了解复制构造函数的工作
Understanding working of copy constructor
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
除了第二次调用复制构造函数时存储在地址 0x61ff08
的内容外,一切正常?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
。那么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;
}
但是收到错误如下:
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)
^~~
我真的不明白哪里出了问题。
您的第一个代码 returns 来自 myFunction
的 Frisky
的副本,这是额外副本的来源。
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 non-const 引用,并且您无法将从 myFunction
返回的临时值传递给 non-const 引用。
与您的问题没有直接关系,但您应该遵守 rule of three 并实施赋值运算符。
Cat(Cat &);
这里使用const
。
Cat(const Cat &);
不相关,但考虑使用这样的初始化。
Cat Sweety{myFunction(Mani)};
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
除了第二次调用复制构造函数时存储在地址 0x61ff08
的内容外,一切正常?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
。那么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;
}
但是收到错误如下:
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)
^~~
我真的不明白哪里出了问题。
您的第一个代码 returns 来自 myFunction
的 Frisky
的副本,这是额外副本的来源。
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 non-const 引用,并且您无法将从 myFunction
返回的临时值传递给 non-const 引用。
与您的问题没有直接关系,但您应该遵守 rule of three 并实施赋值运算符。
Cat(Cat &);
这里使用const
。
Cat(const Cat &);
不相关,但考虑使用这样的初始化。
Cat Sweety{myFunction(Mani)};