C编程中的简单链接列表结构

Question

This is my first link list program in C, I am trying to initialize the values of the nodes and trying to print it. however, its not giving me the intended output. Can anyone let me know where am I going wrong?

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

typedef struct node NODE;

int main ()
{
    NODE a,b,c;
    NODE *list;
    list=&a;
    list->key = 10;
    list->next = &b;
    list->next->key=20;
    list->next->next=&c;
    list->next->next->key=30;
    list->next->next->next=NULL;
    printf("%d  %d  %d", a,b,c);
   return 0;
}

It prints 10 and 20 with some garbage in between.

Answer 1

You really shoulnd't be passing entire structures (the variables a , b and c ) to printf() like that, did that even compile?

You want to pass the integer data:

printf("%d %d %d\n", a.key, b.key, c.key);

but of course that totally ignores the links between the nodes.

It would more "interesting", in this context, to have something like:

static void print_list(const NODE *head)
{
  const NODE *prev = NULL;

  for(; head != NULL; prev = head, head = head->next)
    printf("%d ", head->key);
  puts(prev != NULL ? "\n" : "");
}

And then call that from main() after setting list :

print_list(list);  /* or print_list(&a); */

You can also simplify the creation of the linked list:

a.key = 10;
a.next = &b;
b.key = 20;
b.next = &c;
c.key = 30;
c.next = NULL;
list = &a;

This more clearly uses the fact that all nodes are directly available, and drops the hysterical link-following.

Answer 2

You want to print the values of the 3 structures, that are in the field key of each of them. So you need to change the line

printf("%d  %d  %d", a,b,c);

with the line

printf("%d  %d  %d", a.key,b.key,c.key);

It is actually strange that you do not get a warning from the compiler, like this one:

main.c:20:31: warning: format specifies type 'int' but the argument has type
  'NODE' (aka 'struct node') [-Wformat]