Issue 393 in memcached: why did we execute function of stop_assoc_maintenance_thread() before the process exit.

2015-01-18 Thread memcached

Status: New
Owner: 
Labels: Type-Defect Priority-Medium

New issue 393 by cfrees...@gmail.com: why did we execute function of  
stop_assoc_maintenance_thread() before the process exit.

https://code.google.com/p/memcached/issues/detail?id=393

In the main function, why did we execute function of  
stop_assoc_maintenance_thread() before the process exit?




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Re: Issue 393 in memcached: why did we execute function of stop_assoc_maintenance_thread() before the process exit.

2015-01-18 Thread memcached

Updates:
Status: Invalid

Comment #1 on issue 393 by dorma...@rydia.net: why did we execute function  
of stop_assoc_maintenance_thread() before the process exit.

https://code.google.com/p/memcached/issues/detail?id=393

y u file bugs instead of mailing list :(

That's old and relatively incomplete code. If memcached has a graceful way  
of stopping itself (and more/all of the global variables are removed) it  
can be embedded/etc. So it tries to be graceful in a few places.


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Alignment in the binary protocol

2015-01-18 Thread Martin Martin
I'm looking to implement a client for the binary protocol.  I take it there 
is no padding between memcached requests, and that the keys are values are 
arbitrary lengths.  So, if I send two requests, and the key for the first 
one is an odd number of bytes, that means the second request will start on 
an odd byte, and the header will be unaligned?

For example, the memcached binary protocol description here uses 'Hello' as 
the key in the example for GET, 

https://code.google.com/p/memcached/wiki/MemcacheBinaryProtocol

so the entire request is 29 bytes long.  This means, if I want to send a 
second request in the same buffer, I need to start the second header at 
offset 29 in my buffer, correct?

The NFS protocol, for example, specifies that any variable length field is 
padded with 0-3 zeros, so that the next field can start on a 32 bit 
boundary.

This isn't a problem for me, but could be for some:

http://blogs.msdn.com/b/oldnewthing/archive/2015/01/16/10586692.aspx


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