使用基类函数指针访问派生类成员函数

问题描述

我在我的项目中使用函数指针，面对问题，创建一个测试用例来显示它...下面代码失败与下面的错误MSVC2005（简单的话，我想访问dervied类函数通过基类函数指针）

  class ClassA {
 public：
 virtual void foo（）
 {
 printf（Foo Parent）; 
} 
}; 
 
 typedef void（ClassA :: * ClassAFoo）（）; 
 
 class ClassB：public ClassA {
 public：
 virtual void foo（）
 {
 printf（Foo Derived）; 
} 
}; 
 
 int main（）{
 ClassAFoo fPtr; 
 fPtr =& ClassB :: foo; 
} 
  

我的问题是

1. 是C ++的行为，我不能通过基类函数指针或其编译器错误访问派生类函数？


2. 我一直在玩上面的case ，如果我注释掉 ClassB :: foo ，这段代码编译正常，没有任何进一步的修改，为什么这样，应该不 fPtr =& ClassB :: foo; 再次导致编译时错误？

解决方案

ol>

这是正确的行为。以这种方式考虑： ClassB 的所有实例都具有成员 ClassA :: foo ，但不是所有 ClassA 有成员 ClassB :: foo ;只有那些实际是 ClassB 实例的基类子对象的 ClassA 实例才有它。因此，将 ClassB :: foo 分配到 ClassAFoo ，然后使用 ClassAFoo 结合纯 ClassA 对象将试图调用不存在的函数。

如果从 ClassB 中删除​​ foo ，表达式 ClassB :: foo acutally是指在 ClassB 中继承的 ClassA :: foo ，因此没有问题。

要详细说明1. further：指向成员的指针实际上与正常指针相反。使用正常指针，可以将 ClassB * 分配到 ClassA * 中（因为类B 也是 ClassA 的实例，但反之亦然。使用成员指针，可以将 ClassA :: * 分配到 ClassB :: * （因为 ClassB 包含 ClassA 的所有成员，但反之不行。

I am using function pointer in my project, facing problem, created a test case to show it... below code fail with below error on MSVC2005 (in simple words i want to access dervied class function through base class function pointer)

class ClassA {
 public:
    virtual void foo()
    {
       printf("Foo Parent");
    }
};

typedef void (ClassA::*ClassAFoo)();

class ClassB : public ClassA {
 public:
    virtual void foo()
    {
        printf("Foo Derived");
    }
};

int main() {
    ClassAFoo fPtr;
    fPtr = &ClassB::foo;
}

My questions are

1. Is it C++ behavior that I cant access derived class function through a base class function pointer or its a compiler bug?
2. I have been playing with above case, if i comment out ClassB::foo, this code compile fine, without any further modification, Why is this so, should not fPtr = &ClassB::foo; again result in compile time error?

解决方案

1. It's correct behaviour. Think of it this way: all instances of ClassB have the member ClassA::foo, but not all instances of ClassA have the member ClassB::foo; only those instances of ClassA which are actually the base class subobject of a ClassB instance have it. Therefore, assigning ClassB::foo into ClassAFoo and then using ClassAFoo in combination with a "pure" ClassA object would try to call a nonexistent function.

2. If you remove foo from ClassB, the expression ClassB::foo acutally refers to ClassA::foo which is inherited in ClassB, so there's no problem there.

To elaborate on 1. further: pointers to members actually work the opposite way to normal pointers. With a normal pointer, you can assign ClassB* into ClassA* (because all instances of ClassB are also instances of ClassA), but not vice versa. With member pointers, you can assign ClassA::* into ClassB::* (because ClassB contains all the members of ClassA), but not vice versa.

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