Because partial specialization of a templated member function is not possible in C++ I now try to circumvent this through class template inheritance.
In this approach, I derive from the base class, specifying only the template parameters for which I want to specialize, leaving the remainder as template parameters in the derived class:

template <typename X, typename Y, typename Z>
class Base{};

template <typename Z> : public Base<int, int, Z>
class Derived{};

I wonder whether it is possible to access the specified parameters in the inheritance relation from the inside of the derived class?
I can image that you store those parameters as static constexpr in Base, thus making them available in Derived.
However, is there another way without explicitly storing them yourself as static constexpr
attributes?
Here is an example I wish to improve:

#include <cstdint> 
#include <iostream>

template<std::size_t Dimension, std::size_t Size, typename Real>
class BaseDistribution
{
    public:
        static constexpr std::size_t dim_ = Dimension;
        static constexpr std::size_t size_ = Size;
};

template<typename Real>
class DerivedDistribution : public BaseDistribution<2, 9, Real>
{
    public:
        void print_dimension()
        {
            std::cout << "Dimension: " << this->dim_ << std::endl;
        }
        
        void print_size()
        {
            std::cout << "Size: " << this->size_ << std::endl;
        }
};

int main()
{
    DerivedDistribution foo = DerivedDistribution<double>();
    
    foo.print_dimension();
    foo.print_size();
    
    return 0; 
}