如何强制转换优先于其他转换

Question

我有类：

class IntegerVector:
{
    IntegerVector operator * (const int scalar) const;
};
class RealVector:
{
    RealVector(const IntegerVector &other);
    RealVector operator * (const double scalar) const;
};

如何强制表达式：integer_vector*1.5等效于RealVector(integer_vector)*1.5，而不是现在的integer_vector*int(1.5)？

编辑

顺便说一句，有很多这样的运算符，所以定义RealVector IntegerVector::operator * (const double scalar) const不是很令人满意。

Answer 1

在C++11中，您可以利用内置的类型提升，如

#include <type_traits>
class IntegerVector;
class RealVector;
template <class T> struct VectorForType {};
template <> struct VectorForType<int> {typedef IntegerVector type;};
template <> struct VectorForType<double> {typedef RealVector type;};
// This is where we figure out what C++ would do..
template <class X, class Y> struct VectorForTypes
{
  typedef typename VectorForType<decltype(X()*Y())>::type type;
};

class IntegerVector
{
public:
    template <class T> struct ResultVector
    {
      typedef typename VectorForTypes<int, T>::type type;
    };
    template <class T>
    typename ResultVector<T>::type operator*(const T scalar) const;
};
class RealVector
{
public:
    template <class T> struct ResultVector
    {
      typedef typename VectorForTypes<double, T>::type type;
    };
    RealVector();
    RealVector(const IntegerVector &other);
    template <class T>
    typename ResultVector<T>::type operator*(const T scalar) const;
};

int main()
{
  IntegerVector v;
  auto Result=v*1.5;
  static_assert(std::is_same<decltype(Result), RealVector>::value, "Oh no!");
}

如果您需要在没有decltype的情况下这样做，您可能也可以将类型提升结果实现为元函数。我想运营商的实现看起来像这样：

template <class T> inline
typename ResultVector<T>::type IntegerVector::operator*(const T scalar) const
{
  typename ResultVector<T>::type Result(this->GetLength());
  for (std::size_t i=0; i<this->GetLength(); ++i)
    Result[i]=(*this)*scalar;
  return Result;
}

Answer 2

您可以使用这样的东西。。。这是一个解决方案，但很奇怪，但我无法发明更好的东西。

#include <iostream>
#include <type_traits>
class IntegerVector;
class RealVector
{
public:
    RealVector(const IntegerVector &other) { }
    RealVector operator * (const double scalar) const { std::cout << "RealV called" << std::endl;  return *this; }
};
class IntegerVector
{
public:
    IntegerVector operator * (const int scalar) const
    {
       std::cout << "IntV called" << std::endl;
       return *this;
    }
    template<typename T>
    typename std::conditional<std::is_same<T, int>::value, IntegerVector, RealVector>::type
    operator * (const T scalar) const
    {
       decltype(operator *<T>(scalar)) object(*this);  
       return object * scalar;
    }
};
int main()
{
   IntegerVector v;
   v * 1.5;
}

http://liveworkspace.org/code/b72cde05ca287042300f4aff0f185a42