Sheesh. It's not exactly a fair test though. The second hash population has all those calls to &randomstring and &hashval in it as well as the dodgy bucket bits....

I'll give it a play. Cheers!

Update:
Doh. But it does have the if (time() - $lasttime  > 5) test in it.

I'm going to lay in the park and drink beer. Have fun.

If there are 53 key/value pairs, why are there 64 buckets and why did only 42 of them get used?

Do buckets have any use whatsoever to the average perl programmer or should I drink less coffee and stop asking questions at this time on a hot Thursday afternoon?

Here's how a hash works, in a nutshell... You've got a list of buckets, each of which is a list of zero or more items. You also have a hash function that converts a key into a "hash" value used to look up a bucket. (For any given key, the hash function must always return the same result; otherwise you wouldn't be able to find your items again!)

When you want to do something with a specific key, you call the hash function for that key, use the result to find the appropriate bucket, and then search all the items in that bucket for the one you want. So, instead of having to check against every item in the hash, you only have to check against a few items that happen to be in the same bucket.

As you keep adding to the hash, each bucket has to hold more items. As the buckets get bigger, the lookups get slower, because you have to search through more items. That problem is solved by dynamically increasing the number of buckets and redistributing the items as the hash grows in size.

Your hash has 64 buckets (Perl always uses a power of 2), although only 42 are being used to store the 53 items. That just means that, with that combination of hash function, number of buckets, and keys, a few of the keys are sharing buckets.

> That problem is solved by dynamically increasing the number of buckets

How does one go about this "easy-to-say" task?