[Interest] what are the requirements for QMetaType::metaObjectForType(QMetaType::type("QQuickLayout*")) to work?

[Thiago Macieira]

On Friday 3 May 2024 08:51:39 GMT-7 NIkolai Marchenko wrote:
> I am assuming the registration of qquicklayout is deferred somewhere
> but what are the exact timings and conditions?

There's no explicit registration of this type, anywhere in the sources. I 
don't even see anything that would be an automatic registration; maybe it gets 
put into a QVariant somewhere that my git grep didn't see.

But one of the unit tests assumes it exists, like your code does:
https://code.qt.io/cgit/qt/qtdeclarative.git/tree/tests/auto/qml/
qmlcppcodegen/tst_qmlcppcodegen.cpp#n1646

That means it's relying on some automatic registration somewhere, somehow.

That's poor programming practice.

-- 
Thiago Macieira - thiago.macieira (AT) intel.com
  Principal Engineer - Intel DCAI Cloud Engineering
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