一个我研究了很多天，却不能解决的问题！关于不等二维数组之间的赋值（使用运算符重载）
下面是源代码，用vc6.0编译通过，但运行时总是提示赋值操作有问题（我用运算符重载实现的）！
首先是头文件
#ifndef ARRAY_H
#define ARRAY_H
#include <iostream.h>
class DoubleSubscriptedArray
{
friend ostream &operator < < (ostream &, const DoubleSubscriptedArray &);
friend istream &operator> > (istream &, DoubleSubscriptedArray &);
int **ptr;
int size;
int y;
int x;
public:
DoubleSubscriptedArray( int , int );
~DoubleSubscriptedArray();
int access(int ,int) const;
const DoubleSubscriptedArray &operator= (const DoubleSubscriptedArray &);
int operator== (const DoubleSubscriptedArray &) const;
int operator!= (const DoubleSubscriptedArray &) const;
int &operator [] (int);
};
#endif
//然后是函数的定义部分
//注意：重载的=有问题，关键问题就在 重载的‘=’上。
#include <iostream.h>
#include <stdlib.h>
#include <assert.h>
#include "array.h "
DoubleSubscriptedArray:: DoubleSubscriptedArray(int m,int n)
{
x=m;
y=n;
size =x*y;
ptr=new int *[x];
for(int j=0;j <y;j++)
ptr[j]=new int[y];
/* assert(ptr!=0);
for(int k=0;k <m;k++)
for(int l=0;l <n;l++)
ptr [k][l] =0;
*/
}
DoubleSubscriptedArray::~DoubleSubscriptedArray()
{
for(int h =0; h < x; h++)
delete []ptr[h];
delete []ptr;
}
int DoubleSubscriptedArray:: access(int a,int b) const
{
return ptr[a][b];
}
int DoubleSubscriptedArray:: operator== (const DoubleSubscriptedArray &right) const
{
if(size!=right.size || x!=right.x || y!=right.y)
return 0;
for(int i=0;i <x;i++)
for(int j=0;j <y;j++)
if(ptr[i][j] != right.ptr[i][j])
return 0;
return 1;
}
int DoubleSubscriptedArray::operator != (const DoubleSubscriptedArray &right) const
{
if(size!=right.size || x!=right.x || y!=right.y)
return 1;
for (int i=0;i <x;i++)
for(int j=0; j <y; j++)
if(ptr[i][j]!= right.ptr[i][j])
return 1;
return 0;
}
//问题基本上出在这块里面：
//我苦想研究了许多天也没有解决
const DoubleSubscriptedArray &DoubleSubscriptedArray:: operator= (const DoubleSubscriptedArray &right)
{
if(&right!= this) {
for(int h =0; h < x; h++)
delete [] ptr[h];
delete [] ptr;
x=right.x;
y=right.y;
ptr=new int*[right.x];
for(int k=0; k <right.x; k++)
ptr[k] = new int[right.y];
//assert(ptr!=0);
for(int i=0;i <x;i++)
for(int j=0;j <y;j++)
ptr[i][j] = right.ptr[i][j];
}
return *this;
}
istream &operator> > (istream &input, DoubleSubscriptedArray &a)
{
for(int i=0;i <a.x;i++)
for(int j=0;j <a.y;j++)
input> > a.ptr[i][j];
return input;
}
ostream &operator < < (ostream &output, const DoubleSubscriptedArray &a)
{
for(int i=0;i <a.x;i++)
for(int j=0;j <a.y;j++)
{
output < <a.ptr[i][j] < < ' ';
if(j==(a.y-1))
output < <endl < < " ";
}
return output;
}
//下面就是测试主程序啦
//大家看看！
#include <iostream.h>
#include "array.h "
void main()
{
DoubleSubscriptedArray a1(2,3),a2(2,2);
cout < < "input 10 intgers " < <endl;
cin> > a1> > a2;
cout < < "a1: " < <a1 < <endl;
cout < < "a2: " < <a2 < <endl;
cout < < "a2(0,0)= " < <a2.access(0,0) < <endl;
if(a1!=a2)
cout < < "the are not equal! " < <endl;
a1=a2;
cout < < "a1: " < <a1 < <endl;
cout < < "a2: " < <a2 < <endl;
if(a1==a2)
cout < < "the are equal! " < <endl;
}
各位兄弟，可以将代码运行检查一下，我研究一周！我实在没有办法弄出问题在哪里？
跪求高手！必将分送给解答正确的那位！
[解决办法]
const DoubleSubscriptedArray &operator= (const DoubleSubscriptedArray &);
改成
DoubleSubscriptedArray & operator= (const DoubleSubscriptedArray &);
[解决办法]
在 operator= 中：
x=right.x;
y=right.y;
size=right.size; //增加这个语句
[解决办法]
否则，在operator== 中
if(size!=right.size || x!=right.x || y!=right.y)
第一个条件分支就不会成立
（除非一开始就是 == 的）
[解决办法]
我觉得 size 的概念用成员函数来表达更合适。
int DoubleSubscriptedArray::size() const {
return x * y;
}
[解决办法]
class DoubleSubscriptedArray
{
.......
const DoubleSubscriptedArray &operator= (const DoubleSubscriptedArray &);
////////====> DoubleSubscriptedArray &operator= (const DoubleSubscriptedArray &);
......................
};
.............
DoubleSubscriptedArray:: DoubleSubscriptedArray(int m,int n)
{
x=m;
y=n;
size =x*y;
ptr=new int *[x];
for(int j=0;j <y;j++) //////=====> for (int j=0; j <x; ++j)
ptr[j]=new int[y];
...........................
}
const DoubleSubscriptedArray &DoubleSubscriptedArray:: operator=
/////====> DoubleSubscriptedArray &DoubleSubscriptedArray:: operator=
(const DoubleSubscriptedArray &right)
{
if(&right!= this) {
for(int h =0; h < x; h++)
delete [] ptr[h];
delete [] ptr;
x=right.x;
y=right.y;
//////====> size = rhs.size;
.............
}
