Linked list

+ <stdio.h>
+ <stdlib.h>
## A singly linked list: structs, pointers and malloc are plain C.
s Node
  value: i32
  next: *Node
f push(head: *Node, value: i32) *Node
  n: *Node = malloc(sizeof(Node))
  n->value = value
  n->next = head
  < n
f total(head: *Node) i32
  sum := 0
  for p: *Node = head; p; p = p->next
    sum += p->value
  < sum
f release(head: *Node)
  while head
    next: *Node = head->next
    free(head)
    head = next
f main()
  list: *Node = NULL
  for i := 1; i <= 5; i++
    list = push(list, i * i)
  for p: *Node = list; p; p = p->next
    printf("%d ", p->value)
  printf("\ntotal %d\n", total(list))
  release(list)
  < 0

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

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

Node *push(Node *head, int value) {
    Node *n = malloc(sizeof(Node));
    n->value = value;
    n->next = head;
    return n;
}

int total(Node *head) {
    int sum = 0;
    for (Node *p = head; p; p = p->next) {
        sum += p->value;
    }
    return sum;
}

void release(Node *head) {
    while (head) {
        Node *next = head->next;
        free(head);
        head = next;
    }
}

int main(void) {
    Node *list = NULL;
    for (int i = 1; i <= 5; i++) {
        list = push(list, i * i);
    }
    for (Node *p = list; p; p = p->next) {
        printf("%d ", p->value);
    }
    printf("\ntotal %d\n", total(list));
    release(list);
    return 0;
}