第九周实验报告4

实验目的：建立一个二维数组类Douary，使该类中包含一下数据成员、成员函数及友元函数，完成矩阵的输入、输出、加、减、相等判断等操作。

实验代码：

#include <iostream>using namespace std;class Douary{public:Douary(int m, int n);//构造函数：用于建立动态数组存放m行n列的二维数组（矩阵）元素，并将该数组元素初始化为Douary::Douary(const Douary &d); ~Douary(); //析构函数：用于释放动态数组所占用的存储空间。           //此处增加一个复制构造函数friend istream &operator >> (istream &input, Douary &d); //重载运算符“>>”输入二维数组，其中d为Dousry类对象；friend ostream &operator << (ostream &output, Douary &d); //重载运算符“<<”以m行n列矩阵的形式输出二维数组，其中d为Douary类对象。friend Douary operator + (const Douary &d1, const Douary &d2); //两个矩阵相加，规则：对应位置上的元素相加friend Douary operator - (const Douary &d1,const Douary &d2); //两个矩阵相减，规则：对应位置上的元素相减bool operator == (const Douary &d);//判断两个矩阵是否相等，即对应位置上的所有元素是否相等private:int * Array;      //Array 为动态数组指针。int row;          //row  为二维数组的行数。int col;          //col   为二维数组的列数。};int main(){Douary d1(2,3),d2(2,3);  //两个2行3列的矩阵cout << "输入d1:" << endl;cin >> d1;cout << "输入d2:" << endl;cin >> d2;cout << endl;cout << "d1 = ";cout << d1;cout << endl;cout << "d2 = ";cout << d2;cout << endl;cout << "d1 + d2 = ";cout << (d1 + d2);cout << endl;cout << "d1 - d2 = ";cout << (d1 - d2);cout << endl;cout << "d1" << ((d1 == d2)?" == ":" != ") << "d2" << endl;system("pause");return 0;}Douary::Douary(int m, int n){row = m;col = n;Array = new int[m * n];}Douary :: Douary(const Douary &d){    row = d.row;      col = d.col;      Array = new int [row * col];for(int i = 0; i < row; ++i)          for(int j = 0; j < col; ++j)              Array[i * col + j] = d.Array[i * col + j];} Douary :: ~Douary()  {       delete [] Array;       }istream & operator >> (istream &input, Douary &d){for(int i = 0; i < d.row * d.col; i++){input >> d.Array[i];}       return input;}ostream &operator << (ostream &output, Douary &d){for(int i = 0; i < d.row * d.col; ++ i){if(i % d.col == 0)output << endl;output << d.Array[i] << "  ";}    output << endl;return output;}Douary operator + (const Douary &d1, const Douary &d2)   {      Douary d(d1.row, d1.col);     for(int i = 0; i < d1.row; ++ i)      {          for(int j = 0; j < d1.col; ++ j)              d.Array[i * d1.col + j] = d1.Array[i * d1.col + j] + d2.Array[i * d1.col + j];      }     return d;  }Douary operator - (const Douary &d1, const Douary &d2) //两个矩阵相加，规则：对应位置上的元素相加   {         Douary d(d1.row, d1.col);      for(int i = 0; i < d1.row; ++ i)      {          for(int j = 0; j < d1.col; ++ j)              d.Array[i * d1.col + j] = d1.Array[i * d1.col + j] - d2.Array[i * d1.col + j];      }      return d;  }bool Douary::operator == (const Douary &d) //判断两个矩阵是否相等，即对应位置上的所有元素是否相等{for(int i = 0; i < d.row; ++i)      {          for(int j = 0; j < d.col; ++j)              if(Array[i * d.col + j] != d.Array[i * d.col + j])return false;    }return true;}

实验结果截图：

实验心得：

看来前面时太得意忘形了，这下好了，被难住了吧，活该！大体难点有这么几点，怎样用结构体建立二维数组，怎样找到二维数组中的数据，并实现加、减、相等的比较，当然还有看到就不舒服的指针，以及我个人不太熟的构析函数，反正麻烦事一大堆了。

嗯，不好意思，稍稍参考了一下老师的哟，在构造函数中用数据成员表示行和列，实现构造函数的多维数组化，用Array指针建立动态数组，保存数据，用行列的乘积显示数组中数据个数，并借由指针找到对象数组中的数据实现对象的加、减和相等的比较，理清了这些，代码实现也不是不可能的，个人觉得任务四是本周试验中综合性和技巧性最好的，是道好的试验问题哦。