Building a linked list

Every node is a separate heap block holding a value and a next pointer. push makes a new node, points its next at the current head, and returns it as the new head.

After three pushes, follow the arrows from head: 3, then 2, then 1, then NULL. The list is in reverse order because each push goes on the front.

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

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

struct node *push(struct node *head, int value) {
    struct node *n = malloc(sizeof *n);
    if (!n) exit(1);
    n->value = value;
    n->next = head;
    return n;
}

int main(void) {
    struct node *head = NULL;
    for (int v = 1; v <= 3; v++) {
        head = push(head, v);
    }
    int sum = 0;
    for (struct node *p = head; p != NULL; p = p->next) {
        sum += p->value;
    }
    printf("sum = %d\n", sum);
    while (head) {
        struct node *next = head->next;
        free(head);
        head = next;
    }
    return 0;
}

Output:

sum = 6

From the lesson: Linked data structures