C++ 对 vtable 和继承的未定义引用

文件 A.h

#ifndef A_H_#define A_H_A类{上市:虚拟 ~A();虚空 doWork();};#万一

文件 Child.h

#ifndef CHILD_H_#define CHILD_H_#include "A.h"类孩子:公共 A {私人的:整数 x,y;上市:孩子();~孩子();void doWork();};#万一

和 Child.cpp

#include "Child.h"孩子::孩子(){x = 5;}子::~子(){...}void Child::doWork(){...};

编译器说 A 有一个对 vtable 的未定义引用.我尝试了很多不同的东西，但都没有奏效.

我的目标是让 A 类成为一个接口，并将实现代码与标头分开.

解决方案

为什么会报错&如何解决?

您需要提供定义 用于 A 类 中的所有虚函数.只允许纯虚函数没有定义.

即:在 class A 两个方法:

虚拟~A();虚空 doWork();

应该被定义(应该有一个主体)

例如:

A.cpp

void A::doWork(){}A::~A(){}

<小时>

警告:
如果您希望 class A 充当接口(又名 抽象类(在 C++ 中)，那么您应该使该方法纯虚拟.

virtual void doWork() = 0;

<小时>

好书:

虚拟表"是什么意思外部未解决?
在构建 C++ 时，链接器说我的构造函数、析构函数或虚拟表未定义.>

File A.h

#ifndef A_H_
#define A_H_

class A {
public:
    virtual ~A();
    virtual void doWork();
};

#endif

File Child.h

#ifndef CHILD_H_
#define CHILD_H_

#include "A.h"

class Child: public A {
private:
    int x,y;
public:
    Child();
    ~Child();
    void doWork();
};
#endif

And Child.cpp

#include "Child.h"

Child::Child(){
    x = 5;
}

Child::~Child(){...}

void Child::doWork(){...};

The compiler says that there is a undefined reference to vtable for A. I have tried lots of different things and yet none have worked.

My objective is for class A to be an Interface, and to seperate implementation code from headers.

解决方案

Why the error & how to resolve it?

You need to provide definitions for all virtual functions in class A. Only pure virtual functions are allowed to have no definitions.

i.e: In class A both the methods:

virtual ~A();
virtual void doWork();

should be defined(should have a body)

e.g.:

A.cpp

void A::doWork()
{
}
A::~A()
{
}

---

Caveat:
If you want your class A to act as an interface(a.k.a Abstract class in C++) then you should make the method pure virtual.

virtual void doWork() = 0;

---

Good Read:

What does it mean that the "virtual table" is an unresolved external?
When building C++, the linker says my constructors, destructors or virtual tables are undefined.