为什么派生成员函数不能访问基类对象的受保护成员？ [英] Why can derived member function not access protected member of a baseclass object?

问题描述

如果我通过引用传递基类对象（可能不会在这里通过引用传递

的差异）作为

派生类的参数成员函数，成员函数不允许
访问基础对象的受保护数据成员。这令我惊喜

我。


有人可以解释为什么会这样吗？我怀疑有一个很好的理由和

我只是在一个缓慢的一天不能自己想出来。


Bob

解决方案

这是一个展示我的意思的示例程序：


// Base.h

#ifndef Base_incl

#define Base_incl


class Derived; //前向声明 - 必需因为Method2（）

of Base类引用Derived类

class Base

{

public：

Base（）; //构造函数


virtual void Method1（Base&）;

virtual void Method2（Base&）;

virtual void Method3（Derived&）; //将

//派生类对象作为基类

//类方法的参数是不常见的，但正如您所见，它可以完成。

受保护：

int mBase;

};


#endif


// Derived.h

#ifndef Derived_incl

#define Derived_incl


#include" Base.h"


class派生：公共基地

{

公开：

派生（ ）; //构造函数


virtual void Method2（Base&）; //这个方法

//覆盖基类Method2（）

virtual void Method4（Base&）; //一个新方法

//没有为Base类定义


private：

int mDerived;

};


#endif


// Base.cpp

#include" ; Base.h"

#include" Derived.h"


#include< iostream>

使用std： ：cout;

使用std :: endl;


Base :: Base（）

{

this-> mBase = 2;

}


void Base :: Method1（Base& B_Param）

{

cout<< （B_Param.mBase）*（this-> mBase）<<结束;

}


void Base :: Method2（Base& B_Param）

{

cout<< this-> mBase<<结束;

}


void Base :: Method3（Derived& D_Param）

{

基地& BRef = static_cast< Base&>（D_Param）; //此演员表创建

//对D_Param对象的基类引用。

cout<< BRef.mBase<<结束;

}


//Derived.cpp

#include" Derived.h"

#include< iostream>

使用std :: cout;

使用std :: endl;


派生： ：派生（）

{

this-> mBase = 3;

this-> mDerived = 5;

}


void Derived :: Method2（Base& B_Param）

{

cout<< this-> mDerived<< endl;

}


void Derived :: Method4（Base& B_Param）

{

派生* DPtr = dynamic_cast< Derived *（& B_Param）; //这个演员

//创建一个Derived类指向B_Param的指针，当且仅当，

// B_Param实际上是Derived类对象，否则

//指针将被设置为0


if（DPtr！= 0）//如果B_Param实际上是派生类对象

cout << DPtr-> mDerived<< endl;

else

cout<< B_Param.mBase<< endl;

}


如果我编译Derived.cpp我会收到以下错误：


1> c ：\documents and settings\blangela\desktop\polymorphismtest

\ derived.cpp（27）：错误C2248：''Base :: mBase''：无法访问

受保护的成员在类''Base'中声明'

1 c：\documents and settings\blangela\desktop\polymorphismtest

\ base .h（17）：看到''Base :: mBase'的声明'

1 c：\documents and settings \ blanlala \\\dktop \ polymmorphismtest

\ base.h（7）：看到''基地'的声明'

对不起，这是行：

cout<< B_Param.mBase<<在下面的成员函数中导致错误的
。


void Derived :: Method4（Base& B_Param）

{

Derived * DPtr = dynamic_cast< Derived *（& B_Param）; //这个

cast

//创建一个指向B_Param的派生类指针，如果和

只有，如果，

// B_Param实际上是派生类对象，否则



//指针将被设置为0


if（ DPtr！= 0）//如果B_Param实际上是派生类

对象

cout<< DPtr-> mDerived<< endl;

else

cout<< B_Param.mBase<< endl;

}

blangela写道：

如果我通过基地class引用的对象（可能不会在这里通过引用传递一个

的区别）作为

派生类成员函数的参数，成员函数不是允许

访问基础对象的受保护数据成员。这令我惊喜

我。


有人可以解释为什么会这样吗？我怀疑有一个很好的理由和

我只是在一个缓慢的日子里自己没有想出来。

这不是FAQ中的问题吗？它应该是。


简而言之，原因是阻止访问不同类型的成员

（跨层次结构）。如果D1和D2来自B，那么D1

的实例不允许访问D2的任何非公开成员（当然，除非他们是朋友，否则是b $ b朋友）。当您将对B的引用传递给D1的成员

函数时，对该类成员的访问被阻止，因为

它*可以*是D2的子对象对象，而不一定是另一个

D1。需要一个例子吗？看看档案，过去几年我们已经多次讨论了这个话题

。


V

-

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我没有回复最热门的回复，请不要问

If I pass a base class object by reference (likely does not make a
difference here that it is passed by reference) as a parameter to a
derived class member function, the member function is not allowed to
access the protected data members of the base object. This surprises
me.

