/*
实验内容: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：矩阵的输入、输出、加、减、相等判断
* 作 者： 薛广晨
* 完成日期： 2012 年 4 月 15 日
* 版 本号： x1.0
*/
//【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。
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) //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；
{
row = m;
col = n;
if(m != 0 && n!= 0)
{
Array = new int *[row];
for (int i = 0; i < row; i++)
{
Array[i] = new int[col];
}
}
else
Array = NULL;
}
Douary :: ~Douary()
{
if (Array != NULL)
{
for (int i = 0; i < row; i++)
{
delete Array[i];
}
delete[] Array;
Array = NULL;
}
}
istream &operator>>(istream &input, Douary &d)//重载运算符“>>”输入二维数组，其中d为Douary类对象；
{
//input >> d.row >> d.col;
if(d.row != 0 && d.col != 0)
{
int i,j;
d.Array = new int *[d.row];
for (i = 0; i < d.row; i++)
{
d.Array[i] = new int[d.col];
}
for (i = 0; i < d.row; ++i)
for(j = 0; j < d.col; ++j)
{
input >> d.Array[i][j];
}
}
else
d.Array = NULL;
return input;
}
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。
{
output << d.row << '/' << d.col << endl;
if (d.row != 0 && d.col != 0)
{
int i,j;
for(i = 0; i < d.row; ++i)
{
for (j = 0; j < d.col; ++j)
output << d.Array[i][j] << " ";
output << endl;
}
}
return output;
}
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 *p = new Douary(0,0);
if (d1.row == d2.row && d1.col == d2.col)
{
int i, j;
p -> row = d2.row;
p -> col = d2.col;
p -> Array = new int *[d1.row];
for (i = 0; i < d1.row; i++)
{
p -> Array[i] = new int[d1.col];
}
for (i = 0; i < d2.row; ++i)
for(j = 0; j < d2.col; ++j)
{
p -> Array[i][j] = d1.Array[i][j] + d2.Array[i][j];
}
}
else
{
cout << "不能构成矩阵!" << endl;
}
return *p;
}
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
{
Douary *p = new Douary(0, 0);
if (d1.row == d2.row && d1.col == d2.col)
{
int i, j;
p -> row = d2.row;
p -> col = d2.col;
p -> Array = new int *[d1.row];
for (i = 0; i < d1.row; i++)
{
p -> Array[i] = new int[d1.col];
}
for (i = 0; i < d2.row; ++i)
for(j = 0; j < d2.col; ++j)
{
p -> Array[i][j] = d1.Array[i][j] - d2.Array[i][j];
}
}
else
{
cout << "不能构成矩阵!" << endl;
}
return *p;
}
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
{
bool b = false;
if (row == d.row && col == d.col)
{
int i,j;
for(i = 0; i < row;++i)
for(j = 0; j < col; ++j)
if(d.Array[i][j] != Array[i][j])
return false;
b = true;
}
return b;
}
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=" << endl << d1;
cout << "d2=" << endl << d2;
cout << "d1+d2=" << endl << (d1 + d2);
cout << "d1-d2=" << endl << (d1 - d2);
cout << "d1" << ((d1 == d2) ? "==" : "!=") << "d2" << endl;
system("pause");
return 0;
}
上机感言：这个题有难度，有挑战性，就是数组的地方比较难