CODEWITHSUNDEEP
clinked-list
sherry
sherry

Reputation: 1

Simple Linked List in C

Here is a C code which creates a simple linked list with three nodes. Afterward a function called printList traverses the created list and prints the data of each node.

    // A simple C program for traversal of a linked list
    #include<stdio.h>
    #include<stdlib.h>

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

    // This function prints contents of linked list starting from
    // the given node
    void printList(struct node *n)
    {
        while (n != NULL)
        {
            printf(" %d ", n->data);
            n = n->next;
        }
    }

int main()
{
    struct node* head = NULL;
    struct node* second = NULL;
    struct node* third = NULL;

// allocate 3 nodes in the heap
    head = (struct node*)malloc(sizeof(struct node));
    second = (struct node*)malloc(sizeof(struct node));
    third = (struct node*)malloc(sizeof(struct node));

    head->data = 1; //assign data in first node
    head->next = second; // Link first node with second

    second->data = 2; //assign data to second node
    second->next = third;

    third->data = 3; //assign data to third node
    third->next = NULL;

    printList(head);
    printList(head); //question

    return 0;
}

Resource: http://quiz.geeksforgeeks.org/linked-list-set-1-introduction/

My question is that, since the input argument of function printList is a pointer of type node, the value of the pointer seems to be changed after the function call. In other words, after calling printList(head), it is reasonable to me that the pointer head must now point to a NULL value, therefore the second call to printList should print some irrelevant values. However, I am obviously wrong, since the output of this program is 1 2 3 1 2 3.

Could you please shed some light on this?

Upvotes: 0

Views: 1247

Answers (2)

Stephan Lechner
Stephan Lechner

Reputation: 35154

Variables are passed by value; this is also true for a variable whose value is a pointer:

Suppose variable head of type struct node* points to a node, let's say a node at address 0x12345. When you call printList(head), whereas the signature of this function is void printList(struct node *n), then the value of head is copied to the value of n; Hence, variable n and variable head, though being different variables, will have the value 0x12345. If printList then changes the value of n, e.g. by statement n = n->next, then only the value of variable n is changed; variable head will still have the value 0x12345.

Hope this helps...

Upvotes: 1

4386427
4386427

Reputation: 44256

C passes parameters by value. That means that the value of the variable passed to a function is copied into a new variable which is local to the function. No matter how much you change the local variable inside the function, it will never change the value of the variable passed to the function.

In other words: n inside the function is not the same as head in main. The local variable n is just initialized to the same value as head

Upvotes: 2

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