Question

我有班级狗和班主任。在类Dog中有一个成员，它是对类所有者（育种者）标本的引用。因此，当我至少有一个狗标本并且由其他狗的示例共用并在其构造中使用时，我需要创建这个标本。

class Dog 
    {
       private:
         Owner & m_owner;
         ...
       public:
           static Owner & m_breeder;
         ...
    }

class Owner
    {private:
         ...
     public:
         ...
      }

dog.cpp

  {
    #include "dog_class.h"
    #include <iostream>
    #include <cstdlib>
    #include <string>
    #include <ctime>

    Owner breeder("breeder");
    Owner & Dog::m_breeder=breeder;
    //methods here
    Dog () : m_owner(m_breeder) {}

    ...
  }

这不起作用（错误：使用已删除的功能'Dog＆amp; Dog :: operator =（const Dog＆amp;）; 我试图将所有者包括为“静态所有者饲养员”;在Dog类的公共部分，稍后在dog.cpp中初始化它，但它也不起作用。怎么了，我的目标是如何获得的？

这里有一些代码，基本上是类Dog和Owner的构造函数：

dog_class_methods.cpp

Dog::Dog(std::string name, int size, int status, int loyalty, int sex, int price, std::vector<Dog> * market, Owner & owner) : m_owner(owner)
{
m_name=name;
m_size=size;
m_status=status;
m_loyalty=loyalty;

m_sex=sex;
m_price=price;
m_market=market; //pointer to the market
m_market->push_back(*this);
//break_out();
 }



  Dog::Dog() : m_owner(m_breeder)
{
m_name="Fang";
srand(time(NULL));
m_size=(rand()%10)+1; //small: random number from 1 to 10;
m_status=0; //not owned
m_loyalty=(rand()%10)+1; //from 1 to 10
Owner no_one("no_one");
//m_owner=no_one;
m_sex=rand()%2; //rand 0 or 1;
m_price=rand()%1001; //0 - 1000;
//show_data();
//break_out();
 }

owner_class_methods.cpp

Owner::Owner()
{

m_name="random";
m_goodness=(rand()%10+1);
m_size=(rand()%10+1);
m_money=rand()%1001;
m_agility=(rand()%10+1);
m_int=(rand()%10+1);
//std::cout<<"Hi! I am the "<<m_name<<std::endl;
 }
  Owner::Owner(std::string name)
 {
 if (name=="no one")
 {
m_name="no one";
m_goodness=0;
m_size=0;
m_money=0;
m_agility=0;
m_int=0;
std::cout<<"Hi! I am the "<<m_name<<std::endl;
   }
   else
  {
m_name=name;
m_goodness=(rand()%10+1);
m_size=(rand()%10+1);
m_money=rand()%1001;
m_agility=(rand()%10+1);
m_int=(rand()%10+1);
std::cout<<"The '"<<m_name<<"' owner made"<<std::endl;
    }
  }


 Owner::Owner(std::string name, int goodness, int size, int money, int    agility, std::vector<Dog> doglist)
  {
m_name=name;
m_size=size;
m_money=money;
m_agility=agility;
for (int i=0;  i<doglist.size();i++)
    m_doglist.push_back(doglist[i]);
  }

“使用已删除的功能'Dog＆amp; Dog :: operator =（const Dog＆amp;）”不断出现！请帮助！

Answer 1

class Dog {
    Dog(Owner& owner)
    : m_owner(owner)
    {}
    Owner& m_owner;
};

Owner breeder("breeder");
Dog the_dog(breeder);

Answer 2

出现错误是因为使用了对象，Dog对象可以复制，但是类中有一个引用成员，需要编写operator =：

  Dog Dog::operator=(const Dog & a)
 {    m_owner=a.m_owner;
      m_loyalty=a.m_loyalty;
      m_size=a.m_size;
      m_status=a.m_status;
      m_sex=a.m_sex;
      m_price=a.m_price;
      m_market=a.m_market; //market один для всех собак;
      m_name=a.m_name;
      return *this;
 }

这解决了整个事情。

在另一个类中创建一个静态类变量，使用deleted function = error

2 个答案: