[whatwg] WebSocket bufferedAmount includes overhead or not.

[Alexey Proskuryakov]

On 05.03.2010, at 10:27, Olli Pettay wrote:

>> I was going to mention this as the primary reason why frame bytes  
>> should
>> be included. JavaScript code needs this information for flow control,
> Why?

I assume you are asking why JavaScript code needs this information for  
flow control.

My recollection is that feature was added as a result of discussions  
about implementing flow control. How else are you supposed to know  
that you're streaming too fast without modifying the server? Since  
WebSockets is a match for TCP/IP, and the latter provides ways to  
adaptively change data rate, it's natural that one expects the same  
from WebSockets.

>> and it's raw bytes that are sent over the tubes, not original message
>> strings.
> Right. But this is about the API. I assume the underlying protocol may
> change or the API can eventually support different kinds of protocols
> (some may use UPD, some TCP, some send text, some binary).
> The API usage should be still the same, if possible.

This is something we agree about.

I guess the root of our disagreement is in how one uses the API. I'm  
saying that the interesting question is how many bytes are buffered to  
be sent over the wire, so in order to keep the API usage the same we  
need to include protocol overhead in this number.

>> In WebKit, we'd have to queue
>> unsent messages separately just to implement this quirk (see
>> https://bugs.webkit.org/attachment.cgi?id=50093 for a proof of  
>> concept).
>> It becomes very difficult to implement we decide to add size of data
>> that an underlying network library buffers internally - which I think
>> would be a reasonable thing to do.
> I don't see why that would be difficult. If you know you have just  
> written x bytes to the whatever network method, you know how many  
> bytes
> of those were frame markers.


That's true, but I don't know how many of these have already been sent  
unless I perform lots of additional bookkeeping. Consider sending  
"data" message three times in a row:

\x00data\xFF\x00data\xFF\x00data\xFF

If we are to exclude protocol overhead in bufferedAmount, and we know  
that there are 8 bytes still queued (a\xFF\x00data\xFF), and we know  
that there were three frames sent (with an overhead of 6 bytes) how  
would we know that the answer is 5?

- WBR, Alexey Proskuryakov