c++ - 为什么我会收到此错误 : nomatch for 'operator<<' in C++?

Question

请帮助我解决为什么我在提到评论人员时遇到错误。我真的被困在这里了!我还应该提供哪些其他详细信息？

main.cpp

#include <iostream>
#include<person.h>
#include <sstream>

int main()
{
    cout << "Constructor Overloading Demo !!!" << endl;
    person vish;
    cout << vish.toString() << endl; /* I get an error for this statement though I feel this is correct */
    return 0;
}

person.cpp//构造函数定义文件

#include "person.h"
#include <sstream>

person::person()
{
    name = "vish";
    age = 25;
    //ctor
}

void person::toString()
{
    stringstream st;
    st << "Name: ";
    st << name;
    st << "& Age: ";
    st << age;
    prnt = st.str();
    cout << prnt << endl;
    //return 10;
}

person.h//构造函数声明

#ifndef PERSON_H
#define PERSON_H
#include<iostream>
using namespace std;
class person
{
    public:
        person();
       void toString();

        //virtual ~person();
    protected:
    private:
        string name;
        int age;
        string prnt;
};

#endif // PERSON_H

Answer 1

作为方法的名称 person::toString()建议它应该返回 string , 不是 void (这意味着该方法不返回任何内容):

class person { 
   ...
   std::string toString(); 
    //^--------- type should be changed

当然，您需要相应地更改该方法的实现:

std::string person::toString()
{
    stringstream st;
    st << "Name: ";
    st << name;
    st << "& Age: ";
    st << age;
    return st.str();
}

I realize that my function was returning void. but my question is: why does that matter ?

你的输出到cout这是:

cout << vish.toString(); // endl is omitted for clarity

等于以下之一:

cout.operator<<( vish.toString() );
operator<<( cout, vish.toString() ); 

任何可用的。这意味着函数或方法 operator<<需要调用 toString() 的结果它不能接受 void .如果您只想从方法 toString() 打印写:

vish.toString();

那会工作得很好(除了方法名称仍然会令人困惑，但编译器不关心)。