Question

我是一名c ++初学者。有人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C2679：二进制'＆lt;＆lt;' ：找不到哪个运算符采用'const eMailMsg'类型的右手操作数（或者没有可接受的转换）

更新了ver1 ：

#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;
}

Answer 1

您正在尝试打印eMailMsg类型。 C ++不知道如何做到这一点，你需要告诉它。

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

您可以执行以下操作：

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

Answer 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，转换运算符方法格式化结构，并且c ++编译器决定使用该运算符转换为“char *”，因为它是“＆lt;＆lt;”的有效参数类型流出的运营商。

c ++打印包含结构的向量

2 个答案: