Question

将申明和定义中 ostream&operator<<(ostream&output,const circle&c)const去掉后可成功运行，为何

#include <iostream>
#include <string>
using namespace std;
class point
{

public:

point(float x=0,float y=0);

void set_point(float,float);

friend ostream & operator << (ostream &,const point
&);

protected:

float x;

float y;
};
point::point(float a,float b):x(a),y(b){}
ostream &
operator<<(ostream&output,const point&p)
{

cout<<"point"<<"("<<p.x<<","<<p.y<<")"<<endl;

return output;
}
void point::set_point(float a,float b)
{

x=a;

y=b;
}
class circle:public point
{

public:

circle(float x=0,float y=0,float radius=0);

void set_r(float r);

float arrer();

friend ostream&operator<<(ostream&,const circle&);

float area();

protected:

float radius;
};
circle::circle(float a,float b,float r):

point(a,b),radius(r){}
void circle::set_r(float r)
{

radius=r;
}
ostream&operator<<(ostream&output,const circle&c)//将标记处的两项去掉后可以通过,为什么
{

cout<<"circle"<<"("<<c.x<<","<<c.y<<")
r="<<c.radius<<" area="<<c.area()<<endl;

return output;
}
float circle::area()
{

return 3.14159*radius*radius;
}
int main()
{

point a(2,2);

cout<<a;

circle b(2,3,2);

cout<<b;

circle&c=b;cout<<c;
}

Answer 1

ostream&operator<<(ostream&output,const circle&c)
参数c是个const引用，也就是个常对象。
但是你在函数体中调用了area()函数，常对象只能调用常成员函数，而area并不是常成员函数，所以会编译出错。
解决办法是把area函数定义为常成员函数，需要修改两个地方：
声明处：float area()const;
定义处：
float circle::area()const
{

return 3.14159*radius*radius;
}