G ++编译器错误。什么地球上的任何这是什么意思，如何解决它？ [英] G++ compiler error. What on Earth does any of this mean and how do I fix it?

问题描述

这是错误：

 在/usr/lib/gcc/x86_64-redhat-linux/4.4包含的文件中。 6 /../../../../ include / c ++ / 4.4.6 / ios：39，
来自/usr/lib/gcc/x86_64-redhat-linux/4.4.6/ .. /../../../include/c++/4.4.6/ostream:40，
来自/usr/lib/gcc/x86_64-redhat-linux/4.4.6/../../ ../../include/c++/4.4.6/iostream:40，
从date.h：15，
从date.cpp：13：
 / usr / lib / gcc /x86_64-redhat-linux/4.4.6/../../../../include/c++/4.4.6/bits/ios_base.h：在复制构造函数'std :: basic_ios< char，std： ：char_traits< char> > :: basic_ios（const std :: basic_ios< char，std :: char_traits< char>>&）'：
 /usr/lib/gcc/x86_64-redhat-linux/4.4.6/。 ./../../../include/c++/4.4.6/bits/ios_base.h:790：error：'std :: ios_base :: ios_base（const std :: ios_base&）'is private 
 /usr/lib/gcc/x86_64-redhat-linux/4.4.6/../../../../include/c++/4.4.6/iosfwd:47：错误：在此上下文中
 /usr/lib/gcc/x86_64-redhat-linux/4.4.6/../../../../include/c++/4.4.6/iosfwd：在复制构造函数'std :: basic_ostream< char，std :: char_traits< char> > :: basic_ostream（const std :: basic_ostream< char，std :: char_traits< char>&）;：
 /usr/lib/gcc/x86_64-redhat-linux/4.4.6/。 ./../../../include/c++/4.4.6/iosfwd:56：note：合成方法'std :: basic_ios< char，std :: char_traits< char> > :: basic_ios（const std :: basic_ios< char，std :: char_traits< char>>&）'
 date.cpp：function'std :: ostream operator< std :: ostream&; Date&）'：
 date.cpp：389：note：合成方法'std :: basic_ostream< char，std :: char_traits< char> > :: basic_ostream（const std :: basic_ostream< char，std :: char_traits< char>>&）第一项必须在这里
 make：*** [date.o] Error 1 
  

我认为这可能与我的头文件和源文件是如何编译在一起的，所以这里是代码：

标题：

  #ifndef DATE_H 
 #define DATE_H 
 
 #include< iostream> 
 using namespace std; 
 
 //基本但冗长的类代码
 
 #endif 
  

$ b b

资料来源：

  // #include< iostream& //试过编译有没有这个，但没有改变
 #include< cassert> 
 #include< cstdlib> 
 #includedate.h//这是（date.cpp：13）
 //我试过使用命名空间std，只是看看会发生什么，但没有改变
  
 
 
 
 最后，这里是编译器引用的函数（date.cpp：389）： 
  ostream运算子<<（ostream& out，const Date& date）
 {
 // day 
 out<<日期。 
 switch（date.day）
 {
 case 1：
 case 21：
 case 31：
 out< st; 
 break; 
 case 2：
 case 22：
 out< nd 
 break; 
 case 3：
 case 23：
 out< rd; 
 break; 
默认值：
 out<< th; 
 break; 
} 
 
 // month 
 const char MONTHS [12] [10] = 
 {January，February，March，April ，May，June，
July，August，September，October 
 
 out<< of<< MONTHS [date.month  -  1]<< ，; 
 
 // year 
 out<< date.year; 
 
 return out; 
} 
  
我已经在最后一个小时Google了，但我找不到任何解决我的问题。非常感谢您的帮助！
解决方案
问题是你不能返回一个简单的 code>。您必须返回对您作为参数接收的引用的引用（注意& ）。
  ostream& operator<<<（ostream& out，const Date& date）
  
它不能通过在 return out上复制 out 创建一个新的 ostream  code>。
 
Here's the error:
In file included from /usr/lib/gcc/x86_64-redhat-linux/4.4.6/../../../../include/c++/4.4.6/ios:39,
                 from /usr/lib/gcc/x86_64-redhat-linux/4.4.6/../../../../include/c++/4.4.6/ostream:40,
                 from /usr/lib/gcc/x86_64-redhat-linux/4.4.6/../../../../include/c++/4.4.6/iostream:40,
                 from date.h:15,
                 from date.cpp:13:
/usr/lib/gcc/x86_64-redhat-linux/4.4.6/../../../../include/c++/4.4.6/bits/ios_base.h: In copy constructor ‘std::basic_ios<char, std::char_traits<char> >::basic_ios(const std::basic_ios<char, std::char_traits<char> >&)’:
/usr/lib/gcc/x86_64-redhat-linux/4.4.6/../../../../include/c++/4.4.6/bits/ios_base.h:790: error: ‘std::ios_base::ios_base(const std::ios_base&)’ is private
/usr/lib/gcc/x86_64-redhat-linux/4.4.6/../../../../include/c++/4.4.6/iosfwd:47: error: within this context
/usr/lib/gcc/x86_64-redhat-linux/4.4.6/../../../../include/c++/4.4.6/iosfwd: In copy constructor ‘std::basic_ostream<char, std::char_traits<char> >::basic_ostream(const std::basic_ostream<char, std::char_traits<char> >&)’:
/usr/lib/gcc/x86_64-redhat-linux/4.4.6/../../../../include/c++/4.4.6/iosfwd:56: note: synthesized method ‘std::basic_ios<char, std::char_traits<char> >::basic_ios(const std::basic_ios<char, std::char_traits<char> >&)’ first required here 
date.cpp: In function ‘std::ostream operator<<(std::ostream&, Date&)’:
date.cpp:389: note: synthesized method ‘std::basic_ostream<char, std::char_traits<char> >::basic_ostream(const std::basic_ostream<char, std::char_traits<char> >&)’ first required here 
make: *** [date.o] Error 1
I think it may have to do with how my header and source file are compiling together so here is the code for that:

Header:
#ifndef DATE_H
#define DATE_H

#include <iostream>
using namespace std;

// basic but lengthy class code

#endif
Source:
// #include <iostream>  // tried compiling with and without this, but no change
#include <cassert>
#include <cstdlib>
#include "date.h"       // this is (date.cpp:13)
// I have tried using namespace std, just to see what would happen but nothing changed
Finally, here is the function that the compiler is referring to (date.cpp:389):
ostream operator <<(ostream &out, const Date &date)
{
    // day                                                                      
    out << date.day;
    switch (date.day)
    {
        case 1:
        case 21:
        case 31:
            out << "st";
            break;
        case 2:
        case 22:
            out << "nd";
            break;
        case 3:
        case 23:
            out << "rd";
            break;
        default:
            out << "th";
            break;
    }

    // month                                                                    
    const char MONTHS[12][10] =
    { "January", "February", "March",     "April",   "May",      "June",
        "July",    "August",   "September", "October", "November", "December"};

    out << " of " << MONTHS[date.month - 1] << ", ";

    // year                                                                     
    out << date.year;

    return out;
}
I am completely baffled here.  I have Googled around for  the last hour but I can't find anything that solves my problem.  Thanks for any help in advance!
解决方案
The problem is that you cannot return a plain ostream. You have to return a reference to the one you received as argument (note the &).
ostream & operator <<(ostream &out, const Date &date)
The compiler complains that it cannot create a new ostream object by copying out on the line return out;.

                        
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