[Webkit-unassigned] [Bug 135311] Web Inspector: shown() called on a content view when stepping over an instruction in the debugger

[bugzilla-daemon at webkit.org]

https://bugs.webkit.org/show_bug.cgi?id=135311





--- Comment #3 from Saam Barati <sbarati at apple.com>  2014-07-29 10:56:19 PST ---
It seems like the main problem is that when stepping over an instruction, the debugger on the JSC side skips to the next instruction, and then emits the paused event. When the Inspector sees this pause event, it adds a new backForwardEntry and shows that new backForwardEntry, even if the content view it shows is for the same Resource. So, if stepping over every instruction in a JS file, there will be backForwardEntries for every instruction that's been stepped over. So clicking the back button at the top of the content browser will just show older content views that are really just the same file that we've been stepping over instructions within that file. The problem, then, doesn't seem to be that shown() is called multiple times on the same content view (because new content views for the same resource are repeatedly created), but rather, that we create new content views for every instruction we step over. This also seems bad from the user experience point of view of stepping over te

I'm not sure what the correct solution for this is, but I think the way it is set up now isn't the correct architecture. 

What do you all think?

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