How to return a pointer function?

Question

I'm new to c++, I'm asked for class work to do a function that takes a doubly linked list and copy it's elements into a singly linked list then returns it, however, I was given this function's declaration only, when I made the function it's either not returning anything or it gives me the error below, I understand the linked list code but I don't understand using a pointer and the returning part, can someone explain this to me? and how to make it work? I'm unable to find explanation to this online, or maybe I am searching the wrong keywords. Any help is appreciated.

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

SinglyLinkedList<T> list1;

DLLnode<T>*p = head;

while (p!=NULL) {

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

}

return list1;

}

//error: cannot convert ‘SLL<int>’ to ‘SLL<int>*’ in return

Do you see the difference between SinglyLinkedList<T>* and SinglyLinkedList<T> ? — Yunnosch
– Yunnosch, Commented Apr 6, 2020 at 12:00
Your function claims it will return SinglyLinkedList<T>*, but you're returning SinglyLinkedList<T>. They're not the same, A pointer to an object is not synonymous to the concrete object itself. The compiler is telling you exactly what the problem with the posted code is. Changing the return type to SinglyLinkedList<T> will probably solve your problem (assuming proper rule-of-three etiquette has been followed). — WhozCraig
– WhozCraig, Commented Apr 6, 2020 at 12:00
Yes, I tried making a pointer list1 instead I asked a question here before regarding this issue, stackoverflow.com/questions/61029659/… when I made it as a pointer it returned an address, answers suggested some ideas but I want to know how I can make this function work using a pointer in the function's declaration. — user5789
– user5789, Commented Apr 6, 2020 at 12:05
@WhozCraig Yea I was told to follow the rule of three but I am still unable to do it correctly for linked lists, and I was given this function for class work, I should implement it pointer way but I don't see how — user5789
– user5789, Commented Apr 6, 2020 at 12:19
If that's the case, SinglyLinkedList<T> list1; should be SinglyLinkedList<T> *list1 = new SinglyLinkedList<T>();, and all list1. in the succeeding code should be list1->. It's syntactically correct, and dreadfully wrong in concept. Wouldn't be the first time academia told it's student body to code-stupid, so I'm sorry for that. — WhozCraig
– WhozCraig, Commented Apr 6, 2020 at 12:22

john · Accepted Answer · 2020-04-06 12:28:22Z

1

1) Using a pointer for this is stupid. But that's what you've been told to do.

2) If you use a pointer then this function will return an address. That's what pointers are. You cannot change that. The trick is to dereference the pointer when you try to print. That way it won't print an address but will instead print what the pointer is pointing at. If you need help with this then post your printing code.

3) Here's how you do it with pointers

template <class T>
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;
}

This is the second occaision in recent days when posters have been told to do something stupid by their university professors. Oh well.

answered Apr 6, 2020 at 12:28

john

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1 Comment

WhozCraig Over a year ago

"second occasion in recent days" - lolz. I can assure you it happens far, far more often than that. Just imagine all the asinine times we don't see on this site because the lemmings blindly walked off the cliff. Sad, to be sure.

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