i have done the <

Question

i have overloaded the '<<' operator in the linked class implementation, and i declared the .cpp file in my main function, but it's not working, and this error occurs: undefined reference to operator<& list)

this is the declaration of the ostream function in linkedlist.h file:

  friend std::ostream& operator<< (std::ostream& os, LinkedList<T>& list);

and this is the implementation of the ostream function:

 template <typename T>
   std::ostream& operator<< (std::ostream& os, LinkedList<T> list) 
 {
  list.current = list.start;
  while(list.current != NULL)
    {
        os<< list.current->info<<" -> ";
        list.current = list.current->next;
    }
    os<<"NULL"<<endl;
    return os;
 }

In the main function, I have created a list that has object from SavingAccount class

LinkedList <SavingAccount> list;

and the error occurs in the main function in this line:

 cout << list <<endl;

well.. this is the LinkedList class implementation:

#include "LinkedList.h"
#include "SavingAccount.h"
#include <cstddef>
using namespace std;

template <typename T>
LinkedList<T>::LinkedList()
{
   start = NULL;
   current = NULL;
}
template <typename T>
LinkedList<T>::~LinkedList()
  {
    // Add code.
  }

  template <typename T>
   std::ostream& operator<< (std::ostream& os,const LinkedList<T>& list) {
    list.current = list.start;
    while(list.current != NULL)
    {
        os<< list.current->info<<" -> ";
        list.current = list.current->next;
    }
    os<<"NULL"<<endl;
    return os;
  }

and this is the header file of LinkedLins class:

#ifndef LINKEDLIST_H
#define LINKEDLIST_H
#include<iostream>

using namespace std;


template <typename T>
struct Node{
    T info;
    Node<T> *next;
};

template <typename T>

class LinkedList
{
    Node<T> *start;
    Node<T> *current;
public:
    LinkedList();
    ~LinkedList();
    friend std::ostream& operator<< (std::ostream& os, const LinkedList<T>& list);

  };

#endif // LINKEDLIST_H

i hope you guys can help me with that, your help is much appreciated

Compare your declaration and your definition (the second parameter). — lapk, Aug 16 '13 at 12:39
Also, if you are asking a question about compiler/linker errors, it always help by you posting the actual errors in the question as well. As complete and unedited as possible please. — Some programmer dude, Aug 16 '13 at 12:41
@juanchopanza after i read your comment i have done the implementation in the header file, but i faced a lot of other errors like:"endl was not declared" — Alladin, Aug 16 '13 at 12:45
**`endl` was not declared** - use fully qualified `std::endl` ( or `using namespace std;`) — lapk, Aug 16 '13 at 12:53
@Alladinsaoudi Generally, you cannot split template declaration and implementation between header and source file (unless it's a special case of explicit instantiation). — lapk, Aug 16 '13 at 13:06
@PetrBudnik , so what should i do?? cuz i'm totally confused, and i'm not familiar with these stuffs, — Alladin, Aug 16 '13 at 13:10

Oleksiy · Answer 1 · 2013-08-16T13:35:16.513

1

I'm not sure if this is what causing the error, but your function definition is different from your function declaration:

// you left out a & (reference sign) - LinkedList<T>& list in declaration
std::ostream& operator<< (std::ostream& os, LinkedList<T> list)

EDIT

Petr Budnik is right, you need to keep your template implementation in the same file as your definition.

There is one thing he didn't mention (and it's probably causing you problems): you need to make it explicit that you are using a template T above your << overload:

template < typename T>
friend std::ostream& operator<< (std::ostream& os, LinkedList<T>& list);

Also, remove const if you are planning to modify the LinkedList you pass in.

Here's (roughly) what your LinkedList.h should look like:

template <typename T>
struct Node{
    T info;
    Node<T> *next;
};

template <typename T>
class LinkedList
{
    Node<T> *start;
    Node<T> *current;
public:
    LinkedList();
    ~LinkedList();

    template < typename T>
    friend std::ostream& operator<< (std::ostream& os, LinkedList<T>& list);


};

template <typename T>
LinkedList<T>::LinkedList()
{
    start = NULL;
    current = NULL;
}

template <typename T>
LinkedList<T>::~LinkedList()
{
    // Add code.
}

template <typename T>
std::ostream& operator<<(std::ostream& os, LinkedList<T>& list) {
    list.current = list.start;
    while (list.current != NULL)
    {
        os << list.current->info << " -> ";
        list.current = list.current->next;
    }
    os << "NULL" << endl;
    return os;
}

P.S.

May I suggest that you use nullptr instead of 0 or NULL? It is a new C++11 syntax and has a few benefits (read about it if you want).

