My problem is this. I have lists of different numeric types, for example:
QList<qreal> mylist;
Now, in my code I have a function that that expects a QVariant argument which is mylist. The only way I've found to do this is by using a for cyle and simply adding all of the data in mylist to second list
QList<QVariant> temp,
for example, and passing temp as a paramter.
I was wondering if there was any other way to do this.
Thank you very much.
I've had to do this several times, and to my knowledge the only way is to create a new list.
If this is something you have to do frequently with varying types of lists you could do something a little fancier:
template <typename T>
QVariantList toVariantList( const QList<T> &list )
{
QVariantList newList;
foreach( const T &item, list )
newList << item;
return newList;
}
so that when you call your function that takes in the QVariant list, you can just call
myFunction( toVariantList(myList) );
Note that the given function will only work for types that can be implicitly converted to QVariants.
来源:https://stackoverflow.com/questions/9265288/casting-a-list-as-qvariant-or-qvariant-list