求指导!!!急求！！！

[s12rhxgc]

定义一个二维方阵类，通过重载实现二元运算“=”、“+”、 “-”、 “*”、 “~”， 分别实现方阵的赋值、矩阵加、矩阵减，矩阵乘及矩阵的转置
要求如下：
 定义几个矩阵类，包含一个二维数组，或者二维指针（高级，可以加分）
 矩阵C=A*B乘定义为 A的第i行乘以B的第j列的值的累加为C的第i行的第j元素，
 转置矩阵的定义为，A=~B B的第i行第j列的元素为A的第j行第i列元素。
[color=#FF0000]求高手帮修改下 急求
#include "stdafx.h"
#include"iostream.h"

const int r=3; const int c=3 const int k=3
class matrix
{
int a[r][c];
public:
matrix();
matrix(int t[r][c])
{
a[i][j]=t[i][j];
}

friend matrix operator+ (matrix & a, matrix & b); //矩阵加
friend matrix operator- (matrix & a, matrix & b); //矩阵减
friend matrix operator* (matrix & a, matrix & b); //矩阵乘
friend matrix operator~ (matrix & a); //矩阵转置
void display();
}
matrix operator +(matrix & a, matrix & b);
{
for (i=0;i<r;i++)
for(j=0;j<c;j++)
c[i][j]=a[i][j]+b[i][j];
}
matrix operator -(matrix & a, matrix & b);
{
for (i=0;i<r;i++)
for(j=0;j<c;j++)
c[i][j]=a[i][j]-b[i][j];
}
matrix operator+ (matrix & a, matrix & b);
{
for(i=0;i<r;i++)
for(j=0;j<c;j++)
{
c[i][j]=0
for(l=0;l<k;l++)
c[i][j]=a[i][j]+b[i][j];
}
}

matrix operator~ (matrix & a);
{
for(i=0;i<k;i++)
for(j=0;j<c;j++)
c[j][i]=a[i][j];
}
matrik::void display()
{
for(i=0;i<r;i++)
for(j<0;j<c;j++)
cout<<c[i][j]<<endl;
}
void main()
{
int a[3][3]={1,2,3,4,5,6,7,8,9};
int b[3][3]={1,2,3, 0,1,2, -1,0,1}
matrix x(a), y(b);
cout<<"-------- x= ------------"<<endl;
x.display(); //按设定格式显示出第一个矩阵
x cout<<"-------- y= ------------"<<endl;
y.display() //按设定格式显示出第二个矩阵
y cout<<"--------- x+y= ---------"<<endl;
(x+y).display(); //输出结果矩阵 x+y
cout<<"--------- x-y= ---------"<<endl;
(x-y).display(); //输出结果矩阵x-y
cout<<"--------- x*y= ---------"<<endl;
(x*y).display(); //输出结果矩阵x*y
cout<<"--------- ~x= ---------"<<endl;
(~x).display(); //输出结果矩阵~x
}

[s12rhxgc]

有问题 实现不了 很纠结 帮忙改改吧 新手 很困难啊

[高性能架构探索]

#include <iostream>
 #include <stdio.h>
#include <iostream>

using namespace std;

const int r=3; const int c=3; const int k=3;
class matrix
{
int a[r][c];
public:
	matrix(){}
matrix(int t[3][3])
{
for (int i=0;i<r;i++)
for(int j=0;j<c;j++)
	a[i][j] = t[i][j];
}

friend matrix operator+ (matrix & a, matrix & b); //矩阵加
friend matrix operator- (matrix & a, matrix & b); //矩阵减
friend matrix operator* (matrix & a, matrix & b); //矩阵乘   
friend matrix operator~ (matrix & a); //矩阵转置
  void display();
};


matrix operator +(matrix & a, matrix & b)
{
	int d[3][3];
for (int i=0;i<r;i++)
for(int j=0;j<c;j++)
	d[i][j]=a[i][j]+b[i][j];

return matrix(d);
}
matrix operator -(matrix & a, matrix & b)
{ 
	int d[3][3];
for (i=0;i<r;i++)
for(j=0;j<c;j++)
d[i][j]=a[i][j]-b[i][j];

return matrix(d);
}
matrix operator+ (matrix & a, matrix & b)
{   

	int d[3][3];
for(int i=0;i<r;i++)
for(int j=0;j<c;j++) 
{
	d[i][j]=a[i][j]+b[i][j];

