Iñaki Baz Castillo writes:

> And those collisions mean that I could look for a key name and
> retrieve another key (as both key names produce the same hash), am I
> right?

no, the values whose keys hash to the same index of hash table, are
added into a linked list and then list list will be searched until the
correct key is found.

> So, if I want to handle ~ 300 concurrent entries, with *no* collision,
> which "size" value is good enough? how to determine it?

you cannot unless you know the hash algorithm and your keys in advance.

-- juha

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