Question

I’m working on a singly linked list in C. This is what I’ve written so far.

C program
#include<stdio.h>
#include<stdlib.h>

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


struct Node* init()
{
   struct Node* head=NULL;
   head=malloc(sizeof(struct Node));
    head->value=-1;
    return head;

}



int length(struct Node* head)
{
    struct Node* current=head;
    int length=0;
    while(current!=NULL)
    {
        length++;
        current=current->next;

    }
    return length;

}


void print(struct Node* head)
{
   int i=0;
   int len=length(head);
   for(i=0;i<len;i++)
   {
    printf("%d%d",i,head[i].value);
    printf("\n");


   }


}




 struct Node* insert(int data,struct Node* head)
 {
    struct Node* current=NULL;
    if(length(head) > 0)
    {
        int val=head->value;        
        if (val==-1)
        {
            head->value=data;
            head->next=NULL;

        }
        else
        {
           current=malloc(sizeof(struct Node));
           current->value=data;
           current->next=head;
           head=current;


      }

  }
   else
  {
    printf("List is empty");

  }

  return head;


}

int main() 
{

/* printf("Hello"); */
struct Node *head=init();

head=insert(20,head);
head=insert(30,head);
head=insert(40,head);

print(head);
printf("%d",length(head)); 

return 0;

}

The output values I get are:
Index Value
0 40
1 0
2 0

and for length is 3. I’m not able to grasp what I’m doing wrong here in pointer manipulation.

Answer 1

One obvious problem is not setting next to NULL on init – that would fail when checking length on the empty list

But your real problem is the print function

You can’t use:

head[i].value

That notation is only valid for arrays, you need to use next to find each member