了解复制构造函数的工作

Question

为了理解构造函数、复制构造函数、析构函数的工作原理，我编写了以下代码。

#include <iostream>
 using namespace std; 
 class Cat
{
 public:
  Cat();
  Cat(Cat&);
  ~Cat();
  int itsage;
};
 
 Cat::Cat()
{
 cout << "Constructor called\n";
 cout << this << endl;
 itsage=2;
}
 
 Cat::Cat(Cat& theCat)
{
 cout << "Copy constructor called\n";
 cout << this << endl;
 itsage=theCat.itsage;
}

 Cat::~Cat()
{
 cout << "Destructor called\n";
 cout << this << endl;
}
 
 Cat myFunction(Cat Frisky)
{
 cout << "Inside myFunction\n";
 cout << "Frisky's address : " << &Frisky ; 
 cout << "\nFrisky's age: " << Frisky.itsage << "\n";
 Frisky.itsage=100;
 cout << "Reassigned Frisky's age: "<< Frisky.itsage << "\n";
 return Frisky;
}
 
 int main()
{
 Cat Mani;
 cout << "Mani's address : " << &Mani ;
 cout << "\nMani's age: " << Mani.itsage << "\n";
 myFunction(Mani);
 return 0;
}

我得到如下输出：

Constructor called
0x61ff04
Mani's address : 0x61ff04
Mani's age: 2
Copy constructor called
0x61ff0c
Inside myFunction
Frisky's address : 0x61ff0c
Frisky's age: 2
Reassigned Frisky's age: 100
Copy constructor called
0x61ff08
Destructor called
0x61ff08
Destructor called
0x61ff0c
Destructor called
0x61ff04

除了在第二次调用复制构造函数时存储在地址0x61ff08 的内容之外，一切对我来说都很好？意味着我们可以看到存储在地址0x61ff0c 和0x61ff04 中的内容，它们只不过是Frisky 和Mani。那0x61ff08那个看不见的东西是什么？

我想通过在 main 函数中做一些小改动来使该对象可见。

#include <iostream>
 using namespace std; 
 class Cat
{
 public:
  Cat();
  Cat(Cat&);
  ~Cat();
  int itsage;
};
 
 Cat::Cat()
{
 cout << "Constructor called\n";
 cout << this << endl;
 itsage=2;
}
 
 Cat::Cat(Cat& theCat)
{
 cout << "Copy constructor called\n";
 cout << this << endl;
 itsage=theCat.itsage;
}

 Cat::~Cat()
{
 cout << "Destructor called\n";
 cout << this << endl;
}
 
 Cat myFunction(Cat Frisky)
{
 cout << "Inside myFunction\n";
 cout << "Frisky's address : " << &Frisky ; 
 cout << "\nFrisky's age: " << Frisky.itsage << "\n";
 Frisky.itsage=100;
 cout << "Reassigned Frisky's age: "<< Frisky.itsage << "\n";
 return Frisky;
}
 
 int main()
{
 Cat Mani;
 cout << "Mani's address : " << &Mani ;
 cout << "\nMani's age: " << Mani.itsage << "\n";
 Cat Sweety = myFunction(Mani); 
 cout << "Sweety's age : " << Sweety.itsage ;
 return 0;
}

但收到如下错误：

ppp.cpp: In function 'int main()':
ppp.cpp:47:25: error: invalid initialization of non-const reference of type 'Cat&' from an rvalue of type 'Cat'
  Cat Sweety = myFunction(Mani);
               ~~~~~~~~~~^~~~~~
ppp.cpp:19:2: note:   initializing argument 1 of 'Cat::Cat(Cat&)'
  Cat::Cat(Cat& theCat)
  ^~~

我真的不明白出了什么问题。

Answer 1

您的第一个代码从myFunction 返回Frisky 的副本，这就是额外副本的来源。

您的第二个代码不起作用，因为您的复制构造函数错误地采用了非常量引用，并且您无法将从 myFunction 返回的临时值传递给非常量引用。

与您的问题没有直接关系，但您应该遵守rule of three 并实现赋值运算符。

Answer 2

Cat(Cat &);

在此处使用const。

Cat(const Cat &);

不相关，但考虑使用这样的初始化。

Cat Sweety{myFunction(Mani)};