不能讓超載運算符<<與包含的私有類

Question

有人可以解釋我，如何使私有類變量超載？ 我試圖在超載使用指針，但也許我應該使用的ostream &操作者< <（ostream的&出來，B & b）或ostream的&操作者< <（ostream的&出來，常量乙& B）？

#include <iostream> 
using namespace std; 

class B { 
    const int x = 5; 
public: 
    B(){} 
    ~B(){} 
    int get_x() { 
     return this->x; 
    } 
}; 
ostream& operator<<(ostream& out, B *b) { 
    return out << b->get_x(); 
} 

class A { 
    B b; 
public: 
    A(){} 
    ~A() {} 
    B get_b() { 
     return this->b; 
    } 
}; 
ostream& operator<<(ostream& out, A *a) { 
    return out; 
} 

int main(int argc, const char * argv[]) { 
    A *a = new A(); 
    cout << a << '\n'; 
    return 0; 
} 

Answer 1

你不應該用指針，你需要注意的常量性：

#include <iostream> 
using namespace std; 

class B { 
    const int x = 5; 
public: 
    B(){} 
    ~B(){} 
    int get_x() const { 
     return this->x; 
    } 
}; 
ostream& operator<<(ostream& out, const B & b) { 
    return out << b.get_x(); 
} 

class A { 
    B b; 
public: 
    A(){} 
    ~A() {} 
    B get_b() const { 
     return this->b; 
    } 
}; 
ostream& operator<<(ostream& out, const A & a) { 
    return out << a.get_b(); 
} 

int main() { 
    A a; 
    cout << a << '\n'; 
    return 0; 
}