Error with Linked Lists in C++

Question

I implemented a linked list in C++. I implemented it properly, however when I made a small change to my code, it gave me an error.
I changed
LinkedList l;
to
LinkedList l=new LinkedList();

It gave me the following error:

"conversion from ‘LinkedList*’ to non-scalar type ‘LinkedList’ requested"

Can anyone please tell me why?

Here is my code:

#include<iostream>
using namespace std;

class Node
{
public:
        int data;
        Node *next;

Node(int d)
{
    data=d;
    next=NULL;
}
};

class LinkedList
{
public:
Node *head;
LinkedList()
{
    head=NULL;
}
void add(int data)
{
    Node *temp,*t=head;
    if(head==NULL)
    {
        temp=new Node(data);
        temp->next=NULL;
        head=temp;
    }
    else
    {
        temp=new Node(data);
        while(t->next!=NULL)
            t=t->next;
        t->next=temp;
        temp->next=NULL;
    }
}

void Display()
{
    Node *temp=head;
    cout<<temp->data<<"\t";
    temp=temp->next;
    while(temp!=NULL)
    {
        cout<<temp->data<<"\t";
        temp=temp->next;
    }
}
};

int main()
{
LinkedList l=new LinkedList();
l.add(30);
l.add(4);
l.add(43);
l.add(22);
l.Display();
} 

Answer 1

Try this:

LinkedList *l=new LinkedList();

l->add(30);

When you use 'new' the return value is a pointer, not the object itself, so you have to declare the type as a pointer.

Don't forget to delete l at the bottom. You could also just say:

LinkedList l;

l.add(30);

Answer 2

You want this instead:

LinkedList * l = new LinkedList();

Note that the new operator returns a pointer (Foo *) to an object allocated on the heap.

Alternatively, more efficiently, and simpler for your purposes, you could just allocate the LinkedList as a local stack variable:

LinkedList l;

Then, you wouldn't have to worry about freeing the pointer (using delete), and the following dot operator use could remain.