Needing help with linked lists

[peste19]

I am reviewing material for my c final and have a couple of questions, for example i did this program below to add 1 and 2 to nodes and the print them, if i have them in main, they print if i try to make a function do it, it doesnt print anything, can anyone point to me why it doesnt?

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
struct node
{
    int data;
    struct node *next;
};

void traverse();
struct node *head;
struct node *tail;
struct node *p;
int main()
{
    struct node *head = '\0';
    struct node *tail = '\0';
    struct node *p;
    head = malloc(sizeof(struct node));
    tail = malloc(sizeof(struct node));
    
    
    head -> data = 1;
    head -> next = tail;
    tail -> data = 2;
    tail ->next = '\0';
    traverse();
    
    while(head)
    {
             printf("%d\n",head ->data);
             head=head ->next;
    }
    
    system("pause");
}

void traverse()
{
     struct node *p;
     p=head;
     while(p)
     {
              printf("%d\n",p ->data);
              p=p ->next;
     }
}

[peste19]

nevermind i figured it out, had it as a local and not global

[oogabooga]

You have head, tail, and p defined both globally and in main. The beginner's rule for when to use globals is "never, unless told otherwise". They should just be in main and head needs to be passed to your function. And get rid of the loop in main since it's now in traverse.

Also, this is incorrect:
struct node *head = '\0';
It may work on most machines, but should be
struct node *head = NULL;
Same for tail and next.

[peste19]

that's actually how i learned how to do it and not NULL and i have noticed that everyone does it the way you mentioned bu thanks for the tip, i noticed that i have them as local instead of global that was why the function didnt "see" them.

Got another question, if i wanted to find out how many nodes i have in my program?

[Salem]

> Got another question, if i wanted to find out how many nodes i have in my program?
Write another traverse(), which counts instead of prints.

[oogabooga]

Now that I think about it, using '\0' will work on all machines since it's really just a constant integer 0 and that will be translated into the proper null-pointer data for whatever type by the compiler.

However, it's still incorrect since it signifies the wrong thing. Basically there's 4 kinds of "zero" in C.

0	integer zero (also 0U, 0L, 0UL)
0.0	double floating point zero (or 0.0f for float zero)
'\0'	ascii-null
NULL	pointer-null (usually defined as either 0 or (void*)0)

They're interchangeable (for the most part) but signify different concepts to the programmer.