Re: Preprocessor magic to expand template instantiation ?

"Alf P. Steinbach" <>
Fri, 07 Aug 2009 00:13:15 +0200
* Juha Nieminen:

Alf P. Steinbach wrote:

I fail to understand why the OP, and now you, cannot see the trivial
solution, which is to use a map of functions, or just a simple array.
It's on a par with using a variable where you need a variable. And
actually it is exactly that, using a variable where you need one, that's
the level of the question.

  How do you populate the map without basically replicating his Create()

There are a number of possibilites.

If the type numbers are consecutive then a recursive template can do it.

If the type numbers are more arbitrary then with each specialization of the
Factory<N> type include registration in the map.

If the map isn't used before entry of 'main', then each of the OP's Factory<N>
types can be self-registrating, like this (eror handing omited):

#include <map>

struct Type { virtual ~Type() {} };
struct Type1: Type {};
struct Type2: Type {};

typedef Type* (*FactoryFunc)();
typedef std::map<int, FactoryFunc> FactoryFuncs;

FactoryFuncs& factoryFuncs()
     static FactoryFuncs theFuncs;
     return theFuncs;

template< int N, class T >
struct RegisterFactoryFunc
     static Type* create() { return new T(); }

     static bool registerFactory()
         factoryFuncs()[N] = &create;
         return true;

     static bool const theRegistrar;

template< int N, class T >
bool const RegisterFactoryFunc<N, T>::theRegistrar =
     RegisterFactoryFunc<N, T>::registerFactory();

enum TypeNumber

template struct RegisterFactoryFunc< typeNumber1, Type1 >;
template struct RegisterFactoryFunc< typeNumber2, Type2 >;

Type* create( TypeNumber n )
     return factoryFuncs()[n]();

#include <iostream>
#include <typeinfo>
int main()
     using namespace std;
     Type* const p1 = create( typeNumber1 );
     Type* const p2 = create( typeNumber2 );

     cout << typeid(*p1).name() << endl;
     cout << typeid(*p2).name() << endl;

Cheers & hth.,

- Alf

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