Beijing University C + + programming Coursera course Fourth week in question 3

Questions -31 point Possible (graded)

Total time limit:

1000ms

Memory Limit:

65536kB

Describe

Write a two-dimensional array class Array2, so that the following program output is:

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,

Program:

#include <iostream> #include <cstring>using namespace std;

Add your code here

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;}

Input

No

Output

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,

Sample input

No

Sample output

0,1,2,3,4,5,6,7,8,9,10,11,next0,1,2,3,4,5,6,7,8,9,10,11,

Answer:

#include <iostream> #include <cstring>using namespace std;class array2{int *ptr;　　　　int row;//number of rows int col;//column Count 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]);//convert int* to int}} ARR　　Ay2 () {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]= (re 　　　　Interpret_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;} The core difficulty is how the double [] operator is overloaded, thought is the first [] place address, and then the second [] can use the default addressing operation, so that only the first [] will be overloaded. In the implementation process, note the use of reinterpret_cast, or C + + type conversion.

Peking University C + + programming Coursera course Fourth Week 3