c++ - operator+ 的模糊重载

Question

Boost 数学库中有多项式类:Boost polynomial class .我想通过添加新函数来扩展这个类的能力，我使用继承如下:

#ifndef POLY_HPP
#define POLY_HPP

#include <boost/math/tools/polynomial.hpp>

template <class T>
class Poly : public boost::math::tools::polynomial<T>{
public:

    Poly(const T* data, unsigned order) : boost::math::tools::polynomial<T>(data, order){

    }
};

#endif 

现在我声明了这个类的两个对象，我想添加它们:

    int a[3] = {2, 1, 3};
    Poly<int> poly(a, 2);
    int b[2] = {3, 1};
    Poly<int> poly2(b, 1);
    std::cout << (poly + poly2) << std::endl;

但是编译时出现错误:

main.cpp: In function ‘int main()’:
main.cpp:28:26: error: ambiguous overload for ‘operator+’ in ‘poly + poly2’
/usr/local/include/boost/math/tools/polynomial.hpp:280:22: note: candidates are: boost::math::tools::polynomial<T> boost::math::tools::operator+(const U&, const boost::math::tools::polynomial<T>&) [with U = Poly<int>, T = int]
/usr/local/include/boost/math/tools/polynomial.hpp:256:22: note:                 boost::math::tools::polynomial<T> boost::math::tools::operator+(const boost::math::tools::polynomial<T>&, const U&) [with T = int, U = Poly<int>]
/usr/local/include/boost/math/tools/polynomial.hpp:232:22: note:                 boost::math::tools::polynomial<T> boost::math::tools::operator+(const boost::math::tools::polynomial<T>&, const boost::math::tools::polynomial<T>&) [with T = int]
make[2]: Leaving

operator+定义了三个重载函数。我认为应该采取:

boost::math::tools::polynomial<T> boost::math::tools::operator+(const boost::math::tools::polynomial<T>&, const boost::math::tools::polynomial<T>&)

因为 Poly 类继承自 Boost 多项式并且参数传递最好，但是它不会发生。如何在没有明确定义 operator+ 的情况下添加两个 Poly 类对象？

Answer 1

据我所知这是不可能的，你需要类似的东西

template <class T>
Poly<T> operator + (const Poly<T>& a, const Poly<T>& b) {
    return Poly<T>(static_cast< boost::math::tools::polynomial<T> >(a) +
                   static_cast< boost::math::tools::polynomial<T> >(b));
}

消除调用的歧义(在您的类中需要一个从 boost::math::tools::polynomial<T> 到 Poly<T> 的适当转换构造函数...)