C++:是否有一种不那么冗长的方式来表示自定义数字类型？

Question

我需要代表Volts,Ampers和Watts及其关系(例如W = V*I)..

这就是我的意思,但看起来真的很冗长.有关如何使其更简洁的任何想法？

#include <stdio.h>
#include <iostream>


template<class T>    
class double_val {   
public:
    explicit double_val(double val):_val(val) { };
    double_val(const T& other) {
        _val = other._val;
    }
    T& operator=(const T &other) {
        _val = other._val;
        return *this;
    }
    T operator+ (const T& other) const {
        return T(this->_val + other._val);
    }
    T operator+ (double val) const {
        return T(this->_val + val);
    }
    T operator- (const T& other) const {
        return T(this->_val - other._val);
    }
    T operator- (double val) const {
        return T(this->_val - val);
    }
    T operator* (const T& other) const {
        return T(this->_val * other._val);
    }
    T operator* (double val) const {
        return T(this->_val * val);
    }
    T operator/ (const T& other) const {
        return T(this->_val / other._val);
    }
    T operator/ (double val) const {
        return T(this->_val / val);
    }
    bool operator== (const T& other) const{
        return this->_val == other._val;
    }
    bool operator!= (const T& other) const{
        return this->_val != other._val;
    }
    bool operator > (const T& other) const{
        return this->_val > other._val;
    }
    bool operator >= (const T& other) const{
        return this->_val >= other._val;
    }
    bool operator < (const T& other) const{
        return this->_val < other._val;
    }
    bool operator <= (const T& other) const{
        return this->_val <= other._val;
    }
    void val(double val){
        _val = val;
    }
    double val() const {
        return _val ;
    }
    virtual const char* name() const = 0;

private:
    double _val;
};
template<class T>
std::ostream& operator<<(std::ostream &os, const double_val<T> &t) {
    return os << t.val() <<  " " << t.name();
}


class Amper:public double_val<Amper> {
public:
    Amper(double val):double_val<Amper>(val) {}
    const char* name() const {
        return "Amper";
    }
};

class Volt:public double_val<Volt> {
public:
    Volt(double val):double_val<Volt>(val) {}
    const char* name() const {
        return "Volt";
    }
};

class Watt:public double_val<Watt> {
public:
    Watt(double val):double_val<Watt>(val) {}
    const char* name() const {
        return "Watt";
    }
};

// Watt = I * V
Watt operator* (const Volt &v, const Amper& a) {
    return Watt(a.val() * v.val());
}
Watt operator* (const Amper &a, const Volt& v) {
    return Watt(a.val() * v.val());
}

// Volts = Watts/Ampers
Volt operator / (const Watt &w, const Amper& a) {
    return Volt(w.val() / a.val());
}

// Ampers = Watts/Volts
Amper operator / (const Watt &w, const Volt& v) {
    return Amper(w.val() / v.val());
}



int main(int argc, char **argv) {
    using namespace std;
    Watt w = Volt(66) * Amper(7);
    Amper a = w / (Volt(646) * Volt(444));

    cout << a << endl;

    return 0;
}

(注意:我故意省略了operator double(),因为我想避免像Volt(4)+ Watt(6)这样的禁止操作

Answer 1

在库形式中确实有一种不那么冗长的方式,Boost.Units将SI单位定义为类型安全类模板实例化.

特别是,它定义了单位volt(s),ampere(s)和watt(s).