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Fix tls_cbc_digest_record is slow using SHA-384 and short messages

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The formula used for this is now

kVarianceBlocks = ((255 + 1 + md_size + md_block_size - 1) / md_block_size) + 1

Notice that md_block_size=64 for SHA256, which results on the
magic constant kVarianceBlocks = 6.
However, md_block_size=128 for SHA384 leading to kVarianceBlocks = 4.

CLA:trivial

Reviewed-by: Matt Caswell <matt@openssl.org>
Reviewed-by: Paul Dale <paul.dale@oracle.com>
(Merged from https://github.com/openssl/openssl/pull/7342)
This commit is contained in:
armfazh 2018-10-19 08:26:58 +10:00 committed by Pauli
parent dc5831da59
commit cb8164b05e
1 changed files with 3 additions and 2 deletions
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5
ssl/s3_cbc.c
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@ -256,12 +256,13 @@ int ssl3_cbc_digest_record(const EVP_MD_CTX *ctx,
* of hash termination (0x80 + 64-bit length) don't fit in the final
* block, we say that the final two blocks can vary based on the padding.
* TLSv1 has MACs up to 48 bytes long (SHA-384) and the padding is not
* required to be minimal. Therefore we say that the final six blocks can
* required to be minimal. Therefore we say that the final |variance_blocks|
* blocks can
* vary based on the padding. Later in the function, if the message is
* short and there obviously cannot be this many blocks then
* variance_blocks can be reduced.
*/
variance_blocks = is_sslv3 ? 2 : 6;
variance_blocks = is_sslv3 ? 2 : ( ((255 + 1 + md_size + md_block_size - 1) / md_block_size) + 1);
/*
* From now on we're dealing with the MAC, which conceptually has 13
* bytes of `header' before the start of the data (TLS) or 71/75 bytes