🤩 New Cool Developer Tools for you. Explore →
FREE JavaScript Video Series Start Learning →
FLAT 75% OFF All Interactive courses at flat ₹250 / $3.25 only. HURRRRRY!! Explore now
  Signup/Sign In
Tests
MCQs to test your knowledge.
Forum
Engage with the community.
Compilers
Compilers to execute code in browser.
← Back to Questions
Ask Question
Not satisfied by the Answer? Still looking for a better solution?
Ask Question

Create a Singly Linked List

Problem Statement:

Declare a struct of your choice (Employee, Student, etc). Declare a node for singly link list. Declare an ADT of this linklist. Then Write a function that takes two ADTs of sorted Link Lists and Merge

Plz! write a code for the given problem this is my assignment and his due date is today 12 Pm ..
cpp data-structure linkedlist
by

2 Answers

Bharatgxwzm
Execution of this calculation is given beneath ?
#include <stdio.h>

#include <stdlib.h>



struct node {

   int data;

   struct node next;

};



struct node *head = NULL;

struct node *current = NULL;



//display the list

void printList() {



   struct node *ptr = head;



   printf("\n[head] =>");

   //start from the beginning

   while(ptr != NULL) {        

      printf(" %d =>",ptr->data);

      ptr = ptr->next;

   }



   printf(" [null]\n");

}



//insert link at the first location

void insert(int data) {

   //create a link

   struct node *link = (struct node) malloc(sizeof(struct node));



   //link->key = key;

   link->data = data;



   //point it to old first node

   link->next = head;



   //point first to new first node

   head = link;

}



int main() {

   insert(10);

   insert(20);

   insert(30);

   insert(1);

   insert(40);

   insert(56); 



   printList();

   return 0;

}


Output of the program should be ?

[head] => 56 => 40 => 1 => 30 => 20 => 10 => [null]
Shahlar1vxp
In order to create a singly linked list, you should not incrtement head after 'head = head-> next' in the for loop otheriwse the printlist returns NULL each time as the loop will not stop until and unless the 'head' is NULL. You may use the code-
LIST current = head;

while (current != NULL) {

    printf("%s    %d", current->str, current->count);

    current = current->next;

}

You need to remember that the second malloc allocates memory but the return value for it is not assigned to anything so that the allocated memory gets lost. A *newList* is allocated but is not initialized. So using a *memcpy* for copying memory to *newList ->str fails since newList ->str points to nothing.

Login / Signup to Answer the Question.

LoginSignup