我如何做一个类的接口匹配双，但在哪个模板可以专门？ [英] How do I make a class whose interface matches double, but upon which templates can be specialized?

问题描述

如何创建一个类的接口匹配double，但其模板类型不能动态转换为双精度类？

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

原因是我有一个运行时类型系统，我希望能够有类似double的类型：

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 {};

然后使用模板专门化来获取有关该类型的运行时信息：

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. }

但是，你不能继承双重...

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

推荐答案

您不能从本机类型派生。请改用组合：

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;
}

输出为：

d = -4.2
d_prime = -4.2

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