another template question

Is there any way to be able to do this:

Code:

---

class record : public base_record<record::pair_type> {
public:
        typedef std::pair<std::string, int> pair_type;
        record() { }
};

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I know pair_type typedef should be defined before using it.. Just wondering if this is possible?

Thanks a lot for help

Use typename?

Code:

---

class record : public base_record<typename record::pair_type> {

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I think that would work.

It doesnt.

I fiddled around with Dinkumware a little, but couldn't find a way to do this. Any variations, like this,

Code:

---

#include <iostream>
#include <string>

template <typename t>
class base_record {};

class record;
typedef std::pair<std::string, int> record::pair_type;

class record : public base_record<typename record::pair_type> {
public:
    //typedef std::pair<std::string, int> pair_type;
    record() { }
};

int main() {
    return 0;
}

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cause this sort of error:

Code:

---

sourceFile.cpp(8) : error C2027: use of undefined type 'record'
        sourceFile.cpp(7) : see declaration of 'record'

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I would suspect that you can't do this; however, I'm not sure. Maybe someone more knowledgeable about C++ will be able to tell you.

Impossible, sorry. But if you want to avoid redundant typing, perhaps you can put the pair_type typedef in the base class and pass the full type as a template parameter?

Code:

---

template<class Pair>
class record_base
{
public:
  typedef Pair pair_type;
};

class record : public record_base< std::pair<...> >
{
};

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