不允许使用抽象类类型“Rectangle”的对象

时间:2013-04-18 01:43:14

标签: c++ abstract-class virtual-functions

//QuizShape.h
#ifndef QUIZSHAPE_H
#define QUIZHAPE_H
#include <iostream>
#include <iomanip>
#include <string>

using namespace std;

class QuizShape
{
protected:
    //outer and inner symbols, and label
    char border, inner;
    string quizLabel;
public:
    //base class constructor with defaults
    QuizShape(char out = '*', char in = '+', string name = "3x3 Square")
    {
        border = out;
        inner = in;
        quizLabel = name;
        cout << "base class constructor, values set" << endl << endl;
    };

    //getters
    char getBorder() const
    { return border; }
    char getInner() const
    { return inner; }
    string getQuizLabel() const
    { return quizLabel; }

    //virtual functions to be defined later
    virtual void draw( ) = 0;
    virtual int getArea( ) = 0;
    virtual int getPerimeter( ) = 0;

};


class Rectangle : public QuizShape
{
protected:
    //height and with of a rectangle to be drawn
    int height, width;
public:
    //derived class constructor
    Rectangle(char out, char in, string name,
                int h = 3, int w = 3):QuizShape(out, in, name)
    {
        height = h;
        width = w;
        cout << "derived class constructor, values set" << endl << endl;
    }

    //getters
    int getHeight() const
    { return height; }
    int getWidth() const
    { return width; }

    //*********************************************
    virtual void draw(const Rectangle &rect1)
    {
        cout << "draw func" << endl;
        cout << rect1.height << endl;
        cout << rect1.getWidth() << endl;
        cout << rect1.getQuizLabel() << endl;
    }

    virtual int getArea(const Rectangle &rect2)
    {
        cout << "area func" << endl;
        cout << rect2.getInner() << endl;
        cout << rect2.getBorder() << endl;
    }

    virtual int getPerimeter(const Rectangle &rect3)
    {
        cout << "perim func" << endl;
        cout << rect3.height << endl;
        cout << rect3.getWidth() << endl;
        cout << rect3.getQuizLabel() << endl;   
    }
    //************************************************
};



#endif

到目前为止,这些是类类型。

//QuizShape.cpp
#include "QuizShape.h"

这目前只能桥接文件。

//pass7.cpp
#include "QuizShape.cpp"

int main()
{
    Rectangle r1('+', '-', "lol", 4, 5);

    cout << r1.getHeight() << endl;
    cout << r1.getWidth() << endl;
    cout << r1.getInner() << endl;
    cout << r1.getBorder() << endl;
    cout << r1.getQuizLabel() << endl;

    system("pause");
    return 0;
}

代码将无法编译,因为Rectangle应该是一个抽象类,当将鼠标悬停在r1中的main声明时,我收到错误

  

“抽象类类型的对象”不允许使用“矩形”。

我已经在本网站和其他网站上检查了其他答案,但没有找到解决问题的方法。

注意:我理解虚函数的语句以= 0结尾;使班级成为一个抽象的班级。 QuizShape应该是抽象的。我已经在Rectangle中定义了虚函数,但它仍然是一个抽象类。

如何修改虚函数Rectangle类以使Rectangle不再是抽象的?

2 个答案:

答案 0 :(得分:2)

抽象类QuizShape中的方法是:

virtual void draw( ) = 0;
virtual int getArea( ) = 0;
virtual int getPerimeter( ) = 0;

但在Rectangle中,他们将const Rectangle &rect1作为参数,因此您可以隐藏方法而不会覆盖抽象方法。您需要在Rectangle中使用与抽象基类中的方法具有相同签名的方法。

答案 1 :(得分:0)

重写的方法必须具有完全相同的签名,在派生类中为它们赋予参数。