Question

我已在MS Visual Studio Express 2012中编写此代码以查看rtti行为但它没有按预期工作。
我的代码出了什么问题？

Shape.h

class Shape
{
public:
    Shape(){}
    virtual ~Shape(){}
    virtual double area() = 0;
};

class Square : public Shape
{
    int a;
public:
    ~Square(){}
    Square(int );
    virtual double area();
};

class Rectangle : public Shape
{
    int l;
    int b;
public:
    ~Rectangle(){}
    Rectangle(int,int);
    virtual double area();
};

class Circle : public Shape
{
    int r;
public:
    ~Circle(){}
    Circle(int);
    virtual double area();
};

ShapeMain.cpp

int main()
{
    Shape* c = new Circle(4);
    cout<< "Area of circle:" << c->area() << endl;
    cout << typeid(c).name();

    Shape* s = new Square(4);
    cout<< "Area of square:" << s->area() << endl;
    cout << typeid(s).name();

    Shape* r = new Rectangle(4,5);
    cout<< "Area of rectangle:" << r->area() << endl;
    cout << typeid(r).name();

}

输出

Area of circle:50.24
class Shape *           //Expected class Circle*
Area of square:16
class Shape *           //Expected class Square*
Area of rectangle:20
class Shape *           //Expected class Rectangle*

Answer 1

typeid()仅在传递多态类型的左值时才实际执行RTTI查找。 Shape是一种多态类型，但您没有传递Shape左值，而是传递Shape*。因此，当您将c，s和r传递给typeid()时，它会报告这些表达式的静态类型，即Shape*。

要获得运行时查找，您可以取消引用指针：std::cout << typeid(*r).name() << std::endl;

或者您可以直接保留参考文献：

Circle circle{4};
Shape& c = circle;
cout << "Area of circle:" << c.area() << endl;
cout << typeid(c).name() << endl;

RTTI行为未按预期显示

1 个答案: