How to adapt static polymorphic classes to dynamic polymorphic behavior?

"" <>
Thu, 24 Dec 2009 16:38:42 CST
Dear all,

I have problems in mixing static and dynamic polymorphism.
I need dynamic polymorphism in order to create "heterogeneous"
containers (e.g., std::vector<base*>).
Specifically, I have some "untouchable classes" (i.e., coming from
third party libraries) which represent
* random number generators
--- [code] ---
template <typename IntT, IntT a, IntT c, IntT m>
class linear_congruential
   IntT operator()() {....} // Generate a random number
template <typename UIntT, size_t w, size_t n, size_t m, size_t r,
UIntT a, size_t u, size_t s, UIntT b, size_t t, UIntT c, size_t l>
class mersenne_twister
   IntT operator()() {....} // Generate a random number
--- [/code] ---

* probability distributions
--- [code] ---
template <typename RealT>
class exponential {...};
template <typename RealT, typename GeneratorT>
RealT rand(exponential<RealT> const& d, GeneratorT& g) {...}

template <typename RealT>
class normal {...};
template <typename RealT, typename GeneratorT>
RealT rand(normal<RealT> const& d, GeneratorT& g) {...}
--- [/code] ---

Now since I need both to store different probability distributions
inside a vector and to decide what distribution to use at run-time,
I'm trying to create a set of adaptor and base classes in order to get
dynamic polymorphism.
Something like this:
--- [code] ---
namespace my {

template <typename RealT> class base_distribution {...};
template <typename DistributionT> class distribution_adaptor: public
base_distribution<typename DistributionT::real_type> {...};
} // Namespace my

int main() {
std::vector<my::base_distribution<double>*> distrs;
distrs.push_back( new distribution_adaptor<normal>(normal(0,1)) );
distrs.push_back( new distribution_adaptor<exponential>(exponential
(1)) );
--- [/code] ---

The problem is how to glue with the "rand" free function during the
iteration of distrs vector?

Since "virtual template methods" are not allowed, a possible solution
might be to bind the random generator to the definition of the
base_distribution class (and of the distribution_adaptor as well):
--- [code] ---
namespace my {

template <typename RealT, typename GeneratorT>
class base_distribution
   virtual RealT rand(GeneratorT& rng) = 0;
template <typename DistributionT, typename GeneratorT>
class distribution_adaptor: public base_distribution<typename
DistributionT::real_type, GeneratorT>
   typedef DistributionT distr_type;
   typedef typename distr_type::real_type real_type;

   real_type rand(GeneratorT& rng) { return ::my::rand(rng); }
} // Namespace my
int main()
typedef linear_congruential<...> generator_type:
std::vector<my::base_distribution<double,generator_type>*> distrs;
distrs.push_back( new distribution_adaptor<normal,generator_type>
(normal(0,1)) );
distrs.push_back( new distribution_adaptor<exponential,generator_type>
(exponential(1)) );
generator_type rng(123456);
for (
it = distrs.begin();
   it != distrs.end();
) {
--- [/code] ---

However this solution is very bad to me since it bind the random
generator to the design of the probability distribution; furthermore,
it does not allow to choose the random generator at run-time (actually
not needed, but maybe in the future)

So, do you have a more effective way to solve this problem?

Thank you very much in advance!!

NOTE: if you want I post the entire source code. I didn't do it since
this post is already very long.

Best Regards,

-- Marco

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