有点疑问~~~

[zhangzhongke007]

#include<iostream>
using namespace std;

class point
{
public:
	point()
	{
		x=y=0;
		cout<<"Default constructor called!"<<endl;
	}
	point(int a,int b)
	{
		x=a;
		y=b;
		cout<<"constructor called!"<<endl;
	}
	~point()
	{
		cout<<"Deconstructor called!"<<endl;
	}
	int get_x()
	{
		return x;
	}
	int get_y()
	{
		return y;
	}
	void move(int m,int n)
	{
		x=m;
		y=n;
	}
private:
	int x;
	int y;
};

class arrayofpoints
{
public:
	arrayofpoints(int n)
	{
		numofpoints=n;
		points=new point[n];
	}

	arrayofpoints(arrayofpoints&p)
	{
		numofpoints=p.numofpoints;
		points=new point[numofpoints];
		for(int i=0;i<numofpoints;i++)
		{
			points[i].move(p.element(i).get_x(),p.element(i).get_y());
		}
	}

	~arrayofpoints()
	{
		cout<<"Deleting..."<<endl;
		numofpoints=0;
		delete []points;
	}

	point& element(int a)//这个形式……
	{
		return *(points+a);
	}
private:
	int numofpoints;
	point *points;
};

void main()
{
	arrayofpoints a1(2);
	a1.element(0).move(1,5);
	a1.element(1).move(6,7);

	arrayofpoints a2(a1);
	cout<<"copy of a1:"<<endl;
	cout<<"point_0 of a2:"<<a2.element(0).get_x()<<' '<<a2.element(0).get_y()<<endl;
	cout<<"point_1 of a2:"<<a2.element(1).get_x()<<' '<<a2.element(1).get_y()<<endl;

    cout<<"After the moving of a1:"<<endl;
	a1.element(0).move(7,8);
	a1.element(1).move(35,78);
	cout<<"point_0 of a2:"<<a2.element(0).get_x()<<' '<<a2.element(0).get_y()<<endl;
	cout<<"point_1 of a2:"<<a2.element(1).get_x()<<' '<<a2.element(1).get_y()<<endl;
}

问题：其中

	point& element(int a)
	{
		return *(points+a);
	}

该这个形式如何解释啊？加&于不加&有何区别？

[aaaa3105563]

帮顶·

[surlover]

&返回的是一个引用

[lpms26086]

楼上正解

[liao05050075]

不加&将返回一个对象的拷贝，这样会有拷贝的开销
返回一个引用则没有

[zgjxwl]

不加&会调用拷贝构造函数，会生成一个临时对象。。加&就返回它本身了。不生成临时对象。

[woods2001]

返回引用，开销小，不容易出现零食变量析构等问题