当我初始化一个类并计算它时，它只显示默认构造函数[关闭]

Question

我在练习代码 oop 时遇到了一个问题。 我的 'class vdv' 是一名运动员，有一个默认构造函数，其中包含“姓名”、“运动”、0 年龄、0 身高和 0 体重。 我也为它编写了运算符重载 cout。 但是当 i cin >> my obj 然后 cout 它时，它只显示默认值“”名称，“”运动，0 年龄，0 高度和 0重量。

这是我的代码

#include <string>
using namespace std;
#include "VanDongVien.h"

//------------------------------------------

#pragma once
class vdv {
private:
    int age;
    string name, sport;
    double weight, height;
public:
    vdv() {
        name = sport = "No name";
        age = 0;
        weight = height = 0;
    }
    vdv(string _name, string _sport, int _age, double _height, double _weight) :
        name(_name), sport(_sport), age(_age), height(_height), weight(_weight) {};
    ~vdv() {
        name = sport = " ";
        age = 0;
        height = weight = 0;
    }
    friend istream& operator>>(istream &is, vdv obj);
    friend ostream& operator<<(ostream &os, vdv u);
    bool operator>(vdv obj);
};

//------------------------------------------

istream &operator>>(istream &is, vdv obj) {

    cout << "Nhap ho va ten: "; fflush(stdin); getline(is, obj.name);
    cout << "Nhap mon thi dau: "; fflush(stdin); getline(is, obj.sport);
    cout << "Nhap tuoi: "; cin >> obj.age;
    cout << "Nhap chieu cao: "; cin >> obj.height;
    cout << "Nhap can nang: "; cin >> obj.weight;
    return is;
}
ostream &operator<< (ostream & os, vdv obj) {
    cout << "Ho va ten: " << obj.name << endl;
    cout << "Mon thi dau: " << obj.sport << endl;
    cout << "Tuoi: " << obj.age << endl;
    cout << "Chieu cao: " << obj.height << endl;
    cout << "Can nang: " << obj.weight << endl;`enter code here`
    return os;
}
bool vdv::operator>(vdv obj) {
    if (height > obj.height) return true;
    else if (height < obj.height) return false;
    else if (weight >= obj.weight) return true;
    else return false;
}
void swap(vdv &o1, vdv &o2) {
    vdv temp = o1;
    o1 = o2;
    o2 = temp;
}
void bubblesort(vdv *a, int n) {
    for (int i = 0; i < n - 1; i++)
        for (int j = 1; j < n; j++) {
            if (a[i] > a[j]) swap(a[i], a[j]);
        }
}

//------------------------------------------

int main() {
    vdv o1;
    cin >> o1;
    cout << o1;
    system("pause");
}

vdv（Van dong vien 是运动员）。

感谢您阅读我的代码并提供帮助。

Answer 1

重载operator<< 和operator>> 时，您应该将第二个参数作为reference 传递，如下所示。我已经在我进行修改的地方添加了 cmets。

class vdv {
private:
    int age;
    std::string name, sport;
    double weight, height;
public:
    vdv() {
        name = sport = "No name";
        age = 0;
        weight = height = 0;
    }
    vdv(std::string _name, std::string _sport, int _age, double _height, double _weight) :
        name(_name), sport(_sport), age(_age), height(_height), weight(_weight) {};
    ~vdv() {
        name = sport = " ";
        age = 0;
        height = weight = 0;
    }
    friend std::istream& operator>>(std::istream &is, vdv &obj);//passed second object by reference
    friend std::ostream& operator<<(std::ostream &os, const vdv &u);//passed second object as a referece to const
    bool operator>(vdv obj);
};


std::istream &operator>>(std::istream &is, vdv &obj) {//added & on second argument

    is >> obj.name >> obj.sport >> obj.age >> obj.height >> obj.weight;
    
    //no need for cout here 
    
    //check if input succeded 
    if(is)
    {
        ;//do something here if you need to 
    }
    //otherwise leave the objec in its default state
    else 
    {
        obj = vdv();
    }
    
    
    return is;
}
std::ostream &operator<< (std::ostream & os, const vdv &obj) {//passed 2nd argument as reference to const
    os << obj.name << obj.sport << obj.age << obj.height << obj.weight;
    
    //no need for cout here
    
    return os;
}
bool vdv::operator>(vdv obj) {
    if (height > obj.height) return true;
    else if (height < obj.height) return false;
    else if (weight >= obj.weight) return true;
    else return false;
}

程序的输出可见here。

我所做的一些修改包括：

1. 第二个对象通过引用传递给operator<< 和operator>>。
2. 添加了检查operator>> 内的输入是否成功。
3. 从operator<< 和operator>> 中删除了不必要的cout 语句。
4. 删除了推荐做法的 using namespace std;。

Answer 2

我认为您需要像这样在运算符中传递 vdv 作为参考>>：

istream &operator>>(istream &is, vdv& obj) {

    cout << "Nhap ho va ten: "; fflush(stdin); getline(is, obj.name);
    cout << "Nhap mon thi dau: "; fflush(stdin); getline(is, obj.sport);
    cout << "Nhap tuoi: "; cin >> obj.age;
    cout << "Nhap chieu cao: "; cin >> obj.height;
    cout << "Nhap can nang: "; cin >> obj.weight;
    return is;
}

当您编写 cin >> o1 时，它将表现为 operator>>(cin, o1) 但您通过值传递 o1 因此在 operator>> 函数中，它会创建另一个副本，该副本与您尝试读取的副本无关。