linked list noob - function create new list

Well i have list with names and year of birth. Now I want to pass that list into function and year (for example 1982) so function should return newly formed list containing only names with specified year. This function must create new list inside function. So this is problem for me. How to create new "start" head inside function? how to fill it with data and display them?I tried but no luck. Note, it must be single linked list.

Code:

---

struct student
{
      char name[20];
      int year;
      struct student *next;
};

struct student *head = NULL;

struct student *selection(struct student *head,int year )
{

}

---

There are a few ways of doing it. One way is to leave the first node empty, because it simplifies the addition and deletion.

One problem you may have with your current design is that you're accepting an element called head in the function selection() and the actual head of the linked list is also called head. This means you can't use the actual head of the linked list inside the function. If you don't need to access it, then you're fine.

From the website tutorials:

http://www.cprogramming.com/tutorial/c/lesson15.html

Start slow. The concept of adding a new node to a linked list is just as simple as inserting it into the next available space. So until it finds a NULL element in the list, it keeps looping until it reaches the end. The exception to this is if the first element is NULL, in which case you have an empty linked list. Add a new element at the head.

This came to my mind but I can't work it out.

Code:

---

struct student
{
      char name[20];
      int year;
      struct student *next;
};
struct student *selection(struct student *head,int year)
{
        struct student *temp;
        struct student *head1=NULL;
       

        temp = (struct student *)malloc(sizeof(*temp));
       
        temp = head;

        while (temp != NULL)
        {
                if (temp->year == year)
                {
                       
                        temp->next=head1;
                        head1=temp;
                        temp=temp->next;
                }
        }

       
        temp=head1;

        if (head1 == NULL)
        {
                printf("List is empty");
                return;
        }
        else
        {
       
        while (temp != NULL)
        {
                printf("\n");
                printf("%s %d",temp->name,temp->year);
                printf("\n");
                temp=temp->next;
        }
        }
        return;

}

---

Code:

---

temp = (struct student *)malloc(sizeof(*temp));
new = (struct student *)malloc(sizeof(*new));

---

Why are you making two new elements just to add one? There's no reason to do that. After you do this, then you're just playing around with what pointer points where without actually doing anything.

In addition....

Code:

---

struct student *next;
struct student *next1;

---

This is a singly-linked list you said, so you don't need two pointers, much less two pointers to point forward.

Yes, I don't need them but I was trying to create another copy so it is separately pointing to new head. That's why *next1. More memory it takes but an idea. And as I said I am linked list noob. Now code is in original state.

You don't need to make any copies of nodes. For one, unless you free() the temporary node, you're causing a memory leak.

I still have no idea what next1 is for, and again, it's not needed.

Check this out: http://en.wikipedia.org/wiki/Linked_List

It has some code written in C, in addition to an extremely lengthy amount of information on linked lists.