重载操作符<<可以打印出两种不同的功能？

Question

这是我的一小段代码：

Cord::Cord(int x, int y){ 

    x_ = x; 
    y_ = y; 
} 
Cord::Cord(int x, int y, int z){ 
    x_ = x; 
    y_ = y; 
    z_ = z; 
} 
std::ostream& operator <<(std::ostream& out, Cord& x) { 
    out << x.x_ << " " << x.y_ << " " << x.z_; 
    return out; 
} 

是否有可能使运营商在< <函数重载两个我的函数重载。现在，如果我只使用x和y的函数，它也会打印出z。如果im只有x和y或者是不可能的，是否有办法让打印出的两个函数都不打印出z？

Answer 1

您需要z_的默认值。

Cord::Cord(int x, int y){ 
    x_ = x; 
    y_ = y; 
    z_ = 0; // If 0 is a good default value for z_ 
} 

具有缺省值，其中一个解决方案可能是

std::ostream& operator <<(std::ostream& out, Cord& x) { 
    if(z_!= 0) // If 0 is your default value for z 
     out << x.x_ << " " << x.y_ << " " << x.z_; 
    else 
     out << x.x_ << " " << x.y_; 
    return out; 
} 

请注意，你的代码是不是精心设计的。

UPDATE：设计命题

阅读有关Encapsulation 和Polymorphism

部分解决：

Coord2d.cpp

Coord2d::Coord2d(int x, int y){ 
    x_ = x; 
    y_ = y; 
} 
int Coord2d::getX(){ 
    return x_; 
} 
int Coord2d::getY(){ 
    return y_; 
} 

std::ostream& operator <<(std::ostream& out, Coord2d& coords2d) { 
    out << x.x_ << " " << x.y_; 
} 

Coord3d.cpp

Coord3d::Coord3d(int x, int y, int z){ 
    x_ = x; 
    y_ = y; 
    z_ = z; 
} 
int Coord3d::getX(){ 
    return x_; 
} 
int Coord3d::getY(){ 
    return y_; 
} 
int Coord3d::getZ(){ 
    return z_; 
} 
std::ostream& operator <<(std::ostream& out, Coord3d& coords3d) { 
    out << x.x_ << " " << x.y_ << " " << x.z_; 
} 

Answer 2

随着optional，你可以这样做：

class Coord 
{ 
public: 
    Coord(int x, int y) : x(x), y(y) {} 
    Coord(int x, int y, int z) : x(x), y(y), z(z) {} 

    friend std::ostream& operator <<(std::ostream& out, const Coord& c) 
    { 
     out << c.x << " " << c.y; 
     if (c.z) { out << " " << *c.z; } 
     return out; 
    } 

private: 
    int x; 
    int y; 
    boost::optional<int> z; 
}; 

Live Demo