c++ - 如何解决此错误的调用不匹配功能？

Question

#include <iostream>
#include <cmath>
#include <cstdlib>
using namespace std;

class Shape3D
{
public:
    virtual double volumn() ; 
    virtual double surfaceArea() ;
};

class RightRectangularPyramid : public Shape3D
{
private:
    double width;
    double length;
    double height;

public:
    RightRectangularPyramid(double width, double length, double height) : Shape3D(width, length, height) {
        this -> width = 0.0;
        this -> length = 0.0;
        this -> height = 0.0;
    }

    double volumn() override{
        return width * length * height ;
    }

    double surfaceArea() override{
        return (length * width) + (length * sqrt(height*height + (width/2.0)*(width/2.0))) + (width * sqrt(height*height + (length/2.0)*(length/2.0)));
    }
};

class Sphere : public Shape3D
{
private:
    double radius;

public:
    Sphere(double radius){
        this -> radius = radius;
    }   

    double volumn() override{
        return 4.0/3.0 * M_PI * pow(radius,2);
    }

    double surfaceArea() override{
        return 4 * M_PI * pow(radius,2);
    }

};

int main()
{
    int tc;
    cin >> tc;
    if (tc == 1) {
        Shape3D **shapes = new Shape3D*[3];
        shapes[0] = new RightRectangularPyramid(1, 5.2, 7);
        shapes[1] = new Sphere(10.6);
        shapes[2] = new RightRectangularPyramid(3, 3.5, 4.13);
        for (int i = 0; i < 3; i++) {
            cout << "Volumn: " << (*shapes)->volumn() << endl;
            cout << "Surface Area: " << (*shapes)->surfaceArea() << endl;
            cout << "--------------------\n";
            *shapes++;
        }
    } else if (tc == 2) {
        RightRectangularPyramid rectangle;
        Sphere sphere;
        cout << rectangle.volumn() << endl;
        cout << rectangle.surfaceArea() << endl;
        cout << sphere.volumn() << endl;
        cout << sphere.surfaceArea() << endl;
    } else if (tc == 3) {
        RightRectangularPyramid rectangle(171.5, 45.33, 31.02);
        cout << rectangle.volumn() << endl;
        cout << rectangle.surfaceArea() << endl;
    } else if (tc == 4) {
        Sphere sphere(34.25);
        cout << sphere.volumn() << endl;
        cout << sphere.surfaceArea() << endl;
    } else if (tc == 5) {
        RightRectangularPyramid rectangle(171.5, 45.33, 31.02);
        Sphere sphere(34.25);
        Shape3D *shape = &rectangle;
        cout << shape->volumn() << " " << shape->surfaceArea() << endl;
        shape = &sphere;
        cout << shape->volumn() << " " << shape->surfaceArea() << endl;
    } else if(tc == 6) {
    }
}

任何错误

错误:没有匹配的函数调用'Shape3D::Shape3D(double＆，double＆，double＆)'
RightRectangularPyramid(双倍宽度，双倍长度，双倍高度):Shape3D(宽度，长度，高度)

错误:没有匹配的函数来调用“RightRectangularPyramid::RightRectangularPyramid()”
RightRectangularPyramid矩形；

错误:没有匹配的函数可以调用'Sphere::Sphere()'
球体

还有这个更多的错误
https://ibb.co/nrV3H3p

Answer 1

这里

RightRectangularPyramid(double width, double length, double height) : Shape3D(width, length, height) {

由于某种原因，您试图将宽度的长度和高度传递给不存在的Shape3D构造函数。只需删除构造函数调用

RightRectangularPyramid(double width, double length, double height) {

这里

RightRectangularPyramid rectangle;

您试图创建一个没有宽度，长度和高度的金字塔，但是您编写了需要宽度，长度和高度的 RightRectangularPyramid构造函数。您需要添加这些值，例如

RightRectangularPyramid rectangle(1.0, 2.0, 3.0);

在这里，同样的问题

Sphere sphere;

您试图创建没有半径的球体，但是您编写了需要半径的 Sphere构造函数。因此，您需要提供一个，例如

Sphere sphere(99.9);

有时，编译器错误消息是含糊不清的，但这些错误消息似乎很明显，编译器会告诉您确切的错误之处。