Where we left off

Time *t = new Time;
t->hour = 5;
do_something_with_time(t);
delete t;

Today's topics

• Intro to linked list structures
• Nodes and node pointers
• Building linked lists
• Traversing linked lists
• Linked Lists vs Arrays

Textbook Sections 13.1

Another list structure? Why?

• Arrays, even dynamically allocated, are a fixed size and order

  int *arr = new int[10];
  

• Growing or shrinking an array is painful

  int *new_arr = new int[20]; // allocate a new bigger array
  for (int i = 0; i < 10; i++) { // copy the old array into the new one
      new_arr[i] = arr[i];
  }
  
  delete [] arr; // free the memory at the old array
  arr = new_arr; // point the original memory to the new array
  new_arr = NULL; // good practice to reset the temporary pointer
  

• This is expensive, particularly with an array of structs

Linked lists overview

• Instead of allocating a whole array, why not string together a bunch of pointers?
• A linked list is a data structure where each element contains a value (or multiple values), as well as a pointer to the next item in the list
• The memory addresses of each item are random, and that's okay!
• To accomplish this, we need a struct that contains a value and a pointer to the next item in the list

  struct Node {
      <data type> data;
      Node *next;
  };
  

That Node structure looks funky

• Node is a struct with a field that's a pointer to another Node
• This is a self-referential structure. The name Node is common for linked lists, but we could have named it anything, for example:

  struct Person {
      string name;
      Person *spouse;
  };
  

• Why is spouse a pointer instead of just a Person?

Pointers so far

So far we've learned how to:

• Allocate memory on the stack by declaring variables
• Allocate memory on the heap by using new
• Define pointers to named memory addresses (on the stack)
• Define pointers to unnamed memory addresses (on the heap)

Linked Lists

• We only need to declare one variable to get started

  Node *head = NULL;
  

• We then use the self-referential pointer to link to the next chunk of data

  head = new Node;
  head->data = 5;
  head->next = NULL;
  

• We can add or remove items from the list dynamically, and only need to keep track of the head pointer

Starting a linked list

• For this simple Node struct (which only holds an int):

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

• We start with the head pointer:

  Node *head = NULL;
  

• This is an empty list
• All we've done so far is allocate space for a pointer to a Node
• Don't lose track of the head pointer!

Building the list

• Unlike arrays, list nodes need to be allocated one at a time

  head = new Node; // allocate a new node on the heap
  head->data = 7; // assign its value
  head->next = NULL; // point to the next one in the list
  

• The last node of the list should always point to NULL
• This is now a singleton list, where the start and end are the same item
• We can add another item to the end by allocating the next pointer:

  head->next = new Node;
  head->next->data = 10;
  head->next->next = NULL;
  

Adding to the front

• We can add a third node to the front of the list, but be careful!

  head = new Node;
  head->data = 1;
  head->next = ???; // uh oh, we lost the old head!
  

• We should instead declare a temporary pointer for the new node:

  Node *temp = new Node;
  temp->data = 1;
  temp->next = head;
  head = temp; // reassign head to the address of temp
  

• Question: should we delete temp?

Adding to the middle (inserting a node)

• At this point we have a list with 3 nodes: 1 -> 7 -> 10
• Let's add a 5 between 1 and 7
• Once more, we'll need a temporary pointer for the new node:

  Node *temp = new Node;
  temp->data = 5;
  temp->next = head->next; // steal the pointer to 7
  head->next = temp; // point 1 to 5
  

Linked list check-in 1/2

We can't just dereference head + 1 to get the next item in the list because...

Linked list check-in 2/2

next needs to be a Node * and not a Node because...

Basic list traversal

• Pointer arithmetic only works because arrays are contiguous in memory
• Since a list is only defined by its head pointer, we need to traverse the list to find any other item - we can't just say head + 1 or head[1]
• The basic algorithm is something like:

  set travelling pointer to head
  while travelling pointer points to something
      do something with the current node
      advance the travelling pointer to the next node
  

• "Do something" can be printing, searching, summing, etc

Traversing a list

In C++ syntax:

Node *current = head;
while (current) { // or while (current != NULL)
    // do something with current->data
    current = current->next;
}

• This is a sentinel loop that stops when current is NULL
  • As always, the LCV update must be the last line of the loop body
  • Very important that the last node point to NULL!
• current does not allocate new memory (other than for the pointer itself)

Inserting a node

Basic approach:

• Find the proper position for insertion (We'll deal with this later)
• Allocate a new node
• Steal the next pointer from the previous node
• Assign the previous node's next pointer to the new node

Handling special cases

We want to insert a node at the "proper" position, which may be:

• At the beginning of the list
• In the middle of the list
• At the end of the list
• In an empty list

Arrays vs Linked Lists

Coming up next

• Lab tomorrow: continue dynamic allocation lab
• Lecture: More linked lists
• Assignment 3: Linked lists

Textbook Chapter 13