因此,我下面有一个完整的程序,该程序创建Book对象,对其进行初始化,并打印在程序执行过程中创建或销毁的所有构造函数/析构函数。
我已经运行了我的代码(并在下面粘贴了输出),并且在理解如何调用析构函数时遇到了麻烦。因此,我知道构造函数是按照与创建它们相反的顺序销毁的。但是我不明白为什么四个析构函数语句的id为4。我假设一个来自“对副本构造函数的显式调用”,另一个来自“从第5本书中声明和初始化第6本书”,另一个来自“声明和初始化书籍1至4”的第一部分。但是我对4的额外ID来自何处感到困惑?
[另外,我想知道为什么没有为创建默认ctor:0的“声明书5”部分打印“-dtor:0”。
非常感谢您的澄清!
main.cc:
#include <iostream>
#include <string>
using namespace std;
#include "Book.h"
void func1(Book);
void func2(Book&);
int main()
{
cout<<endl<<"Declaring and initializing books 1 to 4..."<<endl;
Book b1(1, "Ender's Game", "Orson Scott Card");
Book b2(2, "Dune", "Frank Herbert");
Book b3(3, "Foundation", "Isaac Asimov");
Book b4(4, "Hitch Hiker's Guide to the Galaxy", "Douglas Adams");
cout<<endl<<"Declaring book 5..."<<endl;
Book b5;
b5.print();
cout<<endl<<"Assigning book 4 to 5..."<<endl;
b5 = b4;
b5.print();
cout<<endl<<"Declaring and initializing book 6 from book 5..."<<endl;
Book b6 = b5;
b6.print();
cout<<endl<<"Calling func1()..."<<endl;
func1(b1);
cout<<endl<<"Calling func2()..."<<endl;
func2(b2);
cout<<endl<<"Explicit call to copy constructor..."<<endl;
Book b7(b6);
cout << endl << endl;
return 0;
}
void func1(Book b)
{
b.print();
}
void func2(Book& b)
{
b.print();
}
book.cc:
#include <iostream>
#include <string>
using namespace std;
#include "Book.h"
Book::Book(int i, string t, string a)
{
cout<<"-- default ctor: "<< i <<endl;
id = i;
title = t;
author = a;
}
Book::Book(const Book& other)
{
id = other.id;
title = other.title;
author = other.author;
cout<<"-- copy ctor: "<< id <<endl;
}
Book::~Book()
{
cout<<"-- dtor: "<< id <<endl;
}
void Book::print()
{
cout<<"** "<< title <<" by "<<author<<endl;
}
book.h:
#ifndef BOOK_H
#define BOOK_H
#include <string>
using namespace std;
class Book
{
public:
Book(int=0, string="Unknown", string="Unknown");
Book(const Book&);
~Book();
void print();
private:
int id;
string title;
string author;
};
#endif
输出:
Declaring and initializing books 1 to 4...
-- default ctor: 1
-- default ctor: 2
-- default ctor: 3
-- default ctor: 4
Declaring book 5...
-- default ctor: 0
** Unknown by Unknown
Assigning book 4 to 5...
** Hitch Hiker's Guide to the Galaxy by Douglas Adams
Declaring and initializing book 6 from book 5...
-- copy ctor: 4
** Hitch Hiker's Guide to the Galaxy by Douglas Adams
Calling func1()...
-- copy ctor: 1
** Ender's Game by Orson Scott Card
-- dtor: 1
Calling func2()...
** Dune by Frank Herbert
Explicit call to copy constructor...
-- copy ctor: 4
-- dtor: 4
-- dtor: 4
-- dtor: 4
-- dtor: 4
-- dtor: 3
-- dtor: 2
-- dtor: 1
我不明白为什么四个析构函数语句的id为4
因为在语句中将b4
分配给b5
b5 = b4;
然后复制构造b6 = b5;
和b7(b6);
各自具有id
= 4,因此析构函数打印
-- dtor: 4
[另外,我想知道为什么没有为创建默认ctor:0的“声明书5”部分打印“-dtor:0”。
因为创建Book b5;
时它具有id
= 0,但是当将b4
分配给b5
id
的代码变为4时,因此没有打印“-dtor:0”。