Question

我在test.cpp文件中获得了以下代码来实现：

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（第4列，第3行，第3行）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

我认为Matrix outT(newcol, noOfColumns, newrow, noOfRows);在实现matrix时会给新的信息（即将列和行数组切换）给矩阵对象。转换但它还没有工作。

这是正确的格式Matrix＆amp; Matrix :: Transpose（）用于实现matrix.Transpose（）？

Answer 1

Matrix::Transpose无法返回对本地声明的对象的引用。这将导致许多问题。

请参阅C++ Returning reference to local variable。

必须按副本返回（然后，函数可以是const，因为当前对象未被修改）：

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，你这样使用它：

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

如果返回Matrix&，则需要让您的方法转置当前对象并将其返回（return *this），这对于帮助调用者链接许多运算符非常有用（比如执行m.Transpose().Transpose()实例）。

然后，它可能（未经测试）：

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，你这样使用它：

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

如何为我的非静态转置函数实现matrix.Transpose（）？

1 个答案: