c++ - Visual Studio 编译和检测运算符重载而 g++ 没有

Question

我有一个家庭类。我用 house house; 初始化一个新的房子元素，然后我将数据传递给它，然后我计算它:

cout << house;

Couting house 在 Visual Studio 中工作得很好，但出于某种原因，当我尝试使用 g++ 进行编译时收到此错误:

main.cpp:19:57: error: no match for ‘operator<<’ (operand types are ‘std::basic_ostream<char>::__ostream_type {aka std::basic_ostream<char>}’ and ‘house’)
cout << "\nnext house to be visited:" << endl << endl << house << endl;

即使我的一个头文件中非常清楚地包含了这个:

friend std::ostream& operator<< (std::ostream& out, house);

如果您能提供任何反馈，我将不胜感激，因为我看不出有什么理由让 g++ 无法看到我的运算符重载函数。

编辑:这是我的运算符重载函数:

std::ostream& operator<< (std::ostream& out, const house& house)
{
    out << "Address: " << house.getAddress() << std::endl
        << "Square Feet: " << house.getSqrFt() << std::endl
        << "Bedrooms: " << +house.getBedrooms() << std::endl
        << "Bathrooms: " << house.getBathrooms() << std::endl
        << "Description: " << house.getDescription() << std::endl;
    return out;
}

这是我的家庭类:

#ifndef HOUSE
#define HOUSE
class house
{
public:
    house();
    house(const char[], const unsigned short& sqrFt, const unsigned char& bedrooms, const float& bathrooms, const char[]);
    house(house & obj);
    house(house *& obj);
    ~house();
    char * getAddress() const;
    unsigned short getSqrFt() const;
    unsigned char getBedrooms() const;
    float getBathrooms() const;
    char * getDescription() const;
    void setAddress(const char address[]);
    void setSqrFt(const unsigned short& sqrFt);
    void setBedrooms(const unsigned char& bedrooms);
    void setBathrooms(const float& bathrooms);
    void setDescription(const char description[]);
    void setEqual(house &, house*);
private:
    char * address;
    unsigned short sqrFt;
    unsigned char bedrooms;
    float bathrooms;
    char * description;
};
#endif

这是我的队列类，其中包含我的运算符重载函数的声明:

#ifndef QUEUE
#define QUEUE
#include <ostream>
#include "house.h"

class queue
{
public:
    queue();
    queue(queue & obj);
    ~queue();
    void enqueue(house *& item);
    bool dequeue(house & item);
    void print() const;
    void readIn(const char []);
private:
    struct node
    {
        node();
        house* item;
        node * next;
    }; 
    node * head;
    node * tail;
    void getLine(std::ifstream&, char key[]);
    friend std::ostream& operator<< (std::ostream& out, const char[]);
    //friend std::ostream& operator<< (std::ostream& out, house *&);
    friend std::ostream& operator<< (std::ostream& out, const house&);
};
#endif

Answer 1

问题是您要声明您的 operator<<对于 house在错误的类中:

class queue
{
    friend std::ostream& operator<< (std::ostream& out, const house&);
};

当您在类中声明友元运算符时 X , 只有当我们寻找 X 时，对该运算符的查找才会成功.有了这个声明，我们只会找到 operator<<(std::ostream&, const house&)当我们查找 queue - 但这是不可能的，因为没有一个参数是 queue所以我们永远不会尝试用一个来查找它。

您需要将您的声明移动到正确的类:

class house {
    friend std::ostream& operator<< (std::ostream& out, const house&);
};