Bug 220627

Summary: [WK2] Bypass NetworkProcess for MessagePort communication within a single WebProcess
Product: WebKit Reporter: Chris Dumez <cdumez>
Component: WebKit2Assignee: Chris Dumez <cdumez>
Status: RESOLVED WONTFIX    
Severity: Normal CC: annulen, esprehn+autocc, ews-watchlist, gyuyoung.kim, kangil.han, ryuan.choi, sergio, ysuzuki
Priority: P2    
Version: WebKit Nightly Build   
Hardware: Unspecified   
OS: Unspecified   
See Also: https://bugs.webkit.org/show_bug.cgi?id=181454
Bug Depends on:    
Bug Blocks: 220038    
Attachments:
Description Flags
WIP Patch
none
Patch
ews-feeder: commit-queue-
Patch none

Description Chris Dumez 2021-01-14 11:46:43 PST 
Bypass NetworkProcess for MessagePort communication within a single WebProcess. When the 2 MessagePorts are within the same WebProcess, there is no reason to go via the NetworkProcess to do the communication. Doing so is bad for performance and also loses functionality because SerializedScriptValue is not able to encode everything over IPC (e.g. Wasm modules, see Bug 220038).
Comment 1 Chris Dumez 2021-01-14 11:48:19 PST 
Created attachment 417635 [details]
WIP Patch
Comment 2 Chris Dumez 2021-01-14 11:53:21 PST 
Created attachment 417636 [details]
Patch
Comment 3 Chris Dumez 2021-01-14 12:06:58 PST 
Created attachment 417639 [details]
Patch
Comment 4 Chris Dumez 2021-01-14 13:25:28 PST 
Comment on attachment 417639 [details]
Patch

Sadly, this is not as easy as it looks. Current patch breaks cross-process communication (used by service workers). A given WebProcess does not seem to know if its remote MessagePort is within its process or not.
Comment 5 Chris Dumez 2021-01-14 16:50:44 PST 
I tried a different approach but it is tricky. At the time a message is posted, we may not know yet if the destination MessagePort will be in the same WebProcess or not.

You can send a MessagePort to a worker and call postMessage() on this MessagePort, before the Worker has received the port (so before entanglement has happened). Entanglement is when we know if both ends of the channel are in the same process.