POJ C++程序设计 编程题＃3 编程作业—运算符重载
编程题 ＃3
来源: POJ (Coursera声明：在POJ上完成的习题将不会计入Coursera的最后成绩。)
注意： 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,
1 #include <iostream> 2 #include <cstring> 3 using namespace std; 4 // 在此处补充你的代码 5 class Array2 { 6 private: 7 int * a; 8 int i, j; 9 public: 10 Array2() {a = NULL;} 11 Array2(int i_, int j_) { 12 i = i_; 13 j = j_; 14 a = new int[i*j]; 15 } 16 Array2(Array2 &t){ 17 i = t.i; 18 j = t.j; 19 a = new int[i * j]; 20 memcpy(a, t.a, sizeof(int)*i*j); 21 } 22 Array2 & operator=(const Array2 &t) { 23 if (a != NULL) delete []a; 24 i = t.i; 25 j = t.j; 26 a = new int[i*j]; 27 memcpy(a, t.a, sizeof(int)*i*j); 28 return *this; 29 } 30 ~Array2() {if (a != NULL) delete []a;} 31 // 将返回值设为int的指针，则可以应用第二个【】，不用重载第二个【】操作符 32 int *operator[](int i_) { 33 return a+i_*j; 34 } 35 int &operator()(int i_, int j_) { 36 return a[i_*j + j_]; 37 } 38 }; 39 40 int main() { 41 Array2 a(3,4); 42 int i,j; 43 for( i = 0;i < 3; ++i ) 44 for( j = 0; j < 4; j ++ ) 45 a[i][j] = i * 4 + j; 46 for( i = 0;i < 3; ++i ) { 47 for( j = 0; j < 4; j ++ ) { 48 cout << a(i,j) << ","; 49 } 50 cout << endl; 51 } 52 cout << "next" << endl; 53 Array2 b; b = a; 54 for( i = 0;i < 3; ++i ) { 55 for( j = 0; j < 4; j ++ ) { 56 cout << b[i][j] << ","; 57 } 58 cout << endl; 59 } 60 return 0; 61 }