Hope this helps!

edited Aug 16 '13 at 13:35

answered Aug 16 '13 at 12:39

Oleksiy

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@Alladinsaoudi can you post the code for your LinkedList class? – Oleksiy Aug 16 '13 at 12:50
i have edited the question, u can see the code there. thank you ^^ – Alladin Aug 16 '13 at 12:54
@Alladinsaoudi also, can you post the declaration of LinkedList in your header file? – Oleksiy Aug 16 '13 at 12:56
any idea on how to do it?? because i'm stuck and i can't continue if i don't solve this – Alladin Aug 16 '13 at 13:14
i have done all what you have said, but adding "template " causes the problem of declaring the template again, this error occurs: declaration of 'class T' and when i delete that line, the fiiirst error occurs again (undefined reference to < – Alladin Aug 16 '13 at 13:53
@Alladinsaoudi Are you sure? I just tested it and it works. Make sure you copy+paste the code above to your header and have nothing else (except main) in your source. – Oleksiy Aug 16 '13 at 13:58
In your operator `friend` declaration use a template parameter name different from that of the one used in class itself. For example, `template < typename U > friend std::ostream& operator<< (std::ostream& , LinkedList< U >& );`. Even easier: define the operator together with declaration inside the class (keeping the `friend` spec) - and you won't need to add `template < typename U >` at all. – lapk Aug 16 '13 at 14:12
@PetrBudnik i have defined it inside the class, but now another error occured :( " no match for operator<< in os << list.LinkedList::current->Node::info " it occured in this line : " os << list.current->info << " -> "; " – Alladin Aug 16 '13 at 14:25
Did you keep the `friend` keyword? – lapk Aug 16 '13 at 14:28
You do not have `operator <<` defined for the type of variable `info`. Is `info` of some custom type of yours? – lapk Aug 16 '13 at 14:43
it's of type T also.. it has been declared this way: " struct Node{ T info; Node *next; }; " – Alladin Aug 16 '13 at 14:45
Since it is of template parameter `T`, in `LinkedList` the `info` is of `SavingAccount` type, correct? Then you need to define `operator <<` for `SavingAccount` type. – lapk Aug 16 '13 at 14:49
ohhhhhhhhh ok now i get it. i'll do it, thx a lot @Petr you really helped me – Alladin Aug 16 '13 at 14:51

lapk · Accepted Answer · 2013-08-16T14:58:06.910

Ok, after you posted your code it seems your problem is that you split LinkedList template declaration and implementation between a header and a source file. Unless you are doing explicit instantiation in the source file, you cannot do this: you cannot separate template declaration and implementation, they have to be in one header.

You need to put the content of your LinkedList.cpp in LinkedList.h, get rid of the source file altogether, and just include the updated header where you need it.

P.S. Also, modifying an object through a const reference is not a good idea, generally speaking. Compiler will not like it. You might want to reconsider your design.

P.P.S. If you REALLY want and mean to modify some data members in objects of a class that you access via const reference, you will need to declare those data members as mutable.

P.P.P.S. To elaborate on template function foo friend declarations for a template bar class and avoid lengthy discussions in comments.

If you are defining such function together with its declaration inside the class, it is a straightforward thing - just use friend keyword. For example, friend void foo( bar< T > & ) { ... }.

If you defining it separately outside the class, you need to use slightly different syntax in the declaration and avoid shadowing of template parameters: template< typename U > friend void foo( bar< U > & ) { ... } (assuming U was not used in template parameters in bar class declaration). And then you define it outside the class.

If you want to specify default template arguments for the foo it gets even more trickier. You cannot use default template arguments in friend declaration, so you have to forward declare everything:

template< typename T > class bar;

template< typename T = int >
void foo( bar< T > & );

template< typename T >
class bar
{
  template< typename U >
  friend void foo( bar< U > & );
};

template< typename T >
void foo( bar< T > & )
{
 ...
}

And one last P.P.P.P.S (I hope!). Since you are using operator << for type SavingAccount in your operator << for LinkedList, you need to define operator << for SavingAccount too.

NOTE TO SELF : Don't answer questions in comments' section to another answer ;).

I have pulled the content to my header file and updated everything, and i have taken off the const. now i'm facing one error: 'std::ostream& LinkedListstd::operator<< (std::ostream& os, LinkedList& list) must take exactly one argument' — Alladin, Aug 16 '13 at 13:27

score 0 · Answer 3 · answered Aug 16 '13 at 12:45

Apart very pathological cases, an "ouptut" operation should not alter the value to be output, so the proper function prototype shold be

template<class T>
std::ostream& operator<<(std::ostream& s, const LinkedList<T>& list)
{
...
}

and you should not assign new values to list members (like in list.current = list.start). You had better to rethink the logic as well...

Also, template function must be implemented in headers, not cpp files

i have done the <

3 Answers3

EDIT