Question

我正在尝试将字符串流传递给一个对象（类），该对象具有已声明和定义的重载提取运算符>>。例如，object1中重载的提取运算符的声明是

friend istream& operator >>(istream& in, Object1& input);

在object2中，我的声明几乎相同

friend istream& operator >>(istream& in, Object2& input);

在object1提取函数期间，func。获取一行，将其转换为字符串流并尝试使用Object2的提取（＆gt;＆gt;）运算符。

istream& operator >>(istream& in, Object1& input){
     Object2 secondObj;
     string data;
     string token;
     in>>data;
     in.ignore();
     token = GetToken(data, ' ', someint); //This is designed to take a part of data
     stringstream ss(token); // I copied token into ss
     ss >> secondObj; // This is where I run into problems.
}

我收到错误No match for operator >>。这是因为我需要将stringstream转换为istream吗？如果是这样，我该怎么做？

最小程序看起来像这样：在main.cpp中：

#include "Object1.h"
#include "Object2.h"
#include "dataClass.h"
using namespace std;
int main(){
    Object1<dataClass> firstObj;
    cin>>firstObj;
    cout<<firstObj<<endl;
}

Object1.h中的

：

#ifdef OBJECT1_H_
#define OBJECT1_H_
#include <iostream>
#include <string>
#include <cstddef>
#include "Object2.h"
template<class T>
class Object1{
public:
    //Assume I made the Big 3
    template<class U>friend istream& operator >>(istream& in, Object1<U>& input);
    template<class U>friend ostream& operator <<(ostream& out, const Object1<U>& output);
private:
    Object2<T>* head;
};
template<class T>
istream& operator >>(istream& in, Object1<T>& input){
     Object2 secondObj;
     string data;
     string token;
     in>>data;
     in.ignore();
     token = GetToken(data, ' ', someint); //This is designed to take a part of data
     stringstream ss(token); // I copied token into ss
     ss >> secondObj; // This is where I run into problems.
}
template<class T>
ostream& operator <<(ostream out, const Object1<T>& output){
     Object2<T>* ptr;
     while(GetNextPtr(ptr) != NULL){
         cout<<ptr;
         ptr = GetNextPtr(ptr); //Assume that I have this function in Object2.h
     }
}

Object2.h文件看起来类似于Object1.h，除了：

template<class T>
class Object2{
public:
//similar istream and ostream funcions of Object1
//a GetNextPtr function
private:
    T data;
    Object2<T>* next;
};
template<class T>
istream& operator >>(istream& in, Object2<T>& input){
      in>>data; //data is the private member variable in Object2.
                //it is of a templated class type.
}

Answer 1

以下编译正常：

#include <string>
#include <sstream>
#include <iostream>

using namespace std;

struct X{};

struct Y{};

istream& operator>>(istream&, X&) {}

istream& operator>>(istream&, Y&)
{
    stringstream ss("foo");

    X x;
    ss >> x;
}

int main()
{
    Y y;
    cin >> y;
}

您的问题必须在其他地方

你能发布一个完整的最小的自包含程序来证明这个问题吗？或者仅仅是函数istream& operator >>(istream& in, Object2& input)的声明和定义？它是否在翻译单元中的Object1版本之前声明了吗？

使用运算符＆gt;＆gt;将字符串流传递给istream

1 个答案: