Question

如果它们适合添加，我真的不知道如何检查两个矩阵的大小。这是我的代码，目前正在运行，没有错误：

matrix operator+(const matrix & mat){
    matrix add;
    add.mSize = mat.mSize;
    add.mP = new int[add.mSize * add.mSize];
    for(int i = 0; i < add.mSize * add.mSize; i++){
        add.mP[i] = mP[i] + mat.mP[i];
    }
    return add;                        
}

Answer 1

一般的2d矩阵有两个不同的维度，行和列（除非它是方阵的特殊情况）。

在一般情况下，您的代码可能是：

#include <stdexcept>

class matrix
{
public:
    matrix(int rows, int columns) :
        mRows(rows), mColumns(columns), mP(NULL)
    {
        mP = new double[mRows * mColumns];
        for (int i = 0; i < mRows * mColumns; i++)
            mP[i] = 0.0;
    }

    matrix(const matrix& other) :
        mRows(other.mRows), mColumns(other.mColumns), mP(NULL)
    {
        mP = new double[mRows * mColumns];
        for (int i = 0; i < mRows * mColumns; i++)
            mP[i] = other.mP[i];
    }

    ~matrix()
    {
        delete[] mP;
    }

    const double* operator[] (int row) const {
        return &mP[row * mColumns];
    }

    double* operator[] (int row) {
        return &mP[row * mColumns];
    }

    matrix& operator=(const matrix& other) {
        if (this == &other) return *this;
        if ((mRows != other.mRows) || (mColumns != other.mColumns)) throw std::invalid_argument( "dimensions don't match" );
        for (int r = 0; r < mRows; r++) {
            for (int c = 0; c < mColumns; c++) {
                (*this)[r][c] = other[r][c];
            }
        }
        return *this;
    }

    matrix& operator+=(const matrix& other) {
        if ((mRows != other.mRows) || (mColumns != other.mColumns)) throw std::invalid_argument( "dimensions don't match" );
        for (int r = 0; r < mRows; r++) {
            for (int c = 0; c < mColumns; c++) {
                (*this)[r][c] += other[r][c];
            }
        }
        return *this;
    }

private:
    int mRows;
    int mColumns;
    double *mP;
};

inline matrix operator+(matrix lhs, const matrix& rhs)
{
    lhs += rhs;
    return lhs;
}

检查矩阵大小

1 个答案: