c++ - 不断收到错误 ' no match for ' operator>>' ；

Question

我已经看到许多其他问题出现相同的错误，但这些问题来自没有重载 >> 运算符的人。我在编写的其他一些程序中遇到了这个问题，因为它们都是练习题，所以彼此非常相似。

我还查看了我的教科书并将我的程序与他们的示例程序进行了比较，但看不出我哪里出错了。

完整的错误是(在 main.cpp 中)

第 17 行:错误:“operator>>”不匹配(操作数类型为“std::ifstream {aka std::basic_ifstream}”和“Suitcase()”)|

非常感谢任何和所有建议。

我的标题是

//suitcase.h
#include <iostream>
#include <string>
#include <fstream>
#include <iomanip>
#ifndef SUITCASE_H
#define SUITCASE_H
using namespace std;

class Suitcase
{
public:
    Suitcase();
    Suitcase(double p,double w,double v,string s,string b);
    ~Suitcase();
    double calc_ratio();
    friend bool operator>(const Suitcase s1,const Suitcase s2);
    friend ostream& operator<<(ostream & out,const Suitcase& s);
    friend istream& operator >>(istream& in,Suitcase& s);
private:
    double price,weight,volume;
    string brand,shop;
};
#endif // SUITCASE_H

我的实现是

//suitcaseimplement.cpp
#include <iostream>
#include <string>
#include <fstream>
#include <iomanip>
#include "suitcase.h"
using namespace std;

Suitcase::Suitcase()
{
    price=0;
    weight=0;
    volume=0;
    shop="";
    brand="";
}
Suitcase::Suitcase(double p,double w,double v,string s,string b)
{
    price=p;
    weight=w;
    volume=v;
    shop=s;
    brand=b;
}
Suitcase::~Suitcase()
{

}
double Suitcase::calc_ratio()
{
    return (volume/weight);
}
bool operator>(Suitcase s1,Suitcase s2)
{
    if(s1.calc_ratio()>s2.calc_ratio())
        return true;
}
ostream& operator<<(ostream & out,const Suitcase& s)
{
    out<<"The suitcase with the highest ratio is the "<<s.brand<<endl;
    out<<"It is available at "<<s.shop<<endl;
    out<<"It weighs "<<s.weight<<"kgs, has a volume of "<<s.volume<<"and costs R"<<s.price<<endl;

    return out;
}
istream& operator >>(istream& in,Suitcase& s)
{
    in>>s.price>>s.weight>>s.volume>>s.brand>>s.shop;
    return in;
}

最后是我的主程序。

//main.cpp
#include <iostream>
#include <string>
#include <cstdlib>
#include <fstream>
#include "suitcase.h"
using namespace std;

int main()
{
    ifstream infile;
    infile.open("Suitcase.txt");
    if(infile.fail())
        exit(1);

Suitcase current_suitcase(), best_suitcase();
infile>>best_suitcase;
while(infile>>current_suitcase)
{
    if(current_suitcase>best_suitcase)
    {
        current_suitcase=best_suitcase;
    }
}
infile.close();
cout.setf(ios::fixed);
cout.setf(ios::showpoint);
cout.precision(2);
cout<<best_suitcase;
return 0;
}

Answer 1

这一行

Suitcase current_suitcase(), best_suitcase();

声明两个返回 Suitcase 的函数，而不是定义两个变量。这称为 Most vexing parse规则。删除多余的括号可以解决问题:

Suitcase current_suitcase, best_suitcase;

另外，不要做 using namespace std .