[Development] Changed enum property behaviour in Qt v6.8

[Phil Thompson]

On 04/10/2024 15:32, Thiago Macieira wrote:
> On Friday 4 October 2024 05:43:38 GMT-7 Phil Thompson via Development 
> wrote:
>> For dynamically created meta-objects...
>> 
>> - the results were dependent on the version of Qt
>> - QMetaProperty.metaType().isValid() always returned false
>> - QMetaProperty.isEnumType() returned true for v6.7.1 and false for
>> v6.8.0
> 
> But how was this constructed? Is it setting the QMP and QME's metatype 
> using
> QMetaEnumBuilder::setMetaType and QMetaPropertyBuilder::setMetaType? 
> Uh...
> there is no such function for QMPB.
> 
>> I was surprised by the result of QMetaProperty.isEnumType() for a
>> moc-generated meta-object without Q_ENUM. The docs suggest this should
>> be true for all enums whether or not Q_ENUM is used.
> 
> No, it needs the bit set by moc, so Q_ENUM or Q_FLAG is required. The 
> doc may
> be misleading then ("if the property's type is an enumeration value").
> 
>> Why does QMetaProperty.metaType().isValid() always return true for
>> moc-generated meta-objects and false for dynamically created
>> meta-objects?
> 
> Because meta types for all properties are required since 6.0. There are 
> no
> exceptions. Since 6.6, moc also generates the metatypes for all enums 
> too.
> 
> QMetaObjectBuilder always generates the latest revision of the meta 
> object, so
> all features are available... but also all requirements are mandatory. 
> Meta
> types for all properties, enums, methods , and for gadgets the gadget 
> itself
> must be provided.
> 
>> Is there any way to dynamically register an enum with the meta-type
>> system (so that QMetaEnum.metaType().isValid() returns true)?
> 
> Sure. And registration is automatic anyway.
> 
> All you should need to do is pass the necessary QMetaType

Understood - but there seems to be no obvious way to dynamically create 
and register a valid QMetaType.

Phil