在C ++中，为什么地址在指针转换时改变？ [英] In C++, why is the address changed when the pointer is converted?

问题描述

以下是代码：

  #include< iostream> 
 using namespace std; 
 
 class B1 {
 public：
 
 virtual void f1（）{
 cout< B1\\\
; 
} 
}; 
 
 class B2 {
 public：
 
 virtual void f1（）{
 cout< B2\\\
; 
} 
}; 
 
 class D：public B1，public B2 {
 public：
 void f1（）{
 cout< OK\\\
; 
} 
}; 
 
 int main（）{
 
 D dd; 
 B1 * b1d =& dd; 
 B2 * b2d =& dd; 
 D * ddd =& dd; 
 
 cout<< b1d<< endl; 
 cout<< b2d<< endl; 
 cout<< ddd<< endl; 
 
 b1d  - > f1（）; 
 b2d  - > f1（）; 
 ddd  - > f1（）; 
} 
  

输出为：

  0x79ffdf842ee0 
 0x79ffdf842ee8 
 0x79ffdf842ee0 
 OK 
 OK 
 OK 
  

这看起来让我很困惑，因为我期望 b1d 和 b2d 将与它们都指向 dd 相同。但是，根据结果， b1d 和 b2d 的值不同。我认为这可能与类型铸造有关，但我不知道它是如何工作的。

有没有人有这个想法？

c> c>继承 B1 和 B2 。

由于 B1 $ c> B1 首先构造对象的一部分，然后创建对象的 B2 部分，然后创建 D 。

因此，当您将派生类型的指针转​​换为基本类型时，您所看到的是这些部分在内存中的区别。

b1d 和 ddd 具有相同的地址，因为它们都指向 c>

B2 D 的一部分。

Following is the code:

#include <iostream>
using namespace std;

class B1 {
public:

  virtual void f1() {
      cout << "B1\n";
      }
};

class B2 {
public:

  virtual void f1() {
      cout << "B2\n";
      }
};        

class D : public B1, public B2 {
public:
  void f1() {
      cout << "OK\n" ;
  }
};

int main () {

D dd;    
B1 *b1d = &dd;
B2 *b2d = &dd;
D *ddd = &dd;

cout << b1d << endl;
cout << b2d << endl;
cout << ddd << endl;

b1d -> f1();
b2d -> f1();
ddd -> f1();
}

The output is :

0x79ffdf842ee0
0x79ffdf842ee8
0x79ffdf842ee0
OK
OK
OK

This looks confusing to me, because I expected b1d and b2d would be the same as both of them point to dd. However, the value of b1d and b2d is different according to the result. I thought it may be related to type casting but I'm not sure how it works.

Does anyone have ideas about this?

解决方案

D inherits from B1 andB2.

Since B1 is inherited from first the B1 part of the object is going to be constructed first and then the B2 part of the object will be created then D.

So what you are seeing is the difference of where those parts are in memory when you cast a pointer of the derived type to the base type.

b1d and ddd have the same address as they both point to the beginning of the class in memory.

b2d is offset as it points to the start of the B2 part of D.

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