从带有操作符重载的模板化类继承。 “运算符*”的模棱两可的重载

Question

出于学习目的，我正在尝试使用模板化类和继承。 其中一件事情是，我希望我的类能够使用+= ， + ， -= ， - ， * ， *=等运算符。为此，我开发了一个类，该类基本上是数据：

template<typename T, int N>
class Array
{
public:
    // Constructor
    Array( T value = T(0))
    {
        /* Initialize the array with the same value everywhere */
    }
    template<typename M>
    Array(const Array<M,N>& other)
    {
        /* Copy constructor allowing for type change */
    }

    // Destructor
    ~Array()
    {
        /* no need to call any destructor since it is a smart pointer */    
    }

    // Product
    template<typename M>
    Array<T,N>& operator*=(const M& rhs)
    {
        /* implementation */
    }
    template<typename M>
    friend Array<T,N> operator*(const Array<T,N>& lhs,const M& rhs)
    {
        /* implementation */
    }
    template<typename M>
    friend Array<T,N> operator*( M lhs, const Array<T,N>& rhs)
    {
        /* implementation */
    }
    Array<T,N>& operator*=(const Array<T,N>& rhs)
    {
        /* implementation */
    }
    friend Array<T,N> operator*(const Array<T,N>& lhs,const Array<T,N>& rhs)
    {
        /* implementation */
    }

    // String
    std::string str() const
    {
        /* return a string containing the array */
    }

private:
    std::shared_ptr<T> data_;
    size_t n_;
};

我已经使用以下代码测试了此类，没有任何问题：

// Test friend Array<T,N> operator*(const Array<T,N>& lhs,const Array<T,N>& rhs)
Array<double,2> array1(1);
std::cout<<"array1: "<<array1.str()<<std::endl;
Array<double,2> array2(2);
std::cout<<"array2: "<<array2.str()<<std::endl;
Array<int,2> array3 = array1*array2;
std::cout<<"array3: array1*array2"<<array3.str()<<std::endl;

// Test friend Array<T,N> operator*( M lhs, const Array<T,N>& rhs)
Array<double,2> array4 = 2*array1;
std::cout<<"array4: 2*array1"<<array4.str()<<std::endl;

// Test friend Array<T,N> operator*(const Array<T,N>& lhs,const M& rhs)
Array<double,2> array5 = array1*5;
std::cout<<"array5: array1*3"<<array5.str()<<std::endl;

然后，我从该数组继承以创建具有特定成员函数的大小为3的数组

template<typename T>
class Vector3: public Array<T,3>
{
public:
    // Use Array constructors
    using Array<T,3>::Array;

    // Specific function
    template<typename M>
    Vector3 crossProduct(const Vector3& other)
    {
        /* implementation */
    }
};

我想按以下方式使用它（请注意，它与Array<T,N>之前相同Array<T,N>但与Vector3<T> ）：

// Test friend Array<T,N> operator*(const Array<T,N>& lhs,const Array<T,N>& rhs)
Vector3<double> vector1(1);
std::cout<<"vector1: "<<vector1.str()<<std::endl;
Vector3<double> vector2(2);
std::cout<<"vector2: "<<vector2.str()<<std::endl;
Vector3<int> vector3 = vector1*vector2;
std::cout<<"vector3: vector1*vector2"<<vector3.str()<<std::endl;

// Test friend Array<T,N> operator*( M lhs, const Array<T,N>& rhs)
Vector3<double> vector4 = 2*vector1;
std::cout<<"vector4: 2*vector1"<<vector4.str()<<std::endl;

// Test friend Array<T,N> operator*(const Array<T,N>& lhs,const M& rhs)
Vector3<double> vector5 = vector1*5;
std::cout<<"vector5: vector1*3"<<vector5.str()<<std::endl;

