编译通不过,那个地方错了呢? const后置为什么？

[harderror]

#include <iostream>
class point
{public:
        point(float x=0,float y=0);
        void setPoint(float,float);
        float getX() const {return x;}
        float getY() const {return y;}
        friend ostream & operator<<(ostream &,const point &);
protected:
          float x,y;
          };
point::point(float a,float b)
          {x=a;y=b;}
point::void setPoint(float a,float b)
          {x=a;y=b;}
point::ostream & operator<<(ostream &output,const point &p)
          {output<<"["<<p.x<<","<<p.y<<"]"<<endl;
          return output;}
          
int main()
{Point p(3.5,6.4);
cout<<"x="<<p.getX()<<",y="<<p.getY()<<endl;
p.setPoint(8.5,6.8);
cout<<"p(new):"<<p<<endl;
} 

[翅膀又硬了]

简单改了一下，只是改成能编译通过运行了，没看逻辑

#include <iostream>
using namespace std;
class point
{public:
point(float x=0,float y=0);
void setPoint(float,float);
float getX() const {return x;}
float getY() const {return y;}
friend ostream & operator<<(ostream &,const point &);
protected:
	float x,y;
};
point::point(float a,float b)
{x=a;y=b;}
void point::setPoint(float a,float b)
{x=a;y=b;}
ostream & operator<<(ostream &output,const point &p)
{output<<"["<<p.x<<","<<p.y<<"]"<<endl;
return output;}

int main()
{point p(3.5,6.4);
cout<<"x="<<p.getX()<<",y="<<p.getY()<<endl;
p.setPoint(8.5,6.8);
cout<<"p(new):"<<p<<endl;
} 

[shen_wei]

point::void setPoint(float a,float b)


这样的写法，真实第一次见！！参看1楼的写法。。

[pathuang68]

point::void setPoint(float a,float b)
改为
void point::setPoint(float a,float b)

[lylm]

float getX() const {return x;}
>>const后置表明该函数不修改类成员

[fzamygsd]

把point::setPoint理解为一个完整的函数名吧

在外边不要随便分开

[fzamygsd]

[Quote=引用 4 楼 的回复:]

float getX() const {return x;}
>>const后置表明该函数不修改类成员
[/Quote]
+ 常对象只能直接调用长成员函数

[木子小新]

LZ是初学者把?