I rambled across this while reviewing one of Jon's patches. Here is the
code from DefaultTuple

/**
     * Construct a tuple with a known number of fields. Package level so
that callers cannot directly invoke it.
     * <br>Resulting tuple is filled pre-filled with null elements. Time
complexity: O(N), after allocation
     *
     * @param size
     *            Number of fields to allocate in the tuple.
     */
    DefaultTuple(int size) {
        mFields = new ArrayList<Object>(size);
        for (int i = 0; i < size; i++)
            mFields.add(null);
    }
Why are we walking through the list to add nulls? Wouldn't the initial
creation of ArrayList suffice?
mFields = new ArrayList<Object>(size) should be enough.

Thanks,
Prashant

Here is my SS:  259 71 2451

On May 26, 2012, at 9:13 PM, Jonathan Coveney <[EMAIL PROTECTED]> wrote:

> -user
> +dev
>
> Haha, I made this very same comment somewhere, and noticed the exact same
> thing (I think I mention it in my SchemaTuple benchmarking).
>
> The reason is so that tuple.size() will return the right value. Also, the
> expectation is that if you append, it goes to the end of all of the nulls,
> not the first position. It's a little confusing, and yeah, it surprised me
> too.
>
> You could definitely amortize the cost of creation over the sets that the
> user does by keeping an index, but when I first saw it I decided that the
> (slightly) increased memory footprint and the increase in code complexity
> wasn't worth a very minimal increase.
>
> 2012/5/26 Prashant Kommireddi <[EMAIL PROTECTED]>
>
>> I rambled across this while reviewing one of Jon's patches. Here is the
>> code from DefaultTuple
>>
>> /**
>>    * Construct a tuple with a known number of fields. Package level so
>> that callers cannot directly invoke it.
>>    * <br>Resulting tuple is filled pre-filled with null elements. Time
>> complexity: O(N), after allocation
>>    *
>>    * @param size
>>    *            Number of fields to allocate in the tuple.
>>    */
>>   DefaultTuple(int size) {
>>       mFields = new ArrayList<Object>(size);
>>       for (int i = 0; i < size; i++)
>>           mFields.add(null);
>>   }
>>
>>
>> Why are we walking through the list to add nulls? Wouldn't the initial
>> creation of ArrayList suffice?
>> mFields = new ArrayList<Object>(size) should be enough.
>>
>> Thanks,
>> Prashant
>>

Is this @highponte a spam? It is there for all
mails(hadoop,hbase,pig). Can hadoop mailing lists spammed?

On 5/27/12, highpointe <[EMAIL PROTECTED]> wrote:
> Here is my SS:  259 71 2451
>
> On May 26, 2012, at 9:13 PM, Jonathan Coveney <[EMAIL PROTECTED]> wrote:
>
>> -user
>> +dev
>>
>> Haha, I made this very same comment somewhere, and noticed the exact same
>> thing (I think I mention it in my SchemaTuple benchmarking).
>>
>> The reason is so that tuple.size() will return the right value. Also, the
>> expectation is that if you append, it goes to the end of all of the
>> nulls,
>> not the first position. It's a little confusing, and yeah, it surprised
>> me
>> too.
>>
>> You could definitely amortize the cost of creation over the sets that the
>> user does by keeping an index, but when I first saw it I decided that the
>> (slightly) increased memory footprint and the increase in code complexity
>> wasn't worth a very minimal increase.
>>
>> 2012/5/26 Prashant Kommireddi <[EMAIL PROTECTED]>
>>
>>> I rambled across this while reviewing one of Jon's patches. Here is the
>>> code from DefaultTuple
>>>
>>> /**
>>>    * Construct a tuple with a known number of fields. Package level so
>>> that callers cannot directly invoke it.
>>>    * <br>Resulting tuple is filled pre-filled with null elements. Time
>>> complexity: O(N), after allocation
>>>    *
>>>    * @param size
>>>    *            Number of fields to allocate in the tuple.
>>>    */
>>>   DefaultTuple(int size) {
>>>       mFields = new ArrayList<Object>(size);
>>>       for (int i = 0; i < size; i++)
>>>           mFields.add(null);
>>>   }
>>>
>>>
>>> Why are we walking through the list to add nulls? Wouldn't the initial
>>> creation of ArrayList suffice?
>>> mFields = new ArrayList<Object>(size) should be enough.
>>>
>>> Thanks,
>>> Prashant
>>>
>