Hi, Based on my old experience with "struct crypto_alg" based drivers, the data you receive there, is padded beforehand(in the upper layers); Therefore the plaintext contains integral multiple of AES block size of data and based on the number of blocks, the crypto transform can be computed.
Regards, Hamid On Mon, Oct 24, 2016 at 6:11 PM, Cata Vasile <cata.vas...@nxp.com> wrote: > > Hi, > > I'm trying to understand the code for AES encryption from ARM-CE. > From the aes-glue.S calls I understand that the encryption primitives receive > the number of blocks, but have no way of determining the number of bytes to > encrypt, if for example the plaintext does not have a length of a multiple of > AES block size. > How does, for example, ecb_encrypt() also encrypt the last remaining bytes in > the plaintext if it is not a multiple of AES block size if It can never > deduce the full plaintext size? > > Catalin Vasile-- > To unsubscribe from this list: send the line "unsubscribe linux-crypto" in > the body of a message to majord...@vger.kernel.org > More majordomo info at http://vger.kernel.org/majordomo-info.html -- To unsubscribe from this list: send the line "unsubscribe linux-crypto" in the body of a message to majord...@vger.kernel.org More majordomo info at http://vger.kernel.org/majordomo-info.html