Can someone explain why this is? I suspect there is a good reason and
I am just having a slow day to not come up with it myself.

Bob

解决方案

Here is a sample program to show what I mean:

// Base.h
#ifndef Base_incl
#define Base_incl

class Derived; // forward declaration - required because Method2()
// of the Base class references the Derived class
class Base
{
public:
Base(); // constructor

virtual void Method1(Base &);
virtual void Method2(Base &);
virtual void Method3(Derived &); //it is unusual to have a
// derived class object as a parameter for a base
// class method, but as you see, it can be done.
protected:
int mBase;
};

#endif

// Derived.h
#ifndef Derived_incl
#define Derived_incl

#include "Base.h"

class Derived: public Base
{
public:
Derived(); // constructor

virtual void Method2(Base &); // this method
// overrides Base class Method2()
virtual void Method4(Base &); // a new method
// which is not defined for the Base class

private:
int mDerived;
};

#endif

// Base.cpp
#include "Base.h"
#include "Derived.h"

#include <iostream>
using std::cout;
using std::endl;

Base::Base()
{
this->mBase = 2;
}

void Base::Method1(Base & B_Param)
{
cout << (B_Param.mBase) * (this->mBase) << endl;
}

void Base::Method2(Base & B_Param)
{
cout << this->mBase << endl;
}

void Base::Method3(Derived & D_Param)
{
Base & BRef = static_cast<Base &>(D_Param); // This cast creates
// a Base class reference to the D_Param object.
cout << BRef.mBase << endl;
}

//Derived.cpp
#include "Derived.h"
#include <iostream>
using std::cout;
using std::endl;

Derived::Derived()
{
this->mBase = 3;
this->mDerived = 5;
}

void Derived::Method2(Base & B_Param)
{
cout << this->mDerived << endl;
}

void Derived::Method4(Base & B_Param)
{
Derived * DPtr = dynamic_cast <Derived *(&B_Param); // this cast
// creates a Derived class pointer to B_Param, if and only if,
// B_Param actually is Derived class object, otherwise the
// pointer will be set to 0

if (DPtr != 0) // if B_Param is actually a Derived class object
cout << DPtr->mDerived << endl;
else
cout << B_Param.mBase << endl;
}

If I compile Derived.cpp I get the following errors:

1>c:\documents and settings\blangela\desktop\polymorphismtest
\derived.cpp(27) : error C2248: ''Base::mBase'' : cannot access
protected member declared in class ''Base''
1 c:\documents and settings\blangela\desktop\polymorphismtest
\base.h(17) : see declaration of ''Base::mBase''
1 c:\documents and settings\blangela\desktop\polymorphismtest
\base.h(7) : see declaration of ''Base''

Sorry, it is the line:

cout << B_Param.mBase << endl;

in the member function below that causes the errors.

void Derived::Method4(Base & B_Param)
{
Derived * DPtr = dynamic_cast <Derived *(&B_Param); // this
cast
// creates a Derived class pointer to B_Param, if and
only if,
// B_Param actually is Derived class object, otherwise
the
// pointer will be set to 0

if (DPtr != 0) // if B_Param is actually a Derived class
object
cout << DPtr->mDerived << endl;
else
cout << B_Param.mBase << endl;
}

blangela wrote:

If I pass a base class object by reference (likely does not make a
difference here that it is passed by reference) as a parameter to a
derived class member function, the member function is not allowed to
access the protected data members of the base object. This surprises
me.

Can someone explain why this is? I suspect there is a good reason and
I am just having a slow day to not come up with it myself.

Isn''t this in the FAQ? It should be.

In short, the reason is to prevent access to members of a different type
(across the hierarchy). If D1 and D2 derive from B, an instance of D1
is not allowed to access any non-public members of D2 (unless they are
friends, of course). When you pass a reference to B to a member
function of D1, the access to members of that class is blocked because
it *can* be a subobject of a D2 object, and not necessarily of another
D1. Need an example? Look in the archives, we had that topic discussed
several times over the past years.

V
--
Please remove capital ''A''s when replying by e-mail
I do not respond to top-posted replies, please don''t ask

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