- From: Alexey Proskuryakov <ap@webkit.org>
- Date: Fri, 05 Mar 2010 12:39:14 -0800
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
Received on Friday, 5 March 2010 12:39:14 UTC