Are multiple socketChannel.write() messages applied in order?

["이희승 (Trustin Lee)"]

Yes, it is an excellent answer.  Thanks Kok!

Best regards,
Trustin

On 07/02/2010 01:30 PM, Kok Hoor (GMail) wrote:
> Hi,
> 
>     I try to take a stab at answering. Hope the more senior Netty users can
> help out and clarify if I answer wrongly.
> 
>     Your Option #1 will do what you intend it to do, but Option #2 is useful
> if you send a lot of data because it will only try to send your second
> buffer when the first buffer is written successfully. This will prevent the
> send out queue from being full when you try to send out too many buffers.
> 
>     Am I correct?
> 
> Regards,
>     Kok Hoor
> 
> -----Original Message-----
> From: netty-users-bounces at lists.jboss.org
> [mailto:netty-users-bounces at lists.jboss.org] On Behalf Of Server Dude
> Sent: Friday, July 02, 2010 6:04 AM
> To: netty-users at lists.jboss.org
> Subject: Are multiple socketChannel.write() messages applied in order?
> 
> 
> I have to write multiple buffers to a socket channel in order. i.e. The
> buffer written first should be applied first. 
> 
> Can I just write code like:
> 
> option #1
> ----------
> 
> channel.write(buffer1);
> channel.write(buffer2);
> 
> or do I have to write some thing like:
> 
> option #2
> ----------
> 
> ChannelFuture future = channel.write(buffer1);
> future.addListener(new ChannelFutureListener() {
>     public void operationComplete(ChannelFuture future) throws Exception {
>         if (future.isSuccess()) {
>           future.getChannel().write(buffer2);
>         }
>     }
> });
> 
> The option #1 seem to work for me. Wanted to know if I can rely on that. The
> Socket Channel write javadoc does not tell any thing about the order of
> message application. 
> 
> The above code is simplified. In real life, these buffers are not available
> together but arrive over network. So my logic for write will reside in a
> messageReceived() method. 

-- 
what we call human nature in actuality is human habit
http://gleamynode.net/

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