Member data in class derived from template

[Matthias von Faber]

Hello,

The code below compiles fine on VS 6.0 and .NET 2005, but G++ complains
about mData not having been declared in the Derived class:



template<typename T>
class Base
{
public:
Base() {}
virtual ~Base() {}

protected:
int mData;
};


template<typename T>
class Derived : public Base<T>
{
public:
Derived() {}
virtual ~Derived() {}

void f()
{
mData = 0;
}
};


I have done extensive searches on the problem, including books etc., but I
have found nothing on the topic. So I am beginning to feel that I might be
fundamentally confusing something...

But then again, a Derived<T> should be a Base<T>, and those should come
with an mData member. So what could it be?

Thanks,

 

[Rolf Magnus]

Hello,

The code below compiles fine on VS 6.0 and .NET 2005, but G++ complains
about mData not having been declared in the Derived class:



template<typename T>
class Base
{
public:
Base() {}
virtual ~Base() {}

protected:
int mData;
};


template<typename T>
class Derived : public Base<T>
{
public:
Derived() {}
virtual ~Derived() {}

void f()
{
mData = 0;
}
};


I have done extensive searches on the problem, including books etc., but I
have found nothing on the topic. So I am beginning to feel that I might be
fundamentally confusing something...

But then again, a Derived<T> should be a Base<T>, and those should come
with an mData member. So what could it be?

Two-phase name loopkup. The name has to be a dependant name. Since the Base
template is not instantiated at the point of Derived's definition, the
compiler doesn't know yet that mData is supposed to be a member of the
object.
Try:

void f()
{
this->mData = 0;
}

 

[Pete Becker]

But then again, a Derived<T> should be a Base<T>, and those should come
with an mData member. So what could it be?

But Base<int>, for example, can be specialized, and it isn't required to
have a member named mData. So you have to tell the compiler where mData
comes from. Instead of the unadorned

mData = 0;

you can say

this->mData = 0;
Base<T>::mData = 0;

and I think there's a third one, that isn't popping into my head at the
moment.

 

[Alf P. Steinbach]

* Pete Becker:

But Base<int>, for example, can be specialized, and it isn't required to
have a member named mData. So you have to tell the compiler where mData
comes from. Instead of the unadorned

mData = 0;

you can say

this->mData = 0;
Base<T>::mData = 0;

and I think there's a third one, that isn't popping into my head at the
moment.

You're probably thinking of a using-declaration.

 

[Matthias von Faber]

this->mData = 0;

That fixed it nicely. Well, seems that I should read up on class templates
then. My conception was that all Base specializations would have an mData.

Thanks very much!