在C++中调用成员的重载<<运算符

-2

以下引发编译时错误。我不知道如何调用一个对象重载运算符< <另一个对象内< <运算符？如果有人能指出我做错了什么，那会非常有帮助。感谢您的时间！在C++中调用成员的重载<<运算符

card.cpp：

#include "card.h" 
#include <iostream> 

ostream& operator<<(ostream &out, Card &cCard) 
{ 
    out << cCard.getValueString() 
     << " OF " 
     << cCard.getSuitString() 
     << "S\n"; 
    return out; 
} 

string Card::getValueString() 
{ 
    Value value = getValue(); 
    switch(value) 
    { 
     case TWO:   return "TWO"; 
     case THREE:   return "THREE"; 
     case FOUR:   return "FOUR"; 
     case FIVE:   return "FIVE"; 
     case SIX:   return "SIX"; 
     case SEVEN:   return "SEVEN"; 
     case EIGHT:   return "EIGHT"; 
     case NINE:   return "NINE"; 
     case TEN:   return "TEN"; 
     case JACK:   return "JACK"; 
     case QUEEN:   return "QUEEN"; 
     case KING:   return "KING"; 
     case ACE:   return "ACE"; 
     case NONE_VALUE: return "NONE_VALUE"; 
     default:   return "UNKNOWN VALUE"; 
    }; 
} 

string Card::getSuitString() 
{ 
    Suit suit = getSuit(); 
    switch(suit) 
    { 
     case SPADE:   return "SPADE"; 
     case HEART:   return "HEART"; 
     case CLUB:   return "CLUB"; 
     case DIAMOND:  return "DIAMOND"; 
     case NONE_SUIT:  return "NONE_SUIT"; 
     default:   return "UNKNOWN SUIT"; 
    }; 
}

card.h

#pragma once 
#include <iostream> 
using namespace std; 

enum Suit { 
       SPADE, 
       HEART, 
       CLUB, 
       DIAMOND, 
       NONE_SUIT 
      }; 

enum Value { 
       TWO=2, 
       THREE, 
       FOUR, 
       FIVE, 
       SIX, 
       SEVEN, 
       EIGHT, 
       NINE, 
       TEN, 
       JACK, 
       QUEEN, 
       KING, 
       ACE, 
       NONE_VALUE 
      }; 

    class Card 
    { 

     public: 
      Card() : m_suit(NONE_SUIT), m_value(NONE_VALUE) { }; 
      Card(Suit suit, Value value) : m_suit(suit), m_value(value) { }; 
      Value getValue() { return m_value; }; 
      Suit getSuit() { return m_suit; }; 
      void setValue(Value value) { m_value = value; }; 
      void setSuit(Suit suit) { m_suit = suit; }; 
      string getValueString(); 
      string getSuitString(); 
      friend ostream& operator<<(ostream &out, Card &cCard); 
     private: 
      Suit m_suit; 
      Value m_value;  
    };

hand.cpp：

#include "hand.h" 
#include "card.h" 

ostream& operator<<(ostream &out, Card &cCard); 

Hand::Hand() 
{ 

} 

Card Hand::getCard(int cardIndex) 
{ 
    if(cardIndex == 0 || cardIndex == 1) 
    { 
     return cards[cardIndex]; 
    } 
    else 
    { 
     cout << "Card index out of bounds.\n"; 
    } 

    return Card(NONE_SUIT, NONE_VALUE); 
} 

void Hand::setCard(int cardIndex, Suit suit, Value value) 
{ 
    if(cardIndex == 0 || cardIndex == 1) 
    { 
     cards[cardIndex].setValue(value); 
     cards[cardIndex].setSuit(suit); 
    } 
    else 
    { 
     cout << "Card index out of bounds.\n"; 
    } 
}; 

ostream& operator<<(ostream &out, Hand &cHand) 
{ 
    out << cHand.getCard(0) 
     << "\n" 
     << cHand.getCard(1); 
    return out; 
}

hand.h：

#pragma once 
#include <iostream> 
#include "card.h" 
using namespace std; 

class Hand 
{ 
    public: 
     Hand(); 
     void setCard(int cardIndex, Suit suit, Value value); 
     Card getCard(int cardIndex); 
     friend ostream& operator<<(ostream &out, Hand &cHand); 
    private: 
     Card cards[2];  
};