}


return matrix(d);
}

matrix operator~ (matrix & a)
{

	int d[3][3];
	for(int i=0;i<k;i++)
	for(int j=0;j<c;j++)
	d[j][i]=a[i][j];

return matrix(d);
}

void matrix::display()
{
for(int i=0;i<r;i++)
for(int j=0;j<c;j++)
	cout<<a[i][j]<<endl;
}
void main()
{
int a[3][3]={1,2,3,4,5,6,7,8,9};
int b[3][3]={1,2,3, 0,1,2, -1,0,1};
  matrix x(a), y(b);
  matrix z;
cout<<"-------- x= ------------"<<endl;
x.display(); //按设定格式显示出第一个矩阵
cout<<"-------- y= ------------"<<endl;
y.display(); //按设定格式显示出第二个矩阵   
cout<<"--------- x+y= ---------"<<endl;
z = x+y;
z.display(); //输出结果矩阵 x+y   
cout<<"--------- x-y= ---------"<<endl;
z = x-y;
z.display(); //输出结果矩阵x-y   
cout<<"--------- x*y= ---------"<<endl;
z = x*y;
z.display(); //输出结果矩阵x*y   
cout<<"--------- ~x= ---------"<<endl;
z = ~x;
(~x).display(); //输出结果矩阵~x
}

[高性能架构探索]

#include <iostream>
 #include <stdio.h>
#include <iostream>

using namespace std;

const int r=3; const int c=3; const int k=3;
class matrix
{
int a[r][c];
public:
	matrix(){}
matrix(int t[3][3])
{
for (int i=0;i<r;i++)
for(int j=0;j<c;j++)
	a[i][j] = t[i][j];
}

friend matrix operator+ (matrix & a, matrix & b); //矩阵加
friend matrix operator- (matrix & a, matrix & b); //矩阵减
//friend matrix operator* (matrix & a, matrix & b); //矩阵乘   
friend matrix operator~ (matrix & a); //矩阵转置
  void display();
};


matrix operator +(matrix & m, matrix & n)
{
	int d[3][3];
for (int i=0;i<r;i++)
for(int j=0;j<c;j++)
	d[i][j]=m.a[i][j]+n.a[i][j];

return matrix(d);
}
matrix operator -(matrix & m, matrix & n)
{ 
	int d[3][3];
for (int i=0;i<r;i++)
for(int j=0;j<c;j++)
d[i][j]=m.a[i][j]-n.a[i][j];

return matrix(d);
}
//matrix operator* (matrix & m, matrix & n)
//{   
//
//	int d[3][3];
//for(int i=0;i<r;i++)
//for(int j=0;j<c;j++) 
//{
//	d[i][j]=m.a[i][j]*n.a[i][j];
//
//}




matrix operator~ (matrix & n)
{

	int d[3][3];
	for(int i=0;i<k;i++)
	for(int j=0;j<c;j++)
		d[j][i]=n.a[i][j];

return matrix(d);
}

void matrix::display()
{
for(int i=0;i<r;i++)
for(int j=0;j<c;j++)
	cout<<a[i][j]<<endl;
}
void main()
{
int a[3][3]={1,2,3,4,5,6,7,8,9};
int b[3][3]={1,2,3, 0,1,2, -1,0,1};
  matrix x(a), y(b);
  matrix z;
cout<<"-------- x= ------------"<<endl;
x.display(); //按设定格式显示出第一个矩阵
cout<<"-------- y= ------------"<<endl;
y.display(); //按设定格式显示出第二个矩阵   
cout<<"--------- x+y= ---------"<<endl;
z = x+y;
z.display(); //输出结果矩阵 x+y   
cout<<"--------- x-y= ---------"<<endl;
z = x-y;
z.display(); //输出结果矩阵x-y   
cout<<"--------- x*y= ---------"<<endl;
  
