Deriving and Template typedef

[LuB]

Not sure why this won't compile ... Is 'DataTypeT' not visible to Base?

I'd like to pass a data type to a base class ... with the data type
visible via typedef in a derived class.



#include <iostream>

template<typename T>
struct Base
{

Base(typename T:ataTypeT b)
{
std::cout << "DataValue: " << b << std::endl;
}

};

struct Derived : public Base<Derived>
{
typedef int DataTypeT;

Derived(int d) : Base<Derived>(d) { }

};

int
main(int argc, char** argv)
{
Derived d(5);
return 0;
}


Thanks,

-Luther

 

[benben]

Not sure why this won't compile ... Is 'DataTypeT' not visible to Base?

I'd like to pass a data type to a base class ... with the data type
visible via typedef in a derived class.



#include <iostream>

template<typename T>
struct Base
{

Base(typename T:ataTypeT b)
{
std::cout << "DataValue: " << b << std::endl;
}

};

struct Derived : public Base<Derived>
{
typedef int DataTypeT;

Derived(int d) : Base<Derived>(d) { }

The type, Derived, is not complete up to this point. You can use a
pointer to Derived, but not itself.

};

int
main(int argc, char** argv)
{
Derived d(5);
return 0;
}

But why over complicate the design? Go simple:

template <typename T>
struct Base
{
typedef T data_type;
Base(T b){/*...*/}
};


struct Derived: public Base<int>
{

Thanks,

-Luther

Regards,
Ben

 

[LuB]

The type, Derived, is not complete up to this point. You can use a
pointer to Derived, but not itself.


But why over complicate the design? Go simple:

template <typename T>
struct Base
{
typedef T data_type;
Base(T b){/*...*/}
};


struct Derived: public Base<int>
{


Regards,
Ben

That is indeed what I have decided to do.

In this particular case, I am already passing 3 template parameters to
Base - so just trying to avoid another one and didn't understand why
that was breaking.

Many thanks for your input,

-Luther