap" modes are specialist
modes used for encrypting key material.
Is there any documentation I could read on the reasoning of why a space
for additional block is needed in this case ("(inl + cipher_block_size)
bytes")? I'm trying to understand the differences between OpenSSL and
other cryptographic backends in an OpenPGP library .
EVP_EncryptUpdate() can be called repeatedly, incrementally feeding in
the data to be encrypted. The ECB mode (when used with AES-128) will
encrypt input data 16 bytes at a time, and the output size will also be
16 bytes per input block. If the data that you feed in to
EVP_EncryptUpdate() is not a multiple of 16 bytes then the amount of
data that is over a multiple of 16 bytes will be cached until a
subsequent call where it does have 16 bytes.
Let's say you call EVP_EncryptUpdate() with 15 bytes of data. In that
case all 15 bytes will be cached and 0 bytes will be output.
If you then call it again with 17 bytes of data, then added to the 15
bytes already cached we have a total of 32 bytes. This is a multiple of
16, so 2 blocks (32 bytes) will be output, so:
(inl + cipher_block_size - 1) = (17 + 16 - 1) = 32
Thank you for your time and help!