------- Original message -------
From: Ruediger Pluem
Don't we have the same issue here as with the XOR version of the patch?
If two different keys (key1, key2) result in the same hash value (so
ht->hash_func(key1, &klen1) == ht->hash_func(key2,
&klen2);) doesn't hashfunc_default result in the same hash value for both
keys since the input value to hash
is the same?
For two keys to cause a collision, the modulo (i.e. the index) has to be
the same, not necessarily the whole hash. So, with xor, we don't gain
anything by using seed against the final hash (modulo will change, but in
the same way for both hash values - the collision will move to a different
bucket). With the hash function, as long as the hash is not completely the
same, we will most likely get a different modulo and therefore a different
index into the table.
Make sense?
--
Bojan
