Question

我正在为学校制作一个小程序。我想在类'IObject'，'Square'和'RotatedSquare'中添加虚拟析构函数。但是当我这样做时，它会出错。在这种情况下是不可能的，或者在这里怎么做？这是我的代码：

interface iobject：

#ifndef IOBJECT_H_
#define IOBJECT_H_
#include "Point.h"

class IObject {
public:
    virtual const Point& getPosition()const = 0;
    virtual double getSize()const = 0;
    //virtual ~IObject();

};

#endif

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#ifndef SQUARE_H_
#define SQUARE_H_
#include "IObject.h"
#include "Point.h"
#include <iostream>
class Square : public IObject{
private:
    Point position;
    double size;
public:
    Square(const Point& position, const double size):position(position),size(size){};
    virtual const Point& getPosition() const {return this->position;}
    virtual double getSize() const {return this->size;}
    friend std::ostream& operator<<(std::ostream& out, const Square& s);
    //virtual ~Square();

protected:
    virtual void print(std::ostream& out) const;
};

#endif

类RotatedSquare

#ifndef ROTATEDSQUARE_H_
#define ROTATEDSQUARE_H_
#include "Square.h"

class RotatedSquare : public Square{
private:
    double angle;
public:
    RotatedSquare(const Point& position, const double size, const double angle): Square(position,size),angle(angle){};
    double getAngle() const {return this->angle;}
    //virtual ~RotatedSquare();
protected:
    virtual void print(std::ostream& out) const;
};

#endif

错误：

Undefined symbols for architecture x86_64:
  "Square::~Square()", referenced from:
      _main in Practice.o
  "vtable for RotatedSquare", referenced from:
      RotatedSquare::RotatedSquare(Point const&, double, double) in Practice.o
  NOTE: a missing vtable usually means the first non-inline virtual member function has no definition.
  "vtable for Square", referenced from:
      Square::Square(Point const&, double) in Practice.o
  NOTE: a missing vtable usually means the first non-inline virtual member function has no definition.
  "vtable for IObject", referenced from:
      IObject::IObject() in Practice.o
  NOTE: a missing vtable usually means the first non-inline virtual member function has no definition.

主要班级：

#include <iostream>
#include "Point.h"
#include "IObject.h"
#include "Square.h"
#include "RotatedSquare.h"

int main() {

     IObject* s1 = new Square(Point(1,4),2.2);
     std::cout << s1->getPosition() << std::endl;
     std::cout << s1->getSize() << std::endl;
  Square s2 = Square(Point(4,5),5);
  Square* s3 = new RotatedSquare(Point(5,5),8,60);
  std::cout << s2 << std::endl;
  std::cout << *s3;
return 0;
}

Answer 1

您需要定义您声明的每个虚拟析构函数。假设您不希望析构函数执行超过默认行为，您可以使用声明进行定义：而不是：

virtual ~IObject();

使用

virtual ~IObject() {}

如果你想在析构函数中做大量的工作，或者通常更喜欢cpp而不是hpp中的所有定义，那么在cpp文件中定义析构函数（保持hpp中没有定义的声明）

IObject::~IObject()
{
// Whatever needs to be done
}

Answer 2

您需要在代码中定义虚拟析构函数。正如您的错误中明确提到的那样。

"Square::~Square()", referenced from:
      _main in Practice.o
  "vtable for RotatedSquare", referenced from:
      RotatedSquare::RotatedSquare(Point const&, double, double) in Practice.o

注意：缺少的vtable通常表示第一个非内联虚拟成员功能没有定义。

写virtual ~IObject() {}就足够了。

Answer 3

你的析构函数缺少定义。

如果您不想提供自己的：

，可以使用编译器生成的编译器

virtual ~X() = default;

这比提供一个空体更好：

virtual ~X() { }

因为用户提供的¹析构函数阻止该类成为trivially destructible。

¹ { /* anything */ }使用户提供，= default没有。

使用虚拟析构函数会产生错误

3 个答案: