[Development] Should qMetaTypeId<const T>() == qMetaTypeId<T>()?

[Olivier Goffart]

On Mittwoch, 24. August 2016 16:06:50 CEST Jędrzej Nowacki wrote:
> On onsdag 24. august 2016 15.23.01 CEST Marc Mutz wrote:
> > Hi,
> > 
> > Currently, it's not, which doesn't make much sense, does it?
> > 
> > So,
> > 
> >   template <typename T>
> >   int qMetyTypeId<T>() {
> >     return qMetaTypeIdHelper<typename std::remove_cv<T>::type>();
> >   }
> > ?
> > 
> > (There's of course a lot more involved in this, registration should
> > discard const, too, e.g.).

I think it make sens, but it should be done at the QMetaTypeId2 Level.
We already do this for const T&, See commit 
03b25125986f083a260b421abeed1f1d088d27b3
In fact, the same should be done for const T

> From C++ perspective const T and T are kind of separate types, metatype
> mimics that. QMetaType is used in many places and in some constness
> matters.
> 
> Consider that example:
>  QMetaType::typeName(qMetaTypeId<const QString>())
> and
>  QMetaType::typeName(qMetaTypeId<QString>())
> The function is used in metaobject to compute signatures of invokables.

I can prove you wrong easily,  both have the same result, since
QMetaType::normalizedType("const QString") == "QString"

They are different C++ type, yes,  but from a practical purpose, their use in 
the metatype system is exactly the same.  You allocate, destroy or copy a 
const QString the same way you do a QString.

-- 
Olivier

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