Specifically, Area is not a template class, but a class template. That is, it is a template from which classes can be generated. Area<int> is such a class (it's not an object, but of course you can create an object from that class in the same ways you can create objects from any other class).
A derived class can have only one direct base class. However, inheritance is transitive. If ClassC is derived from ClassB , and ClassB is derived from ClassA , ClassC inherits the members declared in ClassB and ClassA . Structs do not support inheritance, but they can implement interfaces.
Templates provide sortability in a much nicer way, since the sortee doesn't need to know that it's being sorted. Complex inheritance typically results when you work with a forced mindset that everything must be an inheritance hierarchy, which is in fact rarely appropriate.
For understanding templates, it's of huge advantage to get the terminology straight because the way you speak about them determines the way to think about them.
Specifically, Area
is not a template class, but a class template. That is, it is a template from which classes can be generated. Area<int>
is such a class (it's not an object, but of course you can create an object from that class in the same ways you can create objects from any other class). Another such class would be Area<char>
. Note that those are completely different classes, which have nothing in common except for the fact that they were generated from the same class template.
Since Area
is not a class, you cannot derive the class Rectangle
from it. You only can derive a class from another class (or several of them). Since Area<int>
is a class, you could, for example, derive Rectangle
from it:
class Rectangle:
public Area<int>
{
// ...
};
Since Area<int>
and Area<char>
are different classes, you can even derive from both at the same time (however when accessing members of them, you'll have to deal with ambiguities):
class Rectangle:
public Area<int>,
public Area<char>
{
// ...
};
However you have to specify which classed to derive from when you define Rectangle
. This is true no matter whether those classes are generated from a template or not. Two objects of the same class simply cannot have different inheritance hierarchies.
What you can do is to make Rectangle
a template as well. If you write
template<typename T> class Rectangle:
public Area<T>
{
// ...
};
You have a template Rectangle
from which you can get a class Rectangle<int>
which derives from Area<int>
, and a different class Rectangle<char>
which derives from Area<char>
.
It may be that you want to have a single type Rectangle
so that you can pass all sorts of Rectangle
to the same function (which itself doesn't need to know the Area type). Since the Rectangle<T>
classes generated by instantiating the template Rectangle
are formally independent of each other, it doesn't work that way. However you can make use of multiple inheritance here:
class Rectangle // not inheriting from any Area type
{
// Area independent interface
};
template<typename T> class SpecificRectangle:
public Rectangle,
public Area<T>
{
// Area dependent stuff
};
void foo(Rectangle&); // A function which works with generic rectangles
int main()
{
SpecificRectangle<int> intrect;
foo(intrect);
SpecificRectangle<char> charrect;
foo(charrect);
}
If it is important that your generic Rectangle
is derived from a generic Area
you can do the same trick with Area
too:
class Area
{
// generic Area interface
};
class Rectangle:
public virtual Area // virtual because of "diamond inheritance"
{
// generic rectangle interface
};
template<typename T> class SpecificArea:
public virtual Area
{
// specific implementation of Area for type T
};
template<typename T> class SpecificRectangle:
public Rectangle, // maybe this should be virtual as well, in case the hierarchy is extended later
public SpecificArea<T> // no virtual inheritance needed here
{
// specific implementation of Rectangle for type T
};
Are you just trying to derive from Area<int>
? In which case you do this:
class Rectangle : public Area<int>
{
// ...
};
EDIT: Following the clarification, it seems you're actually trying to make Rectangle
a template as well, in which case the following should work:
template <typename T>
class Rectangle : public Area<T>
{
// ...
};
class Rectangle : public Area<int> {
};
Make Rectangle a template and pass the typename on to Area:
template <typename T>
class Rectangle : public Area<T>
{
};
Rectangle
will have to be a template, otherwise it is just one type. It cannot be a non-template whilst its base magically is. (Its base may be a template instantiation, though you seem to want to maintain the base's functionality as a template.)
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