题目
填空题-3
1 point possible (graded)
- 总时间限制:
- 1000ms
- 内存限制:
- 65536kB
- 描述
-
写一个二维数组类 Array2,使得下面程序的输出结果是：0,1,2,3,
4,5,6,7,
8,9,10,11,
next
0,1,2,3,
4,5,6,7,
8,9,10,11,
程序：
-
#include <iostream> #include <cstring> using namespace std;
// 在此处补充你的代码
-
int main() { Array2 a(3,4); int i,j; for( i = 0;i < 3; ++i ) for( j = 0; j < 4; j ++ ) a[i][j] = i * 4 + j; for( i = 0;i < 3; ++i ) { for( j = 0; j < 4; j ++ ) { cout << a(i,j) << ","; } cout << endl; } cout << "next" << endl; Array2 b; b = a; for( i = 0;i < 3; ++i ) { for( j = 0; j < 4; j ++ ) { cout << b[i][j] << ","; } cout << endl; } return 0; } - 输入
- 无
- 输出
- 0,1,2,3,
4,5,6,7,
8,9,10,11,
next
0,1,2,3,
4,5,6,7,
8,9,10,11, - 样例输入
-
无
- 样例输出
-
0,1,2,3, 4,5,6,7, 8,9,10,11, next 0,1,2,3, 4,5,6,7, 8,9,10,11, 答题：
#include <iostream>
#include <cstring>
using namespace std;
class Array2
{
int *ptr;
int row;//行数
int col;//列数
public:
Array2(int a,int b):row(a),col(b)
{
ptr=new int[row];
for(int i = 0 ; i < row; i++)
{
ptr[i] = reinterpret_cast<int>(new int[col]); //把int* 转化成int
}
}
Array2(){ptr=NULL;row=0;col=0;}
~Array2()
{
for(int i = 0 ; i < row; i++)
{
delete []reinterpret_cast<int*>(ptr[i]);
}
delete []ptr;
}
int* operator[](int a)
{
return reinterpret_cast<int*>(ptr[a]);
}
Array2& operator =(const Array2 &k)
{
if(ptr==k.ptr) return *this;
if(ptr!=NULL)
{
for(int i = 0 ; i < row; i++)
{
delete []reinterpret_cast<int*>(ptr[i]);
}
delete []ptr;
}
if(k.ptr==NULL)
{
ptr=NULL;
row=0;
col=0;
return *this;
}
ptr=new int[k.row];
row=k.row;
col=k.col;
for(int i = 0 ; i < row; i++)
{
ptr[i] = reinterpret_cast<int>(new int[col]);
}
for(int i = 0 ; i < row; i++)
for(int j = 0 ; j < col; j++)
{(reinterpret_cast<int*>(ptr[i]))[j]=(reinterpret_cast<int*>(k.ptr[i]))[j];}
cout<<"done"<<endl;
return (*this);
}
int& operator()(int a,int b)
{
return (*this)[a][b];
}
};
int main()
{
Array2 a(3,4);
int i,j;
for( i = 0;i < 3; ++i )
for( j = 0; j < 4; j ++ )
a[i][j] = i * 4 + j;
for( i = 0;i < 3; ++i )
{
for( j = 0; j < 4; j ++ )
{
cout << a(i,j) << ",";
}
cout << endl;
}
cout << "next" << endl;
Array2 b;
b = a;
for( i = 0;i < 3; ++i )
{
for( j = 0; j < 4; j ++ )
{
cout << b[i][j] << ",";
}
cout << endl;
}
return 0;
}
核心难点是双[]运算符怎么重载出来，思想是第一个[]中放地址，然后第二个[]就可以用默认的寻址运算，这样只需要对第一个[]进行重载。
实现的过程中注意运用reinterpret_cast即c++的类型转换。
原文地址：https://www.cnblogs.com/xuhaoforwards/p/9248638.html