因此,我有一个问题是我最近尝试重载<<运算符时遇到的错误。
我有一个名为“structPixels.h”的文件,在其中我定义了一个结构,如下所示:
#pragma once
#include <iostream>
namespace Eng
{
/**
* A structure to represent pixels
*/
typedef struct Pixel{
unsigned int x; ///The x-coordinate
unsigned int y; ///The y-coordinate
bool operator ==(const Pixel& rhs) const
{
return (this->x == rhs.x && this->y == rhs.y);
};
bool operator !=(const Pixel& rhs) const
{
return (this->x != rhs.x || this->y != rhs.y);
};
bool operator <(const Pixel& rhs) const
{
return std::tie(this->x, this->y) < std::tie(rhs.x, rhs.y);
}
friend std::ostream& operator << (std::ostream& out, const Pixel& p);
} Pixel;
}//End of namespace Eng
namespace std {
//Removing the inline does not fix the error. Rather, it fixed another error
//which I had with duplicate symbols
inline ostream& operator<<(ostream &os, const Eng::Pixel &p)
{
os << "{"
<< p.x
<< ","
<< p.y
<< "}";
return os;
}
}//End of namespace std
但是,当我创建它并按如下所示调用cout时:
#include "structPixels.h"
Pixel test = {3, 4};
std::cout << "Test: " << test << std::endl;
我明白了:
error: use of overloaded operator '<<' is ambiguous (with operand types 'basic_ostream<char, std::__1::char_traits<char> >' and 'Eng::Pixel')
std::cout << "Test: " << test << std::endl;
~~~~~~~~~~~~~~~~~~~~~ ^ ~~~~
编辑:
从下面的帮助中,我们移动了运算符,使代码如下所示:
namespace Eng
{
/**
* A structure to represent pixels
*/
typedef struct Pixel{
unsigned int x; ///The x-coordinate
unsigned int y; ///The y-coordinate
bool operator ==(const Pixel& rhs) const
{
return (this->x == rhs.x && this->y == rhs.y);
};
bool operator !=(const Pixel& rhs) const
{
return (this->x != rhs.x || this->y != rhs.y);
};
bool operator <(const Pixel& rhs) const
{
return std::tie(this->x, this->y) < std::tie(rhs.x, rhs.y);
}
} Pixel;
inline std::ostream& operator<<(std::ostream& os, const Pixel& rhs)
{
os << "{" << rhs.x << "," << rhs.y << "}";
return os;
}
}//End of namespace Eng
这解决了错误:)!
最佳答案
首先,您不能将函数重载添加到std
中。这样做是未定义的行为。你想做的就是放
inline std::ostream& operator<<(std::ostream &os, const Eng::Pixel &p)
{
os << "{"
<< p.x
<< ","
<< p.y
<< "}";
return os;
}
在
Eng
命名空间中。这样做allows the code to compile just fine。在 namespace 中使用运算符还可以使ADL工作,因此即使用户的代码中没有using namespace Eng;
也会找到运算符。您所做的工作变得模棱两可的原因是,编译器发现了两个不同的函数。它发现
friend std::ostream& operator << (std::ostream& out, const Pixel& p);
通过ADL,它将找到您在
std
中定义的函数。由于这些函数在不同的命名空间中,因此它们被视为重载,并且由于两者相等,因此会出现歧义错误。
关于c++ - 重载<<运算符导致其模棱两可,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/58963662/