编译时，出现以下错误：

error: ambiguous overload for 'operator*' (operand types are 'Vector3<double>' and 'Vector3<double>')
     Vector3<int> vector3 = vector1*vector2;
                                   ^
note: candidates are:
note: Array<T, N> operator*(M, const Array<T, N>&) [with M = Vector3<double>; T = int; int N = 3]
     friend Array<T,N> operator*( M lhs, const Array<T,N>& rhs)
                       ^
note: Array<T, N> operator*(const Array<T, N>&, const M&) [with M = Vector3<double>; T = int; int N = 3]
     friend Array<T,N> operator*(const Array<T,N>& lhs,const M& rhs)
                       ^
note: Array<T, N> operator*(M, const Array<T, N>&) [with M = Vector3<double>; T = double; int N = 3]
     friend Array<T,N> operator*( M lhs, const Array<T,N>& rhs)
                       ^
note: Array<T, N> operator*(const Array<T, N>&, const M&) [with M = Vector3<double>; T = double; int N = 3]
     friend Array<T,N> operator*(const Array<T,N>& lhs,const M& rhs)
                       ^
note: Array<double, 3> operator*(const Array<double, 3>&, const Array<double, 3>&)
     friend Array<T,N> operator*(const Array<T,N>& lhs,const Array<T,N>& rhs)
                       ^
warning: ISO C++ says that these are ambiguous, even though the worst conversion for the first is better than the worst conversion for the second: [enabled by default]
     Vector3<double> vector4 = 2*vector1;
                                 ^
note: candidate 1: Array<T, N> operator*(M, const Array<T, N>&) [with M = int; T = double; int N = 3]
     friend Array<T,N> operator*( M lhs, const Array<T,N>& rhs)
                       ^
note: candidate 2: Array<T, N> operator*(const Array<T, N>&, const M&) [with M = Vector3<double>; T = int; int N = 3]
     friend Array<T,N> operator*(const Array<T,N>& lhs,const M& rhs)
                       ^
warning: ISO C++ says that these are ambiguous, even though the worst conversion for the first is better than the worst conversion for the second: [enabled by default]
     Vector3<double> vector4 = 2*vector1;
                                 ^
note: candidate 1: Array<T, N> operator*(M, const Array<T, N>&) [with M = int; T = double; int N = 3]
     friend Array<T,N> operator*( M lhs, const Array<T,N>& rhs)
                       ^
note: candidate 2: Array<T, N> operator*(const Array<T, N>&, const M&) [with M = Vector3<double>; T = double; int N = 2]
     friend Array<T,N> operator*(const Array<T,N>& lhs,const M& rhs)
                       ^
warning: ISO C++ says that these are ambiguous, even though the worst conversion for the first is better than the worst conversion for the second: [enabled by default]
     Vector3<double> vector4 = 2*vector1;
                                 ^
note: candidate 1: Array<T, N> operator*(M, const Array<T, N>&) [with M = int; T = double; int N = 3]
     friend Array<T,N> operator*( M lhs, const Array<T,N>& rhs)
                       ^
note: candidate 2: Array<T, N> operator*(const Array<T, N>&, const M&) [with M = Vector3<double>; T = int; int N = 2]
     friend Array<T,N> operator*(const Array<T,N>& lhs,const M& rhs)
                       ^
warning: ISO C++ says that these are ambiguous, even though the worst conversion for the first is better than the worst conversion for the second: [enabled by default]
     Vector3<double> vector4 = 2*vector1;
                                 ^
note: candidate 1: Array<T, N> operator*(M, const Array<T, N>&) [with M = int; T = double; int N = 3]
     friend Array<T,N> operator*( M lhs, const Array<T,N>& rhs)
                       ^
note: candidate 2: Array<T, N> operator*(const Array<T, N>&, const M&) [with M = Vector3<double>; T = double; int N = 3]
     friend Array<T,N> operator*(const Array<T,N>& lhs,const M& rhs)
                       ^
warning: ISO C++ says that these are ambiguous, even though the worst conversion for the first is better than the worst conversion for the second: [enabled by default]
     Vector3<double> vector5 = vector1*5;
                                       ^
note: candidate 1: Array<T, N> operator*(const Array<T, N>&, const M&) [with M = int; T = double; int N = 3]
     friend Array<T,N> operator*(const Array<T,N>& lhs,const M& rhs)
                       ^
note: candidate 2: Array<T, N> operator*(M, const Array<T, N>&) [with M = Vector3<double>; T = int; int N = 3]
     friend Array<T,N> operator*( M lhs, const Array<T,N>& rhs)
                       ^
warning: ISO C++ says that these are ambiguous, even though the worst conversion for the first is better than the worst conversion for the second: [enabled by default]
     Vector3<double> vector5 = vector1*5;
                                       ^
note: candidate 1: Array<T, N> operator*(const Array<T, N>&, const M&) [with M = int; T = double; int N = 3]
     friend Array<T,N> operator*(const Array<T,N>& lhs,const M& rhs)
                       ^
note: candidate 2: Array<T, N> operator*(M, const Array<T, N>&) [with M = Vector3<double>; T = double; int N = 2]
     friend Array<T,N> operator*( M lhs, const Array<T,N>& rhs)
                       ^
warning: ISO C++ says that these are ambiguous, even though the worst conversion for the first is better than the worst conversion for the second: [enabled by default]
     Vector3<double> vector5 = vector1*5;
                                       ^
note: candidate 1: Array<T, N> operator*(const Array<T, N>&, const M&) [with M = int; T = double; int N = 3]
     friend Array<T,N> operator*(const Array<T,N>& lhs,const M& rhs)
                       ^
note: candidate 2: Array<T, N> operator*(M, const Array<T, N>&) [with M = Vector3<double>; T = int; int N = 2]
     friend Array<T,N> operator*( M lhs, const Array<T,N>& rhs)
                       ^
warning: ISO C++ says that these are ambiguous, even though the worst conversion for the first is better than the worst conversion for the second: [enabled by default]
     Vector3<double> vector5 = vector1*5;
                                       ^
note: candidate 1: Array<T, N> operator*(const Array<T, N>&, const M&) [with M = int; T = double; int N = 3]
     friend Array<T,N> operator*(const Array<T,N>& lhs,const M& rhs)
                       ^
note: candidate 2: Array<T, N> operator*(M, const Array<T, N>&) [with M = Vector3<double>; T = double; int N = 3]
     friend Array<T,N> operator*( M lhs, const Array<T,N>& rhs)
                       ^

