How to return a pointer ?

Question

Now I am unsure about how to do below as Iam fairly new to C++

I’m trying to do a function that returns a linked list, however in the method, I am asked to do, the return type includes a pointer

// copying the values from the doubly linked list to the singly linked list then returns the singly linked list.

However, it’s not returning anything, what am I doing wrong?

SinglyLinkedList<T>* DoublyLinkedList<T>:: Function() {

SinglyLinkedList<T> list1;

DLLnode<T>*p = head;

while ( p!= NULL ) {

list1.addtoHead(p->value);
p=p->next;
}

Return list1;
}

For example Original doubly linked list
1 <=> 2 <=> 3

The returned singly linked list
1 -> 2 -> 3

error message given:

*cannot convert SinglyLinkedList<int> to SinglyLinkedList<int>* in return

Answer 1

Can you post your code please?
Because I don't think you would need to return it as pointer.

Answer 2

Mustafa K. thanks for your comment! sure I will edit the post but that might take a little because I am using my phone, however I was given the function declaration in class and told to return the singly linked list but I don’t understand it quite well or how to make this idea work.

Answer 3

Noura Can you try creating your list1 as pointer because your function wants it pointer, or you simply delete asterisk sign here SinglyLinkedList<T>* DoublyLinked<T>::Function()

Answer 4

I tried this now, I had to do some changes to it for example
list1->addtoHead but when I compile it gives me segmentation fault , however when I tried yesterday it gave me an address instead of the list itself

Answer 5

Mustafa K.

Answer 6

Noura If it gives you segmentation fault then that means you are trying to use a place at ram before you reserve it.
You should do something like that in your list in addtohead:
You need to create new node,
node* temp = new node( data );
Then you need to,
head->next = temp;
Do something like this then it does work.

edit:
then of course you need to change the place of tail too

Answer 7

the underlying problem here is that compiler always make a stack for a function call. as soon as the function exits, the function stack get removed which causes all the local variables of functions goes out of scope. ( which means that the local variable "list1" no longer exist ).

so we can fix that by either :

using the new operator ( this is generally a bad idea. If you must do it this way then maybe you can use a smart pointer ):

SinglyLinkedList<T>* DoublyLinkedList<T>:: Function() {

SinglyLinkedList<T>* list1 = new SinglyLinkedList<T>();

DLLnode<T>*p = head;

while ( p!= NULL ) {

list1->addtoHead(p->value);
p=p->next;
}

Return list1;
}

or by making it static. (as static variables have a property of preserving their value even after they are out of their scope):

SinglyLinkedList<T>* DoublyLinkedList<T>:: Function() {

static SinglyLinkedList<T> list1;

DLLnode<T>*p = head;

while ( p!= NULL ) {

list1.addtoHead(p->value);
p=p->next;
}

Return &list1;
}