How to remove all nodes from a circular list?

Question

I’m having a hard time getting the logic of how to remove all nodes from a circular list, if someone could give a brief explanation of concept and / or point out the errors in the code I’ve already done would be very grateful :

int freelist ( Node ** list ) {

    if ( empty ( list ) ) {

        return 0;

    }

    if ( ( * list ) == ( * list ) -> next ) {

        free ( * list );

        ( * list ) = NULL;

        return 1;

    }

    Node * tmp = NULL, * aux = ( * list ) -> next;

    while ( aux != ( * list ) ) {

        tmp = aux;

        aux = aux -> next;

        free ( tmp );

    }

    // ( * list ) = NULL;

    return 1;

}

If the list contains only one item the function works perfectly, already if more than one it does not delete them ... ( Full code here )

Answer 1

I built this method in C++, but it’s easy for you to pass it to C.

    void clean() {

    if (head == NULL) {
        cout << "A lista esta vazia\n";
        return;
    }
    else {

        while (head != tail) {
            Node * aux = head;
            head = head->next;
            delete(aux);
        }

        Node * aux = head;
        head = NULL;
        delete(aux);

        cout << "A lista foi esvaziada\n";

    }

}

The head is responsible for marking where the circular list begins, and the tail is responsible for marking the last one on the list.

Note that I use a repetition structure in which I keep the node I want to displace on an auxiliary pointer, then I modify the head (head points to the next in what will be out of focus), and finally a delete at the address aux guard to shift the memory.

Notice that I use delete instead of free, because I used the operator newto create a no in my list in C++. In C, you will use the malloc to create the ones in your list, and the function free to shift the memory.