我不明白为什么继承类会出现错误，而基类却没有。 我尝试using Array<T,3>::operator* ，但是编译器在此行抱怨说：

error: no members matching 'Array<double, 3>::operator*' in 'class Array<double, 3>'
     using Array<T,3>::operator*;
                               ^

谁能解释我为什么在子类中不能正确解析运算符，而在基类中却无法解决？ 如果我必须做一些using Array<T,3>::operator...事情using Array<T,3>::operator...该怎么写？

谢谢！

Answer 1

之所以会出现这些错误，是因为当编译器尝试匹配运算符参数时，当在基类的实例上调用相同的运算符时，它将强制对基类执行额外的向下强制转换，同时获得完全匹配。 您应该使用::std::enable_if对M类型施加其他限制，如下所示：

template
<
     typename M
,    typename TEnable = typename ::std::enable_if_t
     <
         ::std::is_integral<M>::value
         ||
         ::std::is_floating_point<M>::value
     >
>

当您将*运算符声明为非成员朋友函数模板时， Array<T,3>类中没有operator* 。

Answer 2

在VTT提供的提示之后，我能够修复代码。 由于我不使用c ++ 14，因此我设法在没有std::enable_if_t情况下做到了。 这是最终的运算符：

// Product
template<typename M, typename std::enable_if<!std::is_base_of<Array<T,N>,M>::value>::type>
Array<T,N>& operator*=(const M& rhs)
{
    /* implementation */
}
template<typename M, typename std::enable_if<!std::is_base_of<Array<T,N>,M>::value>::type>
friend Array<T,N> operator*(const Array<T,N>& lhs,const M& rhs)
{
    /* implementation */
}
template<typename M, typename std::enable_if<!std::is_base_of<Array<T,N>,M>::value>::type>
friend Array<T,N> operator*( M lhs, const Array<T,N>& rhs)
{
    /* implementation */
}
Array<T,N>& operator*=(const Array<T,N>& rhs)
{
    /* implementation */
}
friend Array<T,N> operator*(const Array<T,N>& lhs,const Array<T,N>& rhs)
{
    /* implementation */
}

希望这会对其他人有所帮助。