c++ 打印一个包含结构的向量答案

【问题标题】：c++ print a vector that contain a structc++ 打印一个包含结构的向量
【发布时间】：2014-01-05 03:33:09
【问题描述】：

我是 C++ 初学者。谁能告诉我如何打印名为 mailVector 的矢量。这是我的代码：

#include<iostream>
#include<vector>
using namespace std;
struct eMailMsg {
string to; // i.e. "professor@stanford.edu"
string from; // i.e. "student@stanford.edu"
string message; // body of message
string subject; // i.e. "CS106 Rocks!"
int date; // date email was sent
int time; // time email was sent
};

int main(){
    vector <eMailMsg> mailVector;
    eMailMsg professor={"professor@stanford.edu","student@stanford.edu","body of message","CS106 Rocks",4,16};
    mailVector.push_back(professor);
    for( std::vector<eMailMsg>::const_iterator i = mailVector.begin(); i != mailVector.end(); ++i)
    std::cout << *i << ' ';

    return 0;
}

相应的错误是错误 1 错误 C2679: binary '

更新版本 1 ：

#include<iostream>
#include<vector>
#include<iterator>
using namespace std;
struct eMailMsg {
string to; // i.e. "professor@stanford.edu"
string from; // i.e. "student@stanford.edu"
string message; // body of message
string subject; // i.e. "CS106 Rocks!"
int date; // date email was sent
int time; // time email was sent
};

ostream& operator<<(ostream& os, const eMailMsg& rightOp)
    {
   os << rightOp.to << " " << rightOp.from << "etc ...";//error C2679: binary '<<' : no operator found which takes a right-hand operand of type 'const std::string' (or there is no acceptable conversion)

   return os;
   // We're writing std::string here and C++ can do that
}

int main(){
    vector <eMailMsg> mailVector;
    eMailMsg professor={"professor@stanford.edu","student@stanford.edu","body of message","CS106 Rocks",4,16};
    mailVector.push_back(professor);

    for( std::vector<eMailMsg>::const_iterator i = mailVector.begin(); i != mailVector.end(); ++i)
std::cout << *i << ' ';

    return 0;
}

【问题讨论】：

你需要为eMailMsg类型重载operator<<。
No match for 'operator<<' in std::cout的可能重复

标签： c++ vector struct

【解决方案1】：

您正在尝试打印eMailMsg 类型。 C++ 不知道怎么做，你需要告诉它。

重载ostream& operator<<(ostream& os, const eMailMsg& rightOp) 以教它。

你可以这样做：

... {
   os << rightOp.to << " " << rightOp.from << "etc ...";
   return os;
   // We're writing std::string here and C++ can do that
}

【讨论】：

@BenSeedangie 你想告诉我们什么？
@BenSeedangie Ehhh，那个重载（运算符
@ScarletAmaranth 是的，你抓住了我。哈哈。但是我怎么能在 main 中使用这个函数呢？只是cout<<mailVector.to<<endl??
在你的情况下只是cout << *t，“eMailMsg，它会神奇地调用它正在寻找但找不到的operator<<(ostream, eMailMsg)原来，现在有了
@ScarletAmaranth 好的。我知道你在说什么。太感谢了。但是，os<<rightOp.to 有一个错误，请查看我更新的代码。错误 C2679: 二进制 '

【解决方案2】：

试试这个：

#include<iostream>
#include<vector>
using namespace std;
struct eMailMsg {
string to; // i.e. "professor@stanford.edu"
string from; // i.e. "student@stanford.edu"
string message; // body of message
string subject; // i.e. "CS106 Rocks!"
int date; // date email was sent
int time; // time email was sent

    operator char const * () const
    {
        return  to.c_str();
    }

};

int main(){
    vector <eMailMsg> mailVector;
    eMailMsg professor={"professor@stanford.edu","student@stanford.edu","body of message","CS106 Rocks",4,16};
    mailVector.push_back(professor);
    for( std::vector<eMailMsg>::const_iterator i = mailVector.begin(); i != mailVector.end(); ++i)
    std::cout << *i << ' ';

    return 0;
}

请注意 cast 运算符方法格式化结构，c++ 编译器决定使用该运算符强制转换为“char *”，因为这是 outstream 的“

【讨论】：