Question

Is it possible to write a class:

template<typename T, ... name> struct Magic {
  T name;
};

such that:

Magic<int, foo> gives:

Magic<int, foo> {
  int foo;
}

and

Magic<float, bar> gives:

Magic<float, bar> {
  float bar;
}

Basically, I want to be able to specify not only the Type, but also the name of the member variables.

Answer 1

That is not possible, you have to resort to either macro-based solutions or use a predefined set of types that provide named members.

A possible macro-based approach:

#define MAGIC(name_) \
    template<typename T> struct Magic1 { \
        T name_; \
    };

MAGIC(foo);

or:

#define MAGIC(type_, name_) \
    struct Magic1 { \
        type_ name_; \
    };

MAGIC(foo);

Using preprocessor magic, e.g. utilizing Boost.Preprocessor, you should be able to generate n named members in a more convenient way.

Another approach might be using a predefined set of classes providing certain named members from that you inherit:

enum { MemberNameFoo, MemberNameBar };

template<class T, int id>
struct named_member;

template<class T>
struct named_member<T, MemberNameFoo> {
    T foo;
};

template<class T>
struct named_member<T, MemberNameBar> {
    T bar;
};

// holder for the above, just one member for this example:

template<class T, int name>
struct holder : named_member<T, name> {};

// using it:

typedef holder<int, MemberNameFoo> HasFoo;
typedef holder<int, MemberNameBar> HasBar;

Using compile-time lists you could then inherit from n named_member instantiations, Boost.MPL could help here.