C linked list inserting node at the end

Question

I'm having some trouble with my insertion method for a linked list in C. It seems to only add at the beginning of the list. Any other insertion I make fail. And this CodeBlocks debugger is so hard to understand I still don't get it. It never gives me value, just addresses in memory. Anyway this is my function. Do you see any reason why it's failing?

/* function to add a new node at the end of the list */
int addNodeBottom(int val, node *head){

    //create new node
    node *newNode = (node*)malloc(sizeof(node));

    if(newNode == NULL){
        fprintf(stderr, "Unable to allocate memory for new node\n");
        exit(-1);
    }

    newNode->value = val;

    //check for first insertion
    if(head->next == NULL){
        head->next = newNode;
        printf("added at beginning\n");
    }

    else
    {
        //else loop through the list and find the last
        //node, insert next to it
        node *current = head;
        while(current->next != NULL)
        {
            if(current->next == NULL)
            {
                current->next = newNode;
                printf("added later\n");
            }
            current = current->next;
        }
    }
    return 0;
}

Then in main, only 929 is added.

   //testing addNodeBottom function
    addNodeBottom(929, head);
    addNodeBottom(98, head);
    addNodeBottom(122, head);
    addNodeBottom(11, head);
    addNodeBottom(1034, head);

Answer 1

I would like to mention the key before writing the code for your consideration.

//Key

temp= address of new node allocated by malloc function (member od alloc.h library in C )

prev= address of last node of existing link list.

next = contains address of next node

struct node {
    int data;
    struct node *next;
} *head;

void addnode_end(int a) {
    struct node *temp, *prev;
    temp = (struct node*) malloc(sizeof(node));
    if (temp == NULL) {
        cout << "Not enough memory";
    } else {
        node->data = a;
        node->next = NULL;
        prev = head;

        while (prev->next != NULL) {
            prev = prev->next;
        }

        prev->next = temp;
    }
}

Answer 2

This works fine:

struct node *addNode(node *head, int value) {
    node *newNode = (node *) malloc(sizeof(node));
    newNode->value = value;
    newNode->next = NULL;

    if (head == NULL) {
        // Add at the beginning
        head = newNode;
    } else {
        node *current = head;

        while (current->next != NULL) {
            current = current->next;
        };

        // Add at the end
        current->next = newNode;
    }

    return head;
}

Example usage:

struct node *head = NULL;

for (int currentIndex = 1; currentIndex < 10; currentIndex++) {
    head = addNode(head, currentIndex);
}

Answer 3

The new node is always added after the last node of the given Linked List. For example if the given Linked List is 5->10->15->20->25 and we add an item 30 at the end, then the Linked List becomes 5->10->15->20->25->30. Since a Linked List is typically represented by the head of it, we have to traverse the list till end and then change the next of last node to new node.

/* Given a reference (pointer to pointer) to the head
   of a list and an int, appends a new node at the end  */


    void append(struct node** head_ref, int new_data)
    {
    /* 1. allocate node */
         struct node* new_node = (struct node*) malloc(sizeof(struct node));

        struct node *last = *head_ref;  /* used in step 5*/

    /* 2. put in the data  */
        new_node->data  = new_data;

    /* 3. This new node is going to be the last node, so make next 
          of it as NULL*/
        new_node->next = NULL;

    /* 4. If the <a href="#">Linked List</a> is empty, then make the new node as head */
        if (*head_ref == NULL)
        {
       *head_ref = new_node;
       return;
        }  

    /* 5. Else traverse till the last node */
        while (last->next != NULL)
        last = last->next;

    /* 6. Change the next of last node */
        last->next = new_node;
        return;    
}

Answer 4

This code will work. The answer from samplebias is almost correct, but you need a third change:

int addNodeBottom(int val, node *head){

    //create new node
    node *newNode = (node*)malloc(sizeof(node));

    if(newNode == NULL){
        fprintf(stderr, "Unable to allocate memory for new node\n");
        exit(-1);
    }

    newNode->value = val;
    newNode->next = NULL;  // Change 1

    //check for first insertion
    if(head->next == NULL){
        head->next = newNode;
        printf("added at beginning\n");
    }

    else
    {
        //else loop through the list and find the last
        //node, insert next to it
        node *current = head;
        while (true) { // Change 2
            if(current->next == NULL)
            {
                current->next = newNode;
                printf("added later\n");
                break; // Change 3
            }
            current = current->next;
        };
    }
    return 0;
}

Change 1: newNode->next must be set to NULL so we don't insert invalid pointers at the end of the list.

Change 2/3: The loop is changed to an endless loop that will be jumped out with break; when we found the last element. Note how while(current->next != NULL) contradicted if(current->next == NULL) before.

EDIT: Regarding the while loop, this way it is much better:

  node *current = head;
  while (current->next != NULL) {
    current = current->next;
  }
  current->next = newNode;
  printf("added later\n");

Answer 5

After you malloc a node make sure to set node->next = NULL.

int addNodeBottom(int val, node *head)
{    
    node *current = head;
    node *newNode = (node *) malloc(sizeof(node));
    if (newNode == NULL) {
        printf("malloc failed\n");
        exit(-1);
    }    

    newNode->value = val;
    newNode->next = NULL;

    while (current->next) {
        current = current->next;
    }    
    current->next = newNode;
    return 0;
}    

I should point out that with this version the head is still used as a dummy, not used for storing a value. This lets you represent an empty list by having just a head node.