使用QVariant ::在交换机块中键入的用户类型的GCC警告

Question

Basically we're keeping the track of object types in a variable of QVariant::Type and then doing something like that

switch(values[i].type)
{
case QVariant::Bool: logic; break;
case QVariant::Int: logic; break;
case QVariant::LongLong: logic; break;
case QVariant::String:  logic; break;
case QVariant::Double:  logic; break;
case QVariant::DateTime:
case QVariant::Date:
case QVariant::Time:   logic; break;
case QVariant::User+1:
{
    logic;break;
}
case QVariant::User+2:
{
    logic;break;
}
default: break;
}

The problem is: gcc produces warnings along these lines for User+X statements:

 warning: case value ‘1025’ not in enumerated type ‘QVariant::Type’ [-Wswitch]

Now, I could suppress that of course, but is that the recommended way or am I doing something fundamentally wrong here?

PS The question isn't about why the warning is produced: I understand why. The queston is more about using QVariant::Type with user types in a proper way and if simply suppressing is correct or what am I doing here is just plain wrong and warning is an indication of a bigger design decision problem.

Answer 1

If you are going to use user types in the QVariant, you should use the member function userType :

Unfortunately, this will make things uglier:

auto u = [](auto& v){return static_cast<int>(v);};
switch(values[i].userType())
{
case u(QVariant::Bool): logic; break;
case u(QVariant::Int): logic; break;
case u(QVariant::LongLong): logic; break;
case u(QVariant::String):  logic; break;
case u(QVariant::Double):  logic; break;
case u(QVariant::DateTime):
case u(QVariant::Date):
case u(QVariant::Time):   logic; break;
case u(QVariant::User)+1:
{
    logic;break;
}
case u(QVariant::User)+2:
{
    logic;break;
}
default: break;
}

To clarify: Note that the type() member function will always return QVariant::UserType when the type value is greater than QMetaType::User . This means it will never return QVariant::UserType+n ! How could it? Those values are not part of the enumeration.

This is what gcc is warning you about:

warning: case value ‘1025’ not in enumerated type ‘QVariant::Type’ [-Wswitch]

Since 1025 ( QVariant::User+1 ) is not part of QVariant::Type , this case label is essentially dead code.

The userType member function, however, returns int . It will return the same underlying value as type when that value is below QMetaType::User , and also the correct user value when above.

Answer 2

From QVariant::type enum :

QMetaType::User 1024    Base value for user types

this is like:

switch(1024) {
   ...
   case (1024 + 1): break;
   default: break;
}

the governing rule in switch for enum is to be explicit .

(1024 + 1) is not explicit thus the warning.

UPDATE

As per qvariant.html#type

Returns the storage type of the value stored in the variant. Although this function is declared as returning QVariant::Type, the return value should be interpreted as QMetaType::Type. In particular, QVariant::UserType is returned here only if the value is equal or greater than QMetaType::User.

Thus, QVariant::UserType + 1 will be dead code.

Time to refactor