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)
* 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
* 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.
*/
* 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
/*
* 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