Access friend class's private member

Question

I'm trying to write a console application like Twitter. User and UserList classes including each other. I'm trying to access to following user's followers. UserList class is used for linked list.

//User.h

#pragma once

#include <iostream>
#include <string>

using namespace std;

class UserList;
class User
{
   friend class UserList;
private:
   string userName;
   string personalComment;
   UserList *followed;
   UserList *following;
   int followedNumber;
   int followingNumber;
   //TWEET
   UserList *blocked;
   User* next;
public:
   User();
   User(string&,string&);
};

//UserList.h
#pragma once

#include <iostream>
#include <string>

using namespace std;


class User;

class UserList{
private:
    User *root;
public:
    UserList();
    ~UserList();
    void addUser(string&,string&);
    bool checkUser(string&);
    User& findUser(string&);
    void printList();
};

Firstly, I wrote a function to find following user.

//userList.cpp
User& UserList::findUser(string& name)
{
   User *temp=root;
   while(temp!=NULL)
   {
       if(!name.compare(temp->userName))
       {
            return temp;
       }
       temp= temp->next;
    }
    temp= temp->next;
    return temp;
}

For example user1 wants to follow user2. I want to check does user1 already follow user2.( checkUser(username) looks for a user in a list and return bool)

//main.cpp in main()
if(users.findUser(user1name).following->checkUser(user2name))
{
            cout<<"Err: The user '"<< user1name << "' has already followed '"<<user2name<<"'!"<<endl;
}

But there is a "UserList* User::following is private" error and "within this context"

How can I access this user's lists?

@tobi303 Thought the exact same thing. :P Apparently there are people not using a HTML renderer but just reading the tags and all... — cadaniluk, Nov 25 '15 at 19:33
I don't see any *nested* class here, there is no class declaration or definition inside `class User`. — Thomas Matthews, Nov 25 '15 at 19:37
You definitely do not show us the line that causes the error. Now you fixed the typo but in the code you show you are never trying to acces `User::followed` — 463035818_is_not_an_ai, Nov 25 '15 at 19:38
BTW, you could use `if (name == temp->userName)` instead of using the `compare` method; less typing. — Thomas Matthews, Nov 25 '15 at 19:41
Lets assume the error is `User::following` is private. Then it is still hard to tell how to fix without knowing where the last code snippet is placed. In main? — 463035818_is_not_an_ai, Nov 25 '15 at 19:42
I corrected "follower" as a "followed".But I think this is not the problem. errors coming from if line — Merve Pia, Nov 25 '15 at 19:44
tobi303 error line: if(users.findUser(user1name).following->checkUser(user2name)) — Merve Pia, Nov 25 '15 at 19:49
@MervePia That's a train-wreck, post a [MCVE] please. Otherwise it's not possible to diagnose your problems. — πάντα ῥεῖ, Nov 25 '15 at 19:53
sorry, I must have missed the line "//main.cpp in main()". Shouldnt `checkUser(username)` be a method of `UserList`? — 463035818_is_not_an_ai, Nov 25 '15 at 19:54
First, I try to find user1,then I try to look user1's following list for user2. Is findUser returning type wrong? Or is if statements condition wrong? — Merve Pia, Nov 25 '15 at 19:55
Argh forget my last comment... The point is: you cannot access private members outside of the class. Expection: You can access them inside a class that is declared as friend. But you cannot access it in the main. — 463035818_is_not_an_ai, Nov 25 '15 at 19:56
checkUser() is in UserList. I added UserList.h to my question — Merve Pia, Nov 25 '15 at 19:56
The code with the error should be encapsulated in the `UserList` class, this will make the error disappear. — 463035818_is_not_an_ai, Nov 25 '15 at 19:57
oww, how can I encapsulate ? should I write a function which will be return UserLists in User object? — Merve Pia, Nov 25 '15 at 20:03

463035818_is_not_an_ai · Accepted Answer · 2015-11-25T20:08:55.870

0

In general you should not put logic of your classes in the main. Your issue can be solved for example by adding a method to User that tries to add another user instead of doing this in the main. Something like this:

User::FollowOther(std::string other){
    if(this->following->checkUser(other)) {
        cout<<"Err: The user '"<< userName << "' has already followed '"<<other<<"'!"<<endl;
    }
    /*...*/
}

The error you got is because User::following is private in User. Private members can only be accessed from within the class. The exception is: You can access private members from within a different class that is declared as friend. But you cannot access private members in the main. Btw, I am not at all a fan of declaring friends, because it breaks encapsulation and makes the relation between classes less obvious, but thats just my personal opinion.

edited Nov 25 '15 at 20:08

answered Nov 25 '15 at 20:02

463035818_is_not_an_ai

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oh I get it! I'm trying now – Merve Pia Nov 25 '15 at 20:12
@MervePia there are some other smaller issues with your code, e.g. why do you store pointers to the lists? I see no need to do so. And another minor thing: having `followedNumber` and `followingNumber` as members of `User` forces you to write unnecessary code because you will have to keep this numbers up to date. Instead provide a `size()` method for `UserList` that will serve the same purpose. – 463035818_is_not_an_ai Nov 25 '15 at 20:16

Access friend class's private member

1 Answers1