Qt Q_ENUM property from QVariant(qulonglong). - qt

Question

Consider the following Qt code:
class Foo : public QObject {
Q_OBJECT
Q_ENUMS(E)
Q_PROPERTY(E x READ x WRITE set_x)
public:
enum E {
a = 0,
b = 1,
c = 2
};
E x() const { return x_; }
void set_x(E value) { x_ = value; }
private:
E x_;
};
int main (int argc, char **argv) {
QCoreApplication app(argc, argv);
Foo f;
f.setProperty("x", Foo::c);
std::cout << f.property("x").toInt() << std::endl; // 2
f.setProperty("x", QVariant((int)1));
std::cout << f.property("x").toInt() << std::endl; // 1
f.setProperty("x", QVariant((long long)0));
std::cout << f.property("x").toInt() << std::endl; // should be 0. is 1.
}
Why does it work that way?

Answer 1

If you test the return value of setProperty, you will see that the set is failing:
ok = f.setProperty("x", QVariant((long long)0));
std::cout << ok << std::endl; // 0, i.e. false
The relevant part of the Qt code is in qmetaobject.cpp, annotated below:
if (isEnumType()) {
if (v.type() == QVariant::String) {
// ... we won't get here.
} else if (v.type() != QVariant::Int && v.type() != QVariant::UInt) {
// We got here because we didn't provide an int or uint.
// This will be 0...
int enumMetaTypeId = QMetaType::type(qualifiedName(menum));
// ... which means this will return false; the property will not be set.
if ((enumMetaTypeId == 0) ||
(v.userType() != enumMetaTypeId) ||
!v.constData())
return false;
// ... we never get here
}
}
// ... we never get here
So the behavior seems to be by design: enum properties can only be set using QVariant objects with a type of int or uint.