<i18n dev> [15] RFR: 8244459: Optimize the hash map size in LocaleProviderAdapters
naoto.sato at oracle.com
naoto.sato at oracle.com
Tue May 5 21:08:46 UTC 2020
Thanks, all. I didn't see this coming!
If I understand the discussion correctly, Peter's suggestion is the most
optimal (Mark, your formula produces 1 for the expected size is 0,
although it won't be happening in this particular case). And Joe, thank
you for finding my silly mistake :-) So here is the updated webrev:
http://cr.openjdk.java.net/~naoto/8244459/webrev.01/
Naoto
On 5/5/20 11:01 AM, naoto.sato at oracle.com wrote:
> And here is the fix. Please review.
>
> http://cr.openjdk.java.net/~naoto/8244459/webrev.00/
>
> Naoto
>
> On 5/5/20 10:25 AM, naoto.sato at oracle.com wrote:
>> Hi Peter,
>>
>> You are correct. Thanks. I'll remove that initial value of 16.
>>
>> Naoto
>>
>> On 5/5/20 9:37 AM, Peter Levart wrote:
>>> Hi Naoto,
>>>
>>> On 4/30/20 12:18 AM, naoto.sato at oracle.com wrote:
>>>> Hello,
>>>>
>>>> Please review this small fix to the following issue:
>>>>
>>>> https://bugs.openjdk.java.net/browse/JDK-8244152
>>>>
>>>> The proposed changeset is located at:
>>>>
>>>> https://cr.openjdk.java.net/~naoto/8244152/webrev.00/
>>>>
>>>> The hash map used there didn't have initial capacity, even though
>>>> the exact numbers are known.
>>>
>>>
>>> Well, it has to be calculated 1st (countTokens), but I guess this
>>> pays off when HashSet (the backing HashMap) does not have to be
>>> rehashed then.
>>>
>>> The expression you use:
>>>
>>> Math.max((int)(tokens.countTokens() / 0.75f) + 1, 16)
>>>
>>> ...has a minimum value of 16. Why is that? 16 is just HashMap's
>>> default initialCapacity if not specified explicitly. But if you only
>>> want to store say 1 entry in the map, you can specify 2 as
>>> initialCapacity and HashMap will happily work for such case without
>>> resizing.
>>>
>>>
>>> So you could just use:
>>>
>>> (int)(tokens.countTokens() / 0.75f) + 1
>>>
>>> And even this expression is sometimes overshooting the minimal
>>> required value by 1 (when # of tokens is "exact" multiple of 0.75f,
>>> say 6). I think the following could be used to optimally pre-size the
>>> HashMap with default load factor 0.75:
>>>
>>> (tokens.countTokens() * 4 + 2) / 3
>>>
>>>
>>> Regards, Peter
>>>
>>>>
>>>> Naoto
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