How do I make a class whose interface matches double, but upon which templates can be specialized?

Question

How do I make a class whose interface matches double, but whose templated types do not dynamic cast to double?

The reason is that I have a run-time type system, and I want to be able to have a type that works just like double:

template<int min_value, int max_value>
class BoundedDouble: public double {};

And then use template specialization to get run-time information about that type:

template<typename T>
class Type { etc. }

template<int min_value, int max_value>
class Type<BoundedDouble<min_value, max_value>> { int min() const { return min_value; } etc. }

But, you can't inherit from double...

Answer 1

You can't derive from native types. Use composition instead:

#include <cstdlib>
#include <string>
#include <stdexcept>
#include <iostream>
using namespace std;

template<typename Type = double, const Type& Min = -10.0, const Type& Max = 10.0> class Bounded
{
public:
    Bounded() {};
    Bounded(const Type& rhs) : val_(rhs) 
    { 
        if(rhs > Max || rhs < Min) 
            throw logic_error("Out Of Bounds"); 
    }

    operator Type () const 
    { 
        return val_; 
    }

    Type val_;
};


int main()
{
    typedef Bounded<double, -10.0, 10.0> double_10;
    double_10 d(-4.2);
    cout << "d = " << d << "\n";
    double d_prime = d;
    cout << "d_prime = " << d_prime << "\n";
    double_10 d2(-42.0);
    cout << "d2 = " << d << "\n";


    return 0;
}

The output is:

d = -4.2
d_prime = -4.2