Question

我正在实现一个模板化的矢量类（不是数据容器，而是线性代数意义上的矢量），每当我在运算符重载中引用rhs时，我就会遇到很多错误。此外，我的复制构造函数似乎不起作用。

#ifndef __VecXd__VecXd__
#define __VecXd__VecXd__

#define ULL unsigned long long
#include <iostream>

using namespace std;

template <class T>
class VecXd
{

public:

    explicit VecXd(ULL newDimension = 1) { dimension = newDimension; vector = new T[newDimension];}
    VecXd(const VecXd<T> &rhs);
    VecXd<T>& operator=(const VecXd<T> &rhs);
    const VecXd<T> operator+(const VecXd<T> &rhs) const;
    VecXd<T>& operator+=(const VecXd<T> &rhs);
    friend ostream& operator<<(ostream &out, VecXd<T> vec);
    friend istream& operator>>(istream &in, VecXd<T>& vec);
    ~VecXd() { delete[] vector; }

    const ULL getDimension() { return dimension; }
    const T itemAtIndex(ULL index) { if(index >= dimension) throw 1; return vector[index]; }

private:

    ULL dimension;
    T *vector;

};

template <class T>
VecXd<T>::VecXd(const VecXd<T> &rhs)
{
    dimension = rhs.getDimension();
    vector = new T[dimension];
    for (ULL i = 0; i < dimension; ++i)
        vector[i] = rhs.itemAtIndex(i);
}

template <class T>
VecXd<T>& VecXd<T>::operator=(const VecXd<T> &rhs)
{
    if (this != &rhs)
    {
        if (dimension != rhs.getDimension())
        {
            delete [] vector;
            dimension = rhs.getDimension();
            vector = new T[dimension];
        }
        for (ULL i = 0; i < dimension; ++i)
            vector[i] = rhs.itemAtIndex(i);
    }
    return *this;
}

template <class T>
VecXd<T>& VecXd<T>::operator+=(const VecXd<T> &rhs)
{
    if (dimension != rhs.getDimension())
    {
        cout << "\nCannot perform addition. Vectors do not have the same dimensions.\n";
        throw 1;
    }
    else
    {
        for (ULL i = 0; i < dimension; ++i)
            vector[i] += rhs[i];
    }
    return *this;
}

template <class T>
const VecXd<T> VecXd<T>::operator+(const VecXd<T> &rhs) const
{
    VecXd<T> temp = *this;
    temp += rhs;
    return temp;
}

template <class T>
ostream& operator<<(ostream &outs, VecXd<T> vec)
{
    for (ULL i = 0; i < vec.dimension; ++i)
        out << vec.vector[i] << (i+1 < vec.dimension ? " " : "");
    return out;
}

template <class T>
istream& operator>>(istream &in, VecXd<T> &vec)
{
    ULL newDim = 1;
    cin >> newDim;
    if (!cin.good())
    {
        cout << "\nImproper input.\n";
        throw 1;
    }
    else
    {
        delete [] vec.vector;
        vec.dimension = newDim;
        vec.vector = new T[vec.dimension];
        for (ULL i = 0; i < vec.dimension; ++i)
            in >> vec.vector[i];
    }
    return in;
}

#endif /* defined(__VecXd__VecXd__) */

我收到了这种风格的错误：

Member function 'getDimension' not viable: 'this' argument has type 'const VecXd<int>', but function is not marked const

每次rhs调用函数（例如getDimension()或itemAtIndex()）时都会发生这种情况;出现两个错误（一次用于VecXd<int>，另一次用于VecXd<int>）。

此外，此行中的重载+运算符函数无法识别复制构造函数：

VecXd<T> temp = *this;

帮助？

Answer 1

为了能够在const对象上调用函数，您需要向编译器承诺该函数不会修改该对象。为此，您可以在参数列表后面用关键字const标记该函数。例如，要使getDimension成为const成员函数，您可以将其更改为：

const ULL getDimension() const { return dimension; }

（请注意，返回类型中的const绝对没有效果，所以你应该摆脱它）

模板化线性代数向量类中奇怪的“成员函数不可行”错误

1 个答案: