我有一个家庭类。我使用 house house;
初始化一个新的 house 元素,然后将数据传递到其中,然后计算它:
cout << house;
Couting house 在 Visual Studio 中工作得很好,但由于某种原因,当我尝试使用 g++ 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;
尽管我非常清楚地在我的头文件之一中包含此内容:
friend std::ostream& operator<< (std::ostream& out, house);
您能提供的任何反馈都将不胜感激,因为我认为 g++ 没有理由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(class):
#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
最佳答案
问题是您正在声明您的 operator<<
对于 house
在错误的类(class)中:
class queue
{
friend std::ostream& operator<< (std::ostream& out, const house&);
};
当你在类中声明友元运算符时X
,只有当我们查找 X
时,该运算符的查找才会成功。 。通过这个声明,我们只会找到 operator<<(std::ostream&, const house&)
当我们查找queue
时- 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。
您需要将声明移至正确的类:
class house {
friend std::ostream& operator<< (std::ostream& out, const house&);
};
关于c++ - Visual Studio 编译并检测操作符重载,而 g++ 则不然,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/33375770/