CODEWITHSUNDEEP
cdata-structureslinked-list
Beginner
Beginner

Reputation: 759

Insertion in Linked list

I am studying how to insert node at the beginning and I came through this code which I am unable to understand.
I am not getting the print function .

typedef struct node
{
    int data;
    struct node *next;
} Node;
Node *head;
void insert(int x)
{
    Node *temp=(Node*)malloc(sizeof(Node));   
    temp->data=x;
    temp->next=head;
    head=temp;
}
void print()
{
    Node *temp=head;
    printf("List is:");
    while(temp!=NULL) //Unable to understand this statement as how will the loop terminate?                  
    {
        printf("%d ",temp->data);
        temp=temp->next;
    }
    printf("\n");
}

Upvotes: 1

Views: 76

Answers (3)

Persixty
Persixty

Reputation: 8589

Indentation on the Question is a mess.

But the print function is:

  void print() {
      Node *temp=head;             
      printf("List is:");         
      while(temp!=NULL) {         
          printf("%d ",temp->data);
          temp=temp->next;
      }
   }

If head==NULL then temp==NULL at the start and the loop will terminate immediately.

If head!=NULL then temp!=NULL at the start and the data for the head element will be output.

The line temp=temp->next; will then make temp point to the next Node in the list and the loop will advance. If there's only on Node in the list the loop will then terminate.

Otherwise it will print the data for that next element and the line temp=temp->next will move to the third Node (if any).

And so on...

NB: The line Node *head; is not recommended. It will be initialized to NULL because it has static storage duration. But Node *head=NULL; is recommended for future-proofing and readability.

Upvotes: 0

user3767013
user3767013

Reputation: 236

A C-Program need a main function to start with. If you want to test the code above, you need to add a main function to initialize head to NULL at start, call insert() several times and print() to check the results.

Upvotes: 0

Gopi
Gopi

Reputation: 19874

Assume your linked list looks like

a->b->c->d
|
|
head

So now we use a temporary variable 

temp = head;

and 

while(temp != NULL)
{
  //Keep moving temp
  printf("%d\n",temp->x);
  temp = temp->next;
}

So head is never moved and it is just the temporary pointer which is moved till end of the list the end of the list is got when we reach temp = NULL

a->b->c->d
|
|
temp = head

temp = temp->next;

    a->b->c->d
    |  |
    |  |
  head temp

Repeating the above move until temp = NULL which is TRUE when the last node contents are printed and we do temp = temp->next;

Upvotes: 2

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