编程作业: 编程作业—运算符重载

4w3:第四周程序填空题1

描述

下面程序的输出是：

3+4i

5+6i

请补足Complex类的成员函数。不能加成员变量。

#include <iostream>
#include <cstring>
#include <cstdlib>
using namespace std;
class Complex {
private:
    double r,i;
public:
    void Print() {
        cout << r << "+" << i << "i" << endl;
    }

// 在此处补充你的代码

};
int main() {
    Complex a;
    a = "3+4i"; a.Print();
    a = "5+6i"; a.Print();
    return 0;
}

输入无输出3+4i
5+6i样例输入

None

样例输出

3+4i
5+6i

Approach:

#include <iostream>
#include <cstring>
#include <cstdlib>
using namespace std;
class Complex {
private:
    double r,i;
public:
    void Print() {
        cout << r << "+" << i << "i" << endl;
    }
    Complex(): r(0), i(0){};
    Complex& operator=(string s) {
        int pos = s.find('+', 0);
        string sTmp = s.substr(0, pos);
        r = stoi(sTmp);
        sTmp = s.substr(pos+1);
        i = stoi(sTmp);
    }
};
int main() {
    Complex a;
    a = "3+4i"; a.Print();
    a = "5+6i"; a.Print();
    return 0;
}

　　

4w4:第四周程序填空题2

描述

下面的MyInt类只有一个成员变量。MyInt类内部的部分代码被隐藏了。假设下面的程序能编译通过，且输出结果是:

4,1

请写出被隐藏的部分。（您写的内容必须是能全部放进 MyInt类内部的，MyInt的成员函数里不允许使用静态变量）。

#include <iostream>
using namespace std;
class MyInt  {
    int nVal;
    public:
        MyInt(int n) { nVal = n; }
        int ReturnVal() { return nVal; }

// 在此处补充你的代码

};
int main ()  {
    MyInt objInt(10);
    objInt-2-1-3;
    cout << objInt.ReturnVal();
    cout <<",";
    objInt-2-1;
    cout << objInt.ReturnVal();
    return 0;
}

输入无输出4,1样例输入

无

样例输出

4,1

Code:

#include <iostream>
using namespace std;
class MyInt  {
    int nVal;
    public:
        MyInt(int n) { nVal = n; }
        int ReturnVal() { return nVal; }
// 在此处补充你的代码
        MyInt& operator-(int s) {
            nVal -= s;
            return *this;
        }
};
int main ()  {
    MyInt objInt(10);
    objInt-2-1-3;
    cout << objInt.ReturnVal();
    cout <<",";
    objInt-2-1;
    cout << objInt.ReturnVal();
    return 0;
}

　　

4w5:第四周程序填空题3

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;

class Array2 {

// 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

None

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,

Code:

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

class Array2 {
// 在此处补充你的代码
private:
    int row, column;
    int **p;    //定义一个指针，指的是*int类型的东西
public:
    Array2(int row_, int column_):row(row_), column(column_) {
        p = new int*[row];  //左右两边都是int**
        for (int i = 0; i < column; ++i) {
            p[i] = new int[column];
        }
    }
    Array2(){}
    int* operator[](int a) {
        return p[a];    //只要重载第一个[]就可以，返回的是int*的数组指针
    }
    int operator()(int a, int b) {
        return p[a][b];
    }
    Array2(const Array2& a) {   //深拷贝
        row = a.row;
        column = a.column;
        p = new int*[row];
        for (int i = 0; i < column; ++i) {
            p[i] = new int[column];
        }

        for (int i = 0; i < row; ++i) {
            for (int j = 0; j < column; ++j) {
                p[i][j] = a.p[i][j];
            }
        }
    }
    void operator=(const Array2 &a) {   //深拷贝
        row = a.row;
        column = a.column;
        p = new int*[row];
        for (int i = 0; i < column; ++i) {
            p[i] = new int[column];
        }

        for (int i = 0; i < row; ++i) {
            for (int j = 0; j < column; ++j) {
                p[i][j] = a.p[i][j];
            }
        }
    }
};

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;
}