[Development] Qt 6 API (change) review

[Lars Knoll]

> On 7 Sep 2020, at 12:10, Edward Welbourne <edward.welbourne at qt.io> wrote:
> 
> Alex Blasche (7 September 2020 09:58)
>> I suggest we exclude reviews for the modules which are not part of Qt
>> 6. It greatly reduces the effort. When a module comes back
>> (e.g. during 6.2) then we have to run a complete 5.15..6.2 review
>> anyway.
> 
> Sounds fair enough.  At first sight.
> Now I just need to know which modules those are.

Check qt5.git/.gitmodules and only include the repos that don’t have 
"status = ignore”.

Cheers,
Lars

> I dimly remember a wiki page about Qt6-readiness.
> 
> In the course of searching for that, I stumbled on
> https://wiki.qt.io/New_Features_in_Qt_6.0
> Given that FF has passed, please add your features here.
> It is currently blank.
> 
> After a little rummaging I find:
> https://wiki.qt.io/Checklist_for_Qt_6.0_inclusion
> 
> It is in a category, but that category is all on its own, with just this
> one page in it, unconnected to the rest of our category hierarchy.
> Which means this page cannot be found by chasing the categories I
> normally search in.  Please, when adding things to categories, think
> about whether that category can be found: if it can't, then it's no use
> at all for those who want to find the page.  The habit of starting each
> category name with a prefix and double-colon *does not* automagically
> cause it to be made a sub-category of the thing before the last
> double-colon in its name: there is *no* magic structure to wiki
> categories, only the structure you actually create.
> I have connected this category to the existing hierarchy.
> 
> However, this page uses "module" in the sense of libraries, not git
> modules, so it doesn't actually tell me which *git* modules to skip.
> Furthermore, it includes Qt OPCUA, whic isn't in qt5/.gitmodules, so
> doesn't match the content of Qt6.
> 
> So no, I can't find the information necessary to exclude modules based
> on whether they're to be included in Qt 6.  Where is it ?
> 
> 	Eddy.
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