c++第九周【任务4】建立一个二维数组类Douary，使该类中有以下数据成员、成员函数及友员函数，完成矩阵的输入、输出、加、减、相等判断等操作。

/* (程序头部注释开始)
* 程序的版权和版本声明部分
* Copyright (c) 2011, 烟台大学计算机学院学生
* All rights reserved.
* 文件名称：  任务4】建立一个二维数组类Douary
* 作    者： 李洪悬
* 完成日期：   2012      年   4    月   16     日
* 版 本 号：
* 对任务及求解方法的描述部分
* 输入描述：
* 问题描述：建立一个二维数组类Douary，使该类中有以下数据成员、成员函数及友员函数，完成矩阵的输入、输出、加、减、相等判断等操作。
* 程序输出：
* 程序头部的注释结束
*/

【任务4】建立一个二维数组类Douary，使该类中有以下数据成员、成员函数及友员函数，完成矩阵的输入、输出、加、减、相等判断等操作。
#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(); //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)
cout<<"输入d1:"<<endl;
cin>>d1;
cout<<"输入d2:"<<endl;
cin>>d2;
coutt<<"d1="<<endl;
cout<<d1;
coutt<<"d2="<<endl;
cout<<d2;
coutt<<"d1+d2="<<endl;
cout<<(d1+d2);
coutt<<"d1-d2="<<endl;
cout<<(d1-d2);
cout<<"d1"<<((d1==d2)?"==":"!=")<<"d2"<<endl;
system("pause");
return 0;
}

#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(const Douary &d);
~Douary(){delete [] Array;} //析构函数：用于释放动态数组所占用的存储空间。

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。
};
Douary::Douary(int m, int n)
{
row = m, col = n;
Array = new int [row*col];

for (int i = 0; i < row*col; i++)
Array[i] = 0;
}
Douary::Douary(const Douary &d)
{
col = d.col, row = d.row;
Array = new int[row*col];

for (int i = 0; i < row*col; i++)
Array[i] = d.Array[i];
}
istream & operator >> (istream &input, Douary &d)
{
cout << "请输入元素（" << d.row << "行" << d.col <<"列）：" << endl;

for (int i = 0; i < d.col*d.row; i++)
{
input >> d.Array[i];
}

return input;
}
ostream & operator << (ostream &output, Douary &d)
{
for (int i = 0; i < d.col*d.row; i++)
{
if (i % d.col == 0) cout << endl;
output << d.Array[i] <<'\t';
}
cout << endl;
return output;
}
Douary operator + (const Douary &d1, const Douary &d2)
{
Douary D(d1);

for (int i = 0; i < d1.col*d1.row; i++)
D.Array[i] = d1.Array[i] + d2.Array[i];

return D;
}
Douary operator - (const Douary &d1, const Douary &d2)
{
Douary D(d1);

for (int i = 0; i < d1.col*d1.row; i++)
D.Array[i] = d1.Array[i] - d2.Array[i];

return D;
}
bool Douary::operator == (const Douary &d)
{
if (this->col == d.col && this->row == d.row)
{
for (int i = 0; i < d.col*d.row; i++)
if (*(this->Array+i) != *(d.Array+i))
return false;
}
else
return false;

return true;
}
int main()
{
Douary d1(2, 3), d2(2, 3);

cout << "输入d1:" << endl;
cin >> d1;
cout << "输入d2:" << endl;
cin >> d2;

cout << "d1=" ;
cout << d1;
cout << "d2=";
cout << d2;

cout << "d1+d2=";
cout << (d1+d2);
cout << "d1-d2=";
cout << (d1-d2);

cout << "d1" << ((d1==d2)? "==":"!=") << "d2" << endl;

system("pause");
return 0;
}

经验总结：刚开始对矩阵很恐惧！又看了很多同学的代码还是没很透彻的理解矩阵！再看一些代码会更好吧！