Table does not exist yet. I'm expecting random access across the rowkey 
namespace. I also expect bursts of access to a given row, all of which will 
change its contents and also read it.

My question is from reading here: 
https://hbase.apache.org/book/regionserver.arch.html#block.cache
that blockcache wastes memory/CPU/GCs if there aren't many hits to it or if you 
have random writes that you typically don't read. So I was wondering if writing 
on every read had the same effect.

After reading the recent thread "Cache invalidation in Blockcache" I can see 
that during bursts the reads would be coming from the memstore and also that 
reads still need to be made to the underlying HFiles because data from both 
sources are merged together to produce a result. 

In the scenario I have, it sounds like it is useful to have the HFile blocks 
cached in the blockcache, even if the data ultimately is returned from the 
memstore due to a recent write.

Regards,
Carlos



-----Original Message-----
From: Stack [mailto:[email protected]] 
Sent: Saturday, March 29, 2014 10:17 AM
To: Hbase-User
Subject: Re: Blockcache Eviction on Write?

What is your blockcache hit rate like?  Are your gets totally random across the 
namespace or is there any locality at all?  Can you try turning it off on one 
of your regionservers and see how it does?  Why you want to turn off the block 
cache, just because it is not being used?  There are also these set of configs 
to try:
http://hbase.apache.org/xref/org/apache/hadoop/hbase/io/hfile/CacheConfig.html#40
If locality of requests, the cache on write might help?

St.Ack


On Fri, Mar 28, 2014 at 2:23 PM, Rendon, Carlos (KBB) <[email protected]>wrote:

> Hi,
>
> If I have a workload where the access pattern is writing to and 
> reading from the same row, does it make sense to turn the blockcache 
> off? In other words, reads and writes to a given row always go 
> together. Does each write cause eviction of the cached block(s) that 
> the row is stored in? I would think so, but I couldn't find an answer to this 
> anywhere online.
>
> Thanks,
> Carlos
>

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