Switch statement on C++ type in Qt

constexpr-type-label.cpp

constexpr quint64 label(const char *const s, size_t shift = 0)
{
    if (s[0] == '\0')
        return 0; // end of string
    if (shift == 56)
        return 0; // avoid integer overflow

    return (static_cast<quint64>(s[0]) << shift) ^ label(&s[1], shift + 1);
}

template<typename T>
constexpr auto label()
{
    return label(QMetaType::fromType<T>().name());
}

constexpr auto label(QMetaType metaType)
{
    return label(metaType.name());
}

static_assert(label<int>());
static_assert(label<int>() != label<uint>());

void demo(const QObject *obj)
{
    switch (label(obj->metaObject()->metaType())) {
    case label<QLabel>():
        qInfo() << "This is a label";
        break;
        
    case label<QButton>():
        qInfo() << "This is a button";
        break;
    }
}

Why switch instead of the usual if-else-forest of dynamic casts?

• Constant-time execution: It's a single lookup, instead a random number of casts, depending on how late the matching cast can be found in your if-else-forest.
• The compiler detects duplicate in switch statements, which is not supported for if-else-forests.