c++ - 如何为我的非静态转置函数实现 matrix.Transpose()？

Question

我在 test.cpp 文件中得到了以下代码来实现:

cout << "Case 2: the non-static Transpose function" << endl;
{
    double column[4] = {2, 1, 0, -1};
    double row[3] = {2, 0, -1};
    Matrix matrix = Matrix::Toeplitz(column, 4, row, 3);
    cout << "The original Matrix = " << endl;
    cout << matrix << endl;  //This part of the code works


    matrix.Transpose();  //How do I implement this?
    cout << "The transposed version = " << endl;
    cout << matrix << endl;
    cout << "Press any key to continue ..." << flush;
    system("read");
    cout << endl;
}

Matrix::Toeplitz(column, 4, row, 3) 的工作方式如下:

Matrix Matrix::Toeplitz(const double* column, const int noOfRows, const double* row, const int noOfColumns){
    Matrix outT(column, noOfRows, row, noOfColumns);
    return outT;
}

那么我将如何实现 matrix.Transpose()？到目前为止，我的代码如下:

Matrix& Matrix::Transpose () {

double newrow[noOfRows];
for(int i=0; i<noOfRows; i++){
    int index = GetIndex(i,0);
    newrow[i] = data[index];
}

double newcol[noOfColumns];
for(int i=0; i<noOfColumns; i++){
    int index = GetIndex(0,i);
    newcol[i] = data[index];
}

Matrix outT(newcol, noOfColumns, newrow, noOfRows);
}

这对 cout<<matrix<<endl; 没有影响

我在想Matrix outT(newcol, noOfColumns, newrow, noOfRows);应该在实现 matrix.Transpose 时向矩阵对象提供新信息(即切换列和行数组)，但它一直没有工作。

这是实现 matrix.Transpose() 的正确格式 Matrix& Matrix::Transpose () 吗？

Answer 1

Matrix::Transpose 无法返回对本地声明对象的引用。这会导致很多问题。

参见 C++ Returning reference to local variable .

它必须通过复制返回(然后，函数可以是 const，因为当前对象没有被修改):

Matrix Matrix::Transpose() const
{
    double newrow[noOfRows];
    for(int i=0; i<noOfRows; i++){
        int index = GetIndex(i,0);
        newrow[i] = data[index];
    }

    double newcol[noOfColumns];
    for(int i=0; i<noOfColumns; i++){
        int index = GetIndex(0,i);
        newcol[i] = data[index];
    }

    return Matrix(newcol, noOfColumns, newrow, noOfRows);
}

然后，您可以这样使用它:

Matrix transposed = matrix.Transpose(); // does not modify matrix object
cout << "The transposed version = " << endl;
cout << transposed << endl;

如果返回Matrix&，您需要让您的方法转置当前对象并返回它(return *this)，仅用于帮助调用者链接许多运算符(比如做 m.Transpose().Transpose())。

然后，它可以(未测试):

Matrix& Matrix::Transpose() 
{
    // backup old content
    double* backupData = new double[noOfRows*noOfColumns];
    memcpy( backupData, data, sizeof(double)*noOfRows*noOfColumns );

    // change matrix geometry
    int oldRowCount = noOfRows;
    noOfRows = noOfColumns;
    noOfColumns = oldRowCount ;

    // transpose matrix by copying from backup content
    for ( unsigned int line = 0; line < noOfRows ; ++line )
    {
        for ( unsigned int col = line; col < noOfColumns; ++col )
        {
            data[line * noOfColumns + col] = backupData[col * noOfRows  + line];
        }
    }

    delete [] backupData;

    return *this;
}

然后，您可以这样使用它:

matrix.Transpose(); // modifies matrix object
cout << "The transposed version = " << endl;
cout << transposed << endl;