unexpected results while creating a linked list in C

Question

MY function to create a linked list was

 void create(node *list)
 {
    fscanf(fp,"%d\t",&list->number);
      if(list->number==0)
           {
                list=NULL;
           }
     else
           {
           list->next=(node *)malloc(sizeof(node));
           create(list->next);
           }
            return;

      }

INPUT in my txt file was like

1 37 79 164 155 32 87 39 113 15 18 78 175 140 200 0

2 4 160 97 191 100 91 20 69 198 196 0

Now when i am trying to access the 0 through head->next->next.....->number even though i have assigned null pointer in the if statement when the input is 0,the output is being shown as 0

Also i tried changing the if statement to free(list); now i am able to access the elements in the next row just by adding more (next->) b/w head and number

 struct linked_list  //structure of node
        {
       int number;
       struct linked_list *next;
        };

Please explain me why the program is behaving like this

awesoon · Accepted Answer · 2014-06-06 02:45:00Z

2

Here:

list = NULL;

You are assigning NULL to a local pointer list, but the real pointer does not change.

You can only manipulate the data, pointer refers to. If you want to change a pointer, you have to pass pointer to pointer as function parameter. But I suggest you rewrite your function like this:

node* inputNode()
{
    int number = 0;
    fscanf(fp, "%d\t", &number);

    node* n = NULL;

    if(number != 0)
    {
        node = (node *)malloc(sizeof(node));
        node->number = number;
        node->next = NULL;
    }

    return node;
}


node* createList()
{
    node* root = inputNode();
    node* curr = root;

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

    return root;
}

answered Jun 6, 2014 at 2:45

awesoon

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