cout<<"--------- ~x= ---------"<<endl;
z = ~x;
(~x).display(); //输出结果矩阵~x
}

给你个能运行的，不过 矩阵乘那块，你的实现有点问题，我给你注掉了，我刚才看了，这个可以正常运行


楼主给分吧

[ryfdizuo]

#include <iostream>
using namespace std;

const int r=3; 
const int c=3;
const int k=3;

class matrix
{
	int a[r][c];
public:
	matrix() {}
	matrix(int t[r][c])
	{
		for (int i=0; i<r; i++)
			for(int j=0; j<c; j++)
				a[i][j]=t[i][j];		
	}

	friend matrix operator+ (matrix & a, matrix & b);	 //矩阵加
	friend matrix operator- (matrix & a, matrix & b);	 //矩阵减
	friend matrix operator* (matrix & a, matrix & b);	 //矩阵乘  
	friend matrix operator~ (matrix & a);	//矩阵转置
	void display();
};		//注意分号

matrix operator +(matrix & a, matrix & b)			//分号去掉
{  
	matrix ret;
	for (int i=0;i<r;i++)
		for(int j=0;j<c;j++)
			ret.a[i][j] = a.a[i][j] + b.a[i][j];

	return ret;
}
matrix operator -(matrix & a, matrix & b)
{  
	matrix ret;
	for (int i=0;i<r;i++)
		for(int j=0;j<c;j++)
			ret.a[i][j] = a.a[i][j]-b.a[i][j];

	return ret;
}
matrix operator* (matrix & a, matrix & b)
{  
	matrix ret;

	for (int i=0;i<r;i++)
		for(int j=0;j<c;j++)
		{
			ret.a[i][j]=0;
			for(int l=0; l<k; l++)
					ret.a[i][j] = a.a[i][j] + b.a[i][j];
		}

	return ret;
}

matrix operator~ (matrix & a)
{
	matrix ret;

	for (int i=0;i<r;i++)
		for(int j=0;j<c;j++)
			ret.a[j][i] = ~a.a[i][j];

	return ret;
}
void matrix::display()
{
	for(int i=0; i<r; i++)
	{
		for(int j=0; j<c; j++)
			cout<< this->a[i][j] << " ";

		cout << endl;
	}
}

void main()
{
	int a[3][3]={1,2,3,4,5,6,7,8,9};
	int b[3][3]={1,2,3, 0,1,2, -1,0,1};
	matrix x(a);
	matrix y(b);
	cout<<"-------- x= ------------"<<endl;
	x.display();	//按设定格式显示出第一个矩阵
	y.display(); //按设定格式显示出第二个矩阵  
		
	(x+y).display();	//输出结果矩阵 x+y  
	cout<<"--------- x-y= ---------"<<endl;
	(x-y).display();	//输出结果矩阵x-y  
	cout<<"--------- x*y= ---------"<<endl;
	(x*y).display();	//输出结果矩阵x*y  
	cout<<"--------- ~x= ---------"<<endl;
	(~x).display();	//输出结果矩阵~x

	system("PAUSE");
}

[ryfdizuo]

矩阵不能直接a[i][j]取元素，需要重载operator[][]或者直接调用成员数组a

[DAsama]

matrix operator* (matrix & m, matrix & n)
{   
   int d[3][3];
   for(int i=0;i<r;i++)
	for(int j=0;j<c;j++) 
	{
		for(int r=0;r<c;r++)
			d[i][j]=m.a[i][r]*n.a[r][j];
	}
	return matrix(d);
}

补充个乘法的