Fermata "play" property does not work

Another issue: untick "Play" greys out time stretch, but the next time you click the fermata and look in the inspector, time stretch is enabled again. But this happens to other articulations as well, might start a new thread for that.

EDIT: I just checked, seems impossible to fix, that's the way connect(someOption, SIGNAL(toggled(bool)), someOtherOption, SLOT(setEnabled(bool))) works, and changing toggled to clicked won't work either :(

See https://github.com/musescore/MuseScore/pull/5446.

Fixed in branch master, commit 993d705a29

_fix #283628: fermata "play" property does not work

Resolves: https://musescore.org/node/283628._

Fixed in branch master, commit 5acccf8dd8

_Merge pull request #5446 from Howard-C/fermata-articulation

fix #283628: fermata "play" property does not work_

That other issue is not as impossible to fix as you might think. See fb5aed9.

This happens to every case concerning connection between two or more settings in the inspector. I've already searched and thus know which cases are affected, if you are willing to do the job then go ahead, if not then I can try to cover them all and add you as a collaborator ;-)

I don't mind doing it, and I don't mind letting you do it. But if you understand what I have shown you (or at least understand how to apply it to the other cases) and you already know which cases are affected, then I think it makes sense for you to do it. If you need any more help along the way, I am here.

I understand why this works, though it's so much different than what I imagined the fix (if possible) to be.
See #296415: Settings in the inspector that are supposed to be greyed out don't appear as such upon the next click on the element for the issue.

@mattmcclinch, I have a question, why use virtual for a function in the child class InspectorFermata, not the parent class InspectorElementBase?

Because Ui::InspectorFermata f is only available in class InspectorFermata.

Maybe I didn't understand the question at first. The function in the base class is already declared as virtual. Every time it is overridden elsewhere, it is also declared as virtual.

But is it essential to use virtual for functions in the child class, with no inherited class?

Essential? No. But that is how it is done all throughout the MuseScore codebase. I try to follow all aspects of the coding style as much as I can.

Ah! Thanks.

Hi! The PR was created: https://github.com/musescore/MuseScore/pull/5447.

Automatically closed -- issue fixed for 2 weeks with no activity.