c++ - 请求从 ‘Point*’ 到非标量类型 ‘Point’ 的转换

Question

我看了几个类似的帖子，但我要么不明白他们提供的是什么，要么他们似乎不适用。我是新来的，我会尽力遵守规则。

我们在类(class)的最后 2 周学习 c++，期末学习 40 小时 :)，所以我是初学者。我熟悉 C。

这是作业的相关部分:

Part 1 [20 points]
Reading material:
Copy constructor: https://www.geeksforgeeks.org/copy-constructor-in-cpp/
References (the & symbol): https://www.geeksforgeeks.org/references-in-c/
------------------------------------------------------------------------------------------------------
Write a class Point representing a point in 2d.
The class will have the following public methods:
// constructor - sets the coordinates to be x,y
Point( int x, int y)
// copy constructor - creates a new point with the same
coordinates
Point(Point &copy)
// getters
int getX()
int getY()
// setters
void setX()
void setY()
// prints to screen the coordinates in the format (x,y)
void print()

我的实现:

点.hpp

#include <iostream>

using namespace std;

class Point { 
   private: 
   int x, y; 

public: 
    // Parameterized Constructor 
    Point(int x1, int y1); 

    ~Point();

    Point (const Point &p2);

    int getX();

    int getY(); 


    void setX(int x2);


    void setY(int y2);

    void print();
};

点.cpp

#include "Point.hpp"

Point::Point(int x1, int y1)
    { 
        x = x1; 
        y = y1; 
    } 
Point::Point (const Point &p2)
    {
        x = p2.x;
        y = p2.y;
    }

int Point::getX()
    { 
        return x; 
    } 

int Point::getY()
    { 
        return y; 
    } 

void Point::setX(int x2)
    {
      x = x2;
    }

void Point::setY(int y2)
    {
      y = y2;
    }

void Point::print()
    {
        cout << "(" << x << "," << y << ")";
    }

问题中我卡住的部分

Part 2 [20 points]
Reading material:
Abstract classes and pure virtual methods:
https://www.geeksforgeeks.org/pure-virtual-functions-and-abstract-classes/
---------------------------------------------------------------------------------------------------
Write an abstract class GeometricShape representing an abstract geometric
shape. The class will have the following public methods:
// Constructor: gets a coordinate. The purpose of the
coordinate depends on the specific shape
GeometricShape(Point coord)
// returns the area of the shape
// returns 0 as default. To be implemented in each
concrete shape
virtual float getArea() { return 0 ; }
// returns the perimeter of the shape
// returns 0 as default. To be implemented in each
concrete shape
virtual float getPerimeter() { return 0 ; }
// virtual method. To be implemented in each concrete
method
virtual void print() = 0 ;

我的实现:

几何形状.hpp

#include <iostream>
#include "Point.hpp"

using namespace std;

class GeometricShape { 

private:
    Point point;

public: 
    // Parameterized Constructor 
    GeometricShape(Point coord); 

    virtual float getArea();

    virtual float getPerimeter(); 

    virtual void print() = 0;
};

GeometricShape.cpp(无法编译)

#include <iostream>
#include "GeometricShape.hpp"


GeometricShape::GeometricShape(Point coord)

{

Point p = new Point(coord.getX(),coord.getY());


}

int main()
{
  return 0;
}

在 Linux Ubunto 18.04.3 上编译时的错误消息(大学实验室远程访问，需要在实验室的 linux 上编译作业)

错误信息:

gsg27@csil-cpu2:~/sfuhome/cmpt-125/5$ g++ GeometricShape.cpp
GeometricShape.cpp: In constructor ‘GeometricShape::GeometricShape(Point)’:
GeometricShape.cpp:9:11: error: conversion from ‘Point*’ to non-scalar type ‘Point’ requested
 Point p = new Point(coord.getX(),coord.getY());
           ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Answer 1

你的直接问题是这条线

Point p = new Point(coord.getX(),coord.getY());

没有意义。 new 表达式返回指向动态分配对象的指针。所以，你写的代码应该是

Point *p = new Point(coord.getX(),coord.getY());

除非您不需要指针，也不需要动态分配的对象:您只是想初始化一个数据成员。所以实际上你的构造函数应该是

GeometricShape::GeometricShape(Point coord) : point(coord) {}

因为您已经有了数据成员 GeometricShape::point 并且 Point 类有一个复制构造函数。写的会比较平常

GeometricShape::GeometricShape(Point const &coord) : point(coord) {}

因为您不需要制作原件的两份，但这没什么大不了的。