Print a linked list in C - c

Question

I am working to develop a linked list API for a Lisp compiler, but I chose to write this part in C because writing it in assembly wouldn't be very fun. I can successfully make a linked list from a series of elements with the list function below. I have verified that it works, shown by the examples in main below. But why does my function print_list only print the first element, and then zero? I am so confused. Please offer any help if you understand what is going wrong.
#include <stdio.h>
#include <stdarg.h>
typedef struct Node Node;
struct Node {
int head; // make to a union later
Node* tail;
};
//////////
Node cons(int curr, Node* next) {Node cell = {curr, next}; return cell;}
int car(Node node) {return node.head;}
Node* cdr(Node node) {return node.tail;}
//////////
void print_elems(Node node) {
printf("%d", node.head);
if (node.tail == NULL) return;
printf(" ");
print_elems(*node.tail);
}
void print_list(Node node) {
printf("(");
print_elems(node);
printf(")\n");
}
//////////
Node make_pairs(va_list args, int argc) {
Node linked_list;
linked_list.head = va_arg(args, int);
if (argc == 0) {
linked_list.tail = NULL;
return linked_list;
}
Node tail = make_pairs(args, argc - 1);
linked_list.tail = &tail;
return linked_list;
}
Node list(int length, ...) {
va_list elems;
va_start(elems, length);
Node linked_list = make_pairs(elems, length);
va_end(elems);
return linked_list;
}
//////////
int main() {
Node big_list = list(5, 1, 2, 3, 4, 5);
/*
This is NULL, as expected:
big_list.tail -> tail -> tail -> tail -> tail -> head
And this is 1:
big_list.head
*/
// but why does this only print "(1 0)"?
print_list(big_list);
}

Answer 1

You could print linked list in this kind of manner : this is an exemple;
My objective is just to make you notice with pointer you don't have problems with local variable
typedef struct Node Node;
Node* first_element = create_list; // just for understanding "first element" is the linked list
Node* ptr = first_element; // copy in another pointer your list then you will avoid loosing your first element and breaking your list (personnal advice : not needed)
while (ptr->tail){
ptrinf("%d\n", ptr->head); // for exemple
ptr = ptr->tail;
}
//after the while ptr = the last element because ptr->tail == NULL
formated to function it looks like that :
#include <stdio.h>
typedef struct Node Node;
void display_list (Node* first_element)
{
Node* ptr = first_element; // copy in another pointer your list then you will avoid loosing your first element and breaking your list (personnal advice : not needed)
while (ptr->tail){
ptrinf("%d\n", ptr->head); // for exemple
ptr = ptr->tail;
}
//after the while ptr = the last element because ptr->tail == NULL
}
if i didn't answer the question well tell me, i'm new on stack i would improve ^^.
I hope i helped you.