Operator overloading in Linked List

Question

I am moving here from my other post. But this time I am able to obtain some type of output. But I cannot seem to iterate through my nodes and get them to print out individually as how it should look on the output. Here is what I have so far and also a screenshot of what the program should output.

enter image description here

LList.h

#ifndef LList_h
#define LList_h

#include <iostream>
#include "node.h"

class LList
{
    public:
        LList(void);            //constructor
        LList(const LList &);   //copy constructor
        ~LList();               //destructor
        LList *next;            //points to next node
        void push_back(const string &str);
        void push_front(const string &str);
        friend ostream& operator<<(ostream& out, const LList& llist);
        LList &operator=(const LList &l);       

    private:
        Node *_head;
        Node *_tail;
        LList *front;       //points to front of the list
};

inline LList::LList(void)
{
    cerr << "head = tail = 0 at 0024f8d0\n";

    _head = 0;
    _tail = 0;
    front = 0;
}

inline void LList::push_back(const string &str)
{
    Node *p = new Node(str);
    if (_tail == 0)
    {
        _head = _tail = p;
    }
    else
    {
        _tail ->next(p);
        _tail = p;
    }

    if (_head == 0)
    {
        _head = _tail = p;
    }
    else
    {
        _head ->next(p);
        _head = p;
    }
}

inline void LList::push_front(const string &str)
{
    Node *p = new Node(str);
    if (_tail == 0)
    {
        _head = _tail = p;
    }
    else
    {
        _tail ->next(p);
        _tail = p;
    }

    if (_head == 0)
    {
        _head = _tail = p;
    }
    else
    {
        _head ->next(p);
        _head = p;
    }
}

LList & LList::operator=(const LList &l)
{
    _head = l._head;
    _tail = l._tail;
    front = l.front;
    return *this;
}

inline LList::~LList()
{
}


#endif

maind.cpp

#include "LList.h"
#include <iostream>

using namespace std;

ostream& operator<<(ostream& out, const LList& llist);

int main( )
{
    LList a;

    a.push_back(  "30" );
    a.push_front( "20" );
    a.push_back(  "40" );
    a.push_front( "10" );
    a.push_back(  "50" );

    cout << "list a:\n" << a << '\n';
    return 0;
}

ostream &operator <<( ostream &out, const LList & llist )
{
    for( LList *p = llist.front; p != 0; p = p -> next )
        out << p -> next;

    return out;
}

I don't think that I am iterating through my list, and that is why I am not seeing any results being outputted. I am not sure what is wrong with my code. The screenshot is what the output is supposed to be. — beginnerprogrammer
– beginnerprogrammer, Commented Feb 14, 2013 at 4:44
Your push_back looks weird - plays with both ends of the list... — John3136
– John3136, Commented Feb 14, 2013 at 4:45
I was thinking I could use "front" to point to the beginning of the list. I use "front" where I am overloading << — beginnerprogrammer
– beginnerprogrammer, Commented Feb 14, 2013 at 4:45

Karthik T · Accepted Answer · 2013-02-14 04:44:06Z

1

out << p -> next;

This line will skip your first element and cause undefined behavior (possibly segfault) on your last element. This should be out<<p.

answered Feb 14, 2013 at 4:44

Karthik T

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1 Comment

beginnerprogrammer Over a year ago

oh ok. I had no idea that was going to happen

Bill Lynch · Accepted Answer · 2013-02-14 04:46:04Z

1

Your operator<< will print nothing because LList::front is never assigned to. It's always null.

answered Feb 14, 2013 at 4:46

Bill Lynch

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1 Comment

beginnerprogrammer Over a year ago

ok, that would mean I would have to make some changes in the private section of the class definitions correct?

David Schwartz · Accepted Answer · 2013-02-14 04:50:10Z

1

Your push algorithms make no sense. To push something at the back of the list, you only want head to be modified if the list is empty, but you have:

if (_head == 0)
{
    _head = _tail = p;
}
else
{
    _head ->next(p);
    _head = p;
}

Why are you setting _head to p if the list had entries in it already? Your code has a number of similar bugs -- the logic just isn't right.

The end should probably just be:

if (_head == 0)
    _head = p;

If there's already a node at the head, adding an entry to the back doesn't affect head at all.

answered Feb 14, 2013 at 4:50

David Schwartz

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1 Comment

David Schwartz Over a year ago

Well, it's obviously wrong. If you have ten nodes in the list and are adding a new one at the tail, clearly the head pointer should not be changed to point to the added node. The same node is still at the head, not this one.

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