从左侧将一个常数与一个对象相乘 [英] Multiplying an object with a constant from left side

问题描述

我有一个 Matrix 类，它具有重载的 * 运算符，用于标量和矩阵乘法.

I have a Matrix class and it has overloaded * operators for scalar and matrix multiplications.

template <class T> class Matrix
{
    public:
        // ...
        Matrix operator*(T scalar) const;
        // ...
}

// ...

template <class T>
Matrix<T> Matrix<T>::operator*(T RightScalar) const
{
    Matrix<T> ResultMatrix(m_unRowSize, m_unColSize);
    for (uint64_t i=0; i<m_unRowSize; i++)
    {
        for (uint64_t j=0; j<m_unColSize; j++)
        {
            ResultMatrix(i, j) = TheMatrix[m_unColSize * i + j] * RightScalar;
        }
    }
    return ResultMatrix;
}

// ...

我可以从右侧将标量与矩阵对象相乘而没有任何问题:

I can multiply a matrix object with a scalar from right side without any problem:

Matrix<double> X(3, 3, /* ... */);  // Define a 3x3 matrix and initialize its contents
Matrix<double> Y;                   // Define an output matrix
Y = X * 10.0;                       // Do the linear operation

但是，如何以相同的方式从左侧将其相乘?

But, how do I multiply it from left side same way?

Matrix<double> X(3, 3, /* ... */);
Matrix<double> Y;
Y = 10.0 * X;

在算术中，在进行乘法运算时在左侧写常量是一种常见的表示法.我想遵守此规则，以使我的代码更具可读性.

In arithmetic, it is a common notation to write constants on the left side when doing multiplication. I would like to obey this rule to make my code more readable.

可以在C ++中实现吗?
如果可能，如何在代码中修改类方法?

Is it possible to implement this in C++?
If it is possible, how do I modify the class method in my code?

推荐答案

成员函数由其左侧参数this指针匹配.由于本机类型不能具有成员函数，因此您必须通过非成员函数(以及其他您没有写权限的类型)与用户定义的类型添加右乘法.

Member functions are matched by their left-hand-side argument which is the this-pointer. Since native types can't have member functions, you have to add right-multiplication with user-defined types through non-member functions (and also for other types you don't have write-access to).

template<typename T>
Matrix<T> operator*(T const& scalar, Matrix<T> rhs)
{
    // scalar multiplication is commutative: s M = M s
    return rhs *= scalar; // calls rhs.operator*=(scalar);
}

注意:我写了上述非成员 operator * ，它是根据成员 operator * = 实现的.建议将所有乘法写为非成员函数，并使用成员 operator * = 通过lhs Matrix元素实现这些乘法.

NOTE: I wrote the above non-member operator* implemented in terms of a member operator*=. It is recommended to write all multiplications as non-member functions, and use a member operator*= to implement these multiplications with a lhs Matrix element.

这将a)保持类接口最小化，并且b)防止隐藏的转换.例如.如果尺寸为1x1，则可以有一个可以隐式转换为标量的Matrix类，如果不提供直接匹配的单独重载，则这些转换可能会悄无声息地发生.

This will a) keep the class interface minimal, and b) prevent hidden conversions. E.g. you could have a Matrix class that is implicitly convertible to scalar if the dimensions are 1x1, and these conversions could silently happen if you don't provide a separate overload that is a direct match.

template<typename T>
Matrix<T> operator*(Matrix<T> lhs, T const& scalar)
{
    return lhs *= scalar; // calls lhs.operator*=(scalar);
}

template<typename T>
Matrix<T> operator*(Matrix<T> lhs, Matrix<T> const& rhs)
{
    return lhs *= rhs; // calls lhs.operator*=(rhs);
}

关于lhs Matrix如何是副本而不是引用的通知.这使编译器可以进行优化，例如复制省略/移动语义.还要注意，这些运算符的返回类型为 Matrix< T> ，而不是某些旧的C ++书籍中建议的 const Matrix< T> C ++ 11.

Notice on how the lhs Matrix is a copy and not a reference. This allows the compiler to make optimizations such as copy elision / move semantics. Also note that the return type of these operators is Matrix<T> and not const Matrix<T> which was recommended in some old C++ books, but which prevents move semantics in C++11.

// class member 
template<typename T>
Matrix<T>& Matrix<T>::operator*=(Matrix<T> const& rhs)
{
    // your implementation
    return *this;
}

// class member 
template<typename T>
Matrix<T>& Matrix<T>::operator*=(T const& scalar)
{
    // your implementation
    return *this;
}

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