Singly linked list, why is the list empty (head == NULL)?

Question

I created a singly linked list:

#include <iostream>

using namespace std;

struct Node{
    int data;
    Node *next;
};

bool isEmpty(Node *head){
    if (head == NULL){
        return true;
    }
    else{
        return false;
    }
}

void append(Node *head, Node *last, int data){
    Node *newNode = new Node;
    newNode->data = data;
    newNode->next = NULL;
    if (isEmpty(head)){
        head = newNode;
        last= newNode;
    }
    else{
        last->next = newNode;
        last= newNode;
    }

}

void printList(Node *current){
    if (isEmpty(current)){
        cout << "List is empty." << endl;
    }
    else{
        int i = 1;
        while (current != NULL){
            cout << i << ". Node: " << endl;
            cout << current->data << endl;
            current = current->next;
            i++;
        }
    }
}

void main(){
    Node *head = NULL;
    Node *last = NULL;
    append(head, last, 53);
    append(head, last, 5512);
    append(head, last, 13);
    append(head, last, 522);
    append(head, last, 55);
    printList(head);
}

When I compile it, the output is this:

List is empty.

But I don't know why. "head" gets an address, so "head" shouldn't be NULL. But apparently it is NULL. I don't know how to solve this problem.

You've tagged C++11: start using nullptr in place of NULL. — Bathsheba
– Bathsheba, Commented Jun 4, 2015 at 9:40

Some programmer dude · Accepted Answer · 2015-06-04 09:42:03Z

4

You have to remember that arguments to functions are by default passed by value, which means that the arguments value is copied and the function only works on the copy and not the original.

Now take the append function, when you pass the head argument the pointer is copied into the argument, and any changes made to head inside the function is only made to that copy.

To make it work you need to pass the arguments you change by reference, like

void append(Node *&head, Node *&last, int data)

Now head and last are references to the original pointer variables, and changes to the variable are made to the original variables.

answered Jun 4, 2015 at 9:42

Some programmer dude

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Comments

songyuanyao · Accepted Answer · 2015-06-04 09:41:07Z

1

The parameter head and last are passed by value, so the change of them inside the function append has nothing to do with the outside variables, they are independent.

You need to pass it by reference or pointer of pointer, such as:

void append(Node *&head, Node *&last, int data) {

answered Jun 4, 2015 at 9:41

songyuanyao

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bhavesh · Accepted Answer · 2015-06-04 09:41:26Z

1

In the function : append(Node *head, Node *tail, int data)

you should pass head and tail by reference. In your approach when you are assigning head = newNode then it is being assigned to a local copy of head and not the head that you passed.

answered Jun 4, 2015 at 9:41

bhavesh

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Rndp13 · Accepted Answer · 2015-06-08 11:57:25Z

0

Pass the address of the head and last variable and it sholud work.

append(&head, &last, 53);

and changes in the append function as below

void append(Node **head, Node **last, int data)

answered Jun 8, 2015 at 11:57

Rndp13

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