NULL Pointer mystery in C

I have one problem with NULL pointer checking, i am creating one strcut pointer like this

struct node *root;

now i am calling one function and pass the address of this pointer to it as a parameter

insert(&root , key);

in this function i check if root is NULL or not?

struct node *
insert(struct node **n, int d)
{
if (!(*n))
{
if (!((*n) = malloc(sizeof(struct node))))
{
return NULL;
}
(*n)->left = (*n)->right = NULL;
(*n)->d = d;
return *n;
}
if (KEY(d) < KEY((*n)->d)) {
return insert(&(*n)->left, d);
}
if (KEY(d) > KEY((*n)->d)) {
return insert(&(*n)->right, d);
}
return NULL;
}

now in the very first call to this function, *n is not NULL and instead of going into if loop for memory allocation, tries to do recursion and program fails giving memory access error.
now if i haven't assigend anything to root, how come it is not NULL?

An equally valid question would be, "if I haven't assigned anything to root, how come it is not 20?"

It's the same thing, really. Until you actually assign a value to root, its value is undefined. And quite likely anthing but NULL.

you could always try passing the node normally, and then using "this" (without qoutes of course). this is a pointer to any data structure. it is automatically assigned to any data structure, even if it is user defined.


pass the node by reference and then try

if(this == NULL)

you cannot assume that ' struct node *root; ' will initialize root to NULL. since it is an uninitialized pointer you need to explicitly assign NULL to it before going fwd.

hope that helps.

m shah,

You might try a little switch in how you allocate memory for structures. When you allocate a block of memory, instead of using malloc() try calloc(), which initializes it's memory to 0.

I now use calloc exclusively, and I have a lot fewer crashes. After that you only need to make sure that you're getting the right size, and your code will be quite trouble-free.