Initialisation of struct elements - c

Question

Why aren't the struct pointers initialized to NULL using the following code
code
#include <stdio.h>
#include <stdlib.h>
struct list_el
{
int val;
struct list_el * right, * left, *parent;
}item_default={0,NULL,NULL,NULL}; //Default values
typedef struct list_el node;
int main(int argc, char const *argv[])
{
node * new_node = (node*) malloc (sizeof(node));
(new_node == NULL) ? printf("0\n") : printf("1\n");
(new_node->parent == NULL) ? printf("0\n") : printf("1\n");
(new_node->right == NULL) ? printf("0\n") : printf("1\n");
(new_node->left == NULL) ? printf("0\n") : printf("1\n");
(new_node->val == 0) ? printf("0\n") : printf("1\n");
return 0;
}
Output
1
1
1
1
0
Is it some issue regarding the pointer initialisation syntax?

Answer 1

struct list_el
{
int val;
struct list_el * right, * left, *parent;
}item_default={0,NULL,NULL,NULL}; //Default values
This does not do what you think it does. You've basically written...
typename typedefinition variable = initial_value;
You've declared the type struct list_el, defined it as { int val; struct list_el * right, * left, *parent; }, declared a new variable of that type called item_default, and assigned it the value {0,NULL,NULL,NULL}.
Aside from the type definition, this is basically int foo = 0.
We can test that by printing out the parts of item_default.
int main(int argc, char const *argv[])
{
printf("%d\n", item_default.val);
printf("%p\n", item_default.right);
printf("%p\n", item_default.left);
printf("%p\n", item_default.parent);
return 0;
}
And these will be 0, 0x0 (ie. NULL), 0x0, 0x0.
Unfortunately, C has no defaults for types. You always must initialize them. When working with structs, this typically means writing new and destroy functions so initialization and cleanup happens consistently.
// Declare the type and typedef in one go.
// I've changed the name from node to Node_t to avoid clashing
// with variable names.
typedef struct node
{
int val;
struct node *right, *left, *parent;
} Node_t;
// Any functions for working with the struct should be prefixed
// with the struct's name for clarity.
Node_t *Node_new() {
Node_t *node = malloc(sizeof(Node_t));
node->val = 0;
node->right = NULL;
node->left = NULL;
node->parent = NULL;
return node;
}
int main() {
Node_t *node = Node_new();
printf("%d\n", node->val);
printf("%p\n", node->right);
// and so on
return 0;
}
Note that I did not use calloc. calloc fills the memory with zeros, but the machine's representation of a null pointer is not necessarily zero. Using NULL and 0 are safe, the compiler can do the translation from context, but calloc doesn't know what you're going to do with the memory its zeroing. It's a relatively minor portability concern that's probably more of a problem these days for embedded systems.

Answer 2

Structure is a data type, you don't give default values to data types. What you're doing is like give an int the default of 3. What you want is give an instance of your struct a default value, but that is not possible in C.
You could have a function to help you with that:
void init_struct(node* nd) {
if (nd != NULL) {
nd->val = 0;
nd->parent = nd->right = nd->left = NULL;
}
}