C++

Question

所以我在下面有一个完整的程序，它创建 Book 对象，初始化它们，并打印在程序执行过程中创建或销毁的任何 constructors/destructors。

我有运行我的代码（并粘贴了下面的输出），但我无法理解如何调用析构函数。所以我知道构造函数的销毁顺序与创建顺序相反。但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 "explicit call to the copy constructor"，另一个来自 "declaring and initializing book 6 from book 5"，另一个来自第一部分of "declaring and initializing books 1 to 4." 但是我很困惑额外的 id 4 是从哪里来的？

此外，我想知道为什么没有为创建默认 ctor: 0 的 "declaring book 5" 部分打印“-- dtor: 0”。

如有任何说明，我将不胜感激！

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

Answer 1

I don't get why four of the destructor statements have an id of 4

因为你在语句

中将b4赋值给b5

b5 = b4;

然后复制构造 b6 = b5; 和 b7(b6); 它们每个都有 id = 4，所以析构函数打印

-- dtor: 4

Additionally, I was wondering why a "-- dtor: 0" wasn't printed for the "declaring book 5" part where default ctor: 0 was created.

因为当 Book b5; 创建时它有 id = 0 但是当代码分配 b4 到 b5 id 变成 4，因此没有“- - dtor: 0" 已打印。

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