我重载了一个cout和cin opeartor,当我尝试使用它时,它给了我一个这样的错误:
1 IntelliSense: function "std::basic_ostream<_Elem, _Traits>::basic_ostream(const std::basic_ostream<_Elem, _Traits>::_Myt &)
[with _Elem=char, _Traits=std::char_traits<char>]"
(declared at line 84 of "C:\Program Files (x86)\Microsoft Visual Studio 12.0\VC\include\ostream") cannot be referenced -- it is a deleted function
这是我班级的头文件:
#pragma once
#include <iostream>
class Point2D
{
private:
int m_X;
int m_Y;
public:
Point2D(): m_X(0), m_Y(0)
{
}
Point2D(int x, int y): m_X(x), m_Y(y)
{
}
friend std::ostream& operator<< (std::ostream out, const Point2D &point)
{
out << "(" << point.m_X << "," << point.m_Y << ")" << std::endl;
return out;
}
friend std::istream& operator>> (std::istream in, Point2D &point)
{
in >> point.m_X;
in >> point.m_Y;
return in;
}
int getX() const { return m_X; }
int getY() const { return m_Y; }
~Point2D();
};
所以,基本上,它只是一个可以返回并设置X和Y坐标的类。我覆盖了<<
和>>
运算符,使事情变得更容易。
但是当我尝试在主函数中使用它时:
#include "stdafx.h"
#include <iostream>
#include "Point2D.h"
using namespace std;
int main(int argc, char * argv[])
{
Point2D point(7, 7);
cout << point; //there's an error here.
return 0;
}
这一行似乎有错误:
cout << point;
我究竟做错了什么?
答案 0 :(得分:5)
错误消息显示ostream
无法复制。
friend std::ostream& operator<< (std::ostream out, const Point2D &point)
{
...
std::ostream out
是按值传递:当调用operator <<
时,编译器会尝试复制参数。但是,无法复制std::ostream
,因此编译失败。
您应该使用参考。
friend std::ostream& operator<< (std::ostream &out, const Point2D &point)
{
...
friend std::istream& operator>> (std::istream &in, Point2D &point)
{
...
(即使您正在使用某些提供复制的类,您也不应该使用pass-by-val。复制某些对象非常昂贵;)
答案 1 :(得分:2)
也有同样的问题。对于其他人寻找解决方案。错误
function "std::basic_ostream<_Elem, _Traits>::basic_ostream(const std::basic_ostream<_Elem, _Traits> &) [with _Elem=char, _Traits=std::char_traits<char>]" (declared at line 83 of "C:\Program Files (x86)\Microsoft Visual Studio\2017\Community\VC\Tools\MSVC\14.15.26726\include\ostream") cannot be referenced -- it is a deleted function
来自以下行:
std::ostream& operator<< (std::ostream os, const Point2D &point)
对此进行更正,应该没问题:
std::ostream& operator<< (std::ostream &os, const Point2D &point)
答案 2 :(得分:0)
这里的问题在于参数
friend std::ostream& operator<< (std::ostream out, const Point2D &point)
输入参数out
是传递值。当编译器尝试按值复制std::ostream
变量时,它将失败,从而导致错误。如下所示通过引用传递变量out
,它应该清除错误。另外,返回类型是按引用传递的,因此通过执行以下更改,您将返回相同的对象,而不创建新副本。
friend std::ostream& operator<< (std::ostream &out, const Point2D &point)
有用的链接:How to properly overload the << operator for an ostream?