为什么C ++编译器不知道指针指向派生类? [英] Why can't C++ compiler know a pointer is pointing to a derived class?

问题描述

我刚刚开始学习C ++中的OOP.我想知道为什么需要虚拟关键字来指示编译器进行后期绑定?为什么编译器在编译时不知道指针指向派生类?

I've just started learning about OOP in C++. I was wondering why is the virtual keyword needed to instruct the compiler to do late binding ? Why can't the compiler know at compile time that the pointer is pointing to a derived class ?

class A { 
    public: int f() { return 'A';}
};

class B : public A {
    public: int f() { return 'B';}
};

int main() {

    A* pa;
    B b;
    pa = &b; 
    cout << pa->f() << endl;
}

推荐答案

假设您有一个简单的类层次结构，如

Lets say you have a simple class-hierarchy like

class Animal
{
    // Generic animal attributes and properties
};

class Mammal : public Animal
{
    // Attributes and properties specific to mammals
};

class Fish : public Animal
{
    // Attributes and properties specific to fishes
};

class Cat : public Mammal
{
    // Attributes and properties specific to cats
};

class Shark : public Fish
{
    // Attributes and properties specific to sharks
};

class Hammerhead : public Shark
{
    // Attributes and properties specific to hammerhead sharks
};

_{[有点long，但是我想让具体的"类彼此远离]}

现在可以说我们有一个类似的功能

Now lets say we have a function like

void do_something_with_animals(Animal* animal);

最后让我们调用此函数:

And finally let's call this function:

Fish *my_fish = new Hammerhead;
Mammal* my_cat = new Cat;

do_something_with_animals(my_fish);
do_something_with_animals(my_cat);

现在，如果我们想一想，在do_something_with_animals函数中，实际上没有办法确切地了解 参数animal可能指向什么.是Mammal吗? Fish?特定的Fish子类型?

Now if we think a little, in the do_something_with_animals function there is really no way of knowing exactly what the argument animal might point to. Is it a Mammal? A Fish? A specific Fish sub-type?

如果在不同的转换单元，其中甚至可能没有Mammal和Fish类(或其任何子类)的定义.

This is even harder for the compiler if the do_something_with_animals function is defined in a different translation unit, where the definition of the Mammal and Fish classes (or any of its sub-classes) might not even be available.

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