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(); }

Scott Jones · Accepted Answer · 2013-05-15 15:20:13Z

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.

edtheprogrammerguy · Accepted Answer · 2013-05-15 15:20:33Z

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);

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