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Posted to user@commons.apache.org by Henrik Vendelbo <hv...@bluprints.com> on 2003/11/03 19:09:26 UTC

[pool] Keyey pool usage

Browsing the mailing-list archive made me wonder if the pools are inteded
for the use I seek.

I have information stored in a database, and want to retrieve it as rarely
as possible.

When I need a piece of information I want to borrow it from the pool and
return it when I am done with it.

I may borrow the same object multiple times and obtain the same object, be
it in parallel or sequentially.

Suppose I do the following :

Object o1 = borrowObject("somename");
Object o2 = borrowObject("somename");
Object o3 = borrowObject("somename");
returnObject(o1);
returnObject(o2);
returnObject(o3);

will the three objects be the same ?

Object o1 = borrowObject("somename");
returnObject(o1);
Object o2 = borrowObject("somename");
returnObject(o2);

will o1 == o2 ?

Henrik



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Re: [pool] Keyey pool usage

Posted by Dirk Verbeeck <di...@pandora.be>.
See inline for answers but I suspect you really need a cache and not a pool.

related question/answer:
http://nagoya.apache.org/eyebrowse/ReadMsg?listName=commons-user@jakarta.apache.org&msgNo=2693

-- Dirk

Henrik Vendelbo wrote:

> Browsing the mailing-list archive made me wonder if the pools are inteded
> for the use I seek.
> 
> I have information stored in a database, and want to retrieve it as rarely
> as possible.
> 
> When I need a piece of information I want to borrow it from the pool and
> return it when I am done with it.
> 
> I may borrow the same object multiple times and obtain the same object, be
> it in parallel or sequentially.
> 
> Suppose I do the following :
> 
> Object o1 = borrowObject("somename");
> Object o2 = borrowObject("somename");
> Object o3 = borrowObject("somename");
> returnObject(o1);
> returnObject(o2);
> returnObject(o3);
> 
> will the three objects be the same ?
no

> Object o1 = borrowObject("somename");
> returnObject(o1);
> Object o2 = borrowObject("somename");
> returnObject(o2);
> 
> will o1 == o2 ?
not guarantied, but probably yes

> Henrik


Dirk



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