该错误是以下内容：

hand.cpp：在函数 '的std :: ostream的&操作者< <（标准:: ostream的&，手&）'： hand.cpp：40：错误：不对应的'（运营商< <'in'out < <（（Hand *）cHand） - > Hand :: getCard（0）' /usr/include/c++/4.2.1/ostream:112：note：candidates are：std：：basic_ostream < _CharT，_Traits> & std :: basic_ostream < _CharT，_Traits> ::运营商< <（std :: basic_ostream < _CharT，_Traits>（*）（std :: basic_ostream < _CharT，_Traits> &））[with _CharT = char，_Traits = std :: char_traits] /usr/include/c++/4.2.1/ostream:121：note：std：：basic_ostream < _CharT，_Traits> &的std :: basic_ostream < _CharT，_Traits> ::运算< <（标准:: basic_ios < _CharT，_Traits> &（*）（标准:: basic_ios < _CharT，_Traits> &））[与_CharT = char，_Traits = std :: char_traits] /usr/include/c++/4.2.1/ostream:131：note：std :: basic_ostream < _CharT，_Traits> & std :: basic_ostream < _CharT，_Traits>：：运营商< <（std :: ios_base &（*）（std :: ios_base &））[with _CharT = char，_Traits = std :: char_traits] /usr/include/c++/4.2.1/ostream:169：note：std :: basic_ostream < _CharT，_Traits> &的std :: basic_ostream < _CharT，_Traits> ::运算< <（长整型）[与_CharT =炭，_Traits =标准:: char_traits] /usr/include/c++/4.2。 1/ostream：173：note：std :: basic_ostream < _CharT，_Traits> & std :: basic_ostream < _CharT，_Traits> :: operator < <（long unsigned int）[with _CharT = char，_Traits = std :: char_traits] /usr/include/c++/4.2.1/ostream:177：note：std :: basic_ostream < _CharT，_Traits> & std :: basic_ostream < _CharT，_Traits> ::运算符< <（布尔型）[with _CharT = char，_Traits = std :: char_traits] /usr/include/c++/4.2.1/bits/ ostream的。TCC：92：注：标准:: basic_ostream < _CharT，_Traits> &的std :: basic_ostream < _CharT，_Traits> ::运算< <（短整型）[与_CharT =炭，_Traits =标准:: char_traits] /USR /include/c++/4.2.1/ostream:184：note：std :: basic_ostream < _CharT，_Traits> & std :: basic_ostream < _CharT，_Traits> :: operator < <（short unsigned int）[with _CharT = char， _Traits = std :: char_traits] /usr/include/c++/4.2.1/bits/ostream.tcc:106：note：std :: basic_ostream < _CharT，_Traits> & std :: basic_ostream < _CharT，_Traits> ::运算符< <（int）[with _CharT = char，_T raits =标准:: char_traits] /usr/include/c++/4.2.1/ostream:195：注：标准:: basic_ostream < _CharT，_Traits> &的std :: basic_ostream < _CharT，_Traits> ::运算< <（ unsigned int）[with _CharT = char，_Traits = std :: char_traits] /usr/include/c++/4.2.1/ostream:204：note：std :: basic_ostream < _CharT，_Traits> & std :: basic_ostream < _CharT ，_Traits> :: operator < <（long long int）[with _CharT = char，_Traits = std :: char_traits] /usr/include/c++/4.2.1/ostream:208：note：std :: basic_ostream < _CharT ，_Traits> & std :: basic_ostream < _CharT，_Traits> :: operator < <（long long unsigned int）[with _CharT = char，_Traits = std :: char_traits] /usr/include/c++/4.2.1/ostream:213：note：std :: basic_ostream < _CharT，_Traits> & std :: basic_ostream < _CharT，_Traits> ::运算符< <（double）[with _CharT = char，_Traits = std :: char_traits] /usr/include/c++/4.2.1/ostream:217：note：std :: basic_ostream < _CharT，_Traits> &的std :: basic_ostream < _CharT，_Traits> ::运算< <（浮点）[与_CharT =炭，_Traits =标准:: char_traits] /usr/include/c++/4.2.1/ostream ：225：note：std :: basic_ostream < _CharT，_Traits> & std :: basic_ostream < _CharT，_Traits> ::运算符< <（long double）[with _CharT = char，_Traits = std :: char_traits] /usr/include/c++/4.2.1/ostream:229：note：的std :: basic_ostream < _CharT，_Traits> &的std :: basic_ostream < _CharT，_Traits> ::运算< <（常量无效*）[与_CharT =炭，_Traits =标准:: char_traits] 的/ usr /包括/ C++/4.2.1/bits/ostream.tcc：120：note：std :: basic_ostream < _CharT，_Traits> & std :: basic_ostream < _CharT，_Traits> :: operator < <（std :: basic_streambuf < _CharT，_Traits> *） [with _CharT = char，_Traits = std :: char_traits] hand.cpp：4：注意：标准:: ostream的&操作者< <（标准:: ostream的&，卡&） hand.cpp：38：注：标准:: ostream的&操作者< <（标准:: ostream的& ，手&）

关于我缺少什么的任何想法？

来源

2013-06-27 user2525441

从我可以看到的那个看起来确定 - 所以要么是问题出在你所忽略的“...”部分，要么你没有很好地解释出了什么问题（换句话说，其他的是比“不编译”错误 - 例如你没有输出，因为你在'cout'行没有刷新 - 添加一个'endl'可以解决这个问题。 –

我猜想的最可能的事情是在类之后缺少'std ::'s，缺少标题或者没有分号：http://coliru.stacked-crooked.com/view?id=ccead1b951d7ebef1db70f3d9bc387f5-3b440a87a52fe2ae7c853c82f4c5144f – chris

Don'在这里看到很多东西。你在类定义的末尾忘了';'？ 'info'不是'A'的成员？你真的使用'...'？ – aschepler

您的operator <<函数应该采用const引用的右侧参数。它告诉你它不能将由getCard返回的临时变量绑定到操作符的非const引用参数。

令人难以置信的长错误信息是因为在任何时候，范围内的<<都有如此令人难以置信的重载，而且编译器将其全部列为“我试过这个，但它不起作用”。

来源

2013-06-27 16:03:20

@ user2525441您能否接受他的回答（点击勾号）？ – 0x499602D2

在C++中调用成员的重载<<运算符

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