Thread: Understanding pointers in C

I'm learning pointers in C and I'm developing a simple game for my class.

I'm confused where it says return a pointer to the new node. Do I need to "return *node"? Why can't I just "return node"?

Also, why is this procedure defined as "NODE *new_node(char *s)" and not "NODE new_node(char *s)" without the asterisk?

Code:

// This procedure should create a new NODE and copy
// the contents of string s into the 
// question_or_animal field.  It should also initialize
// the left and right fields to NULL.
// It should return a pointer to the new node


NODE *new_node(char *s)
{
  NODE new;
  strcpy(new->question_or_animal, s);
  new->left = NULL;
  new->right = NULL;


  return *new;
}

I should have mentioned that node is defined in a header file as

Code:

typedef struct node {
  char question_or_animal[200];
  struct node *left;
  struct node *right;
} NODE;

Does that change any of what you said?

This is all a little speculative, since you didn't provide a definition of NODE, but like Quzah said, it seems to not be a typedef'ed pointer (probably just a struct).

Originally Posted by monmon_4

I'm confused where it says return a pointer to the new node. Do I need to "return *node"? Why can't I just "return node"?

Because you have to return something of the right type, and a NODE and NODE * are not the same type. One is a thing, the other is a pointer to that thing.

Also, why is this procedure defined as "NODE *new_node(char *s)" and not "NODE new_node(char *s)" without the asterisk?

If they returned the thing pointed to, instead of the pointer (address), you would return a copy of the NODE. That isn't a huge problem, but it isn't terribly efficient. As for why you shouldn't return the address of a local variable, Quzah already said that goes away and becomes invalid when the function exits. Most likely, the new_node function is supposed to allocate memory and pass back the address of what it allocated. If you returned a NODE instead of a NODE *, you would return a copy of what you just allocated, and then you would lose the address of the allocated memory, which would cause a memory leak.

Code:

  return *new;

You are asking for "the thing pointed to by new", but new isn't a pointer, so you can't dereference it (that's what the * does in this context). That doesn't even make sense. What you might have been thinking of is the & (address of) operator. But that would, as has been explained, be returning the address of a local variable, which is a big no-no.

These may help your general pointer understanding:
Pointers in C - Tutorial - Cprogramming.com
Cprogramming.com FAQ > All about pointers
Eternally Confuzzled - All About Pointers

EDIT: I see you just posted the definition of NODE, and no, it doesn't change anything.

Originally Posted by monmon_4

I'm learning pointers in C and I'm developing a simple game for my class.

I'm confused where it says return a pointer to the new node. Do I need to "return *node"? Why can't I just "return node"?

You could but that's not what's been asked of you apparently.

Originally Posted by monmon_4

Also, why is this procedure defined as "NODE *new_node(char *s)" and not "NODE new_node(char *s)" without the asterisk?

The * means that new_node() returns a pointer to NODE instead of an object of type NODE.

Originally Posted by monmon_4

I should have mentioned that node is defined in a header file as

Code:

typedef struct node {
  char question_or_animal[200];
  struct node *left;
  struct node *right;
} NODE;

Does that change any of what you said?

Nope!

This mostly makes sense but doesn't 'NODE *n' have to point to something in your code? Shouldn't there an 'address-of' operator somewhere to denote where *n points to?

Following from the above question, when you declare 'n->left' and 'n->right', there is nothing for it to assign to left/right because n is only a NODE pointer and not a NODE variable and it doesn't point to anything. Even if it did, wouldn't you need to say '*n->left' and '*n->right' respectively in order to dereference the pointer first?

I'm probably wrong on most accounts here but I don't understand why.

Originally Posted by quzah

You can't do either correctly here, because 'new' is a local variable, and I'm assuming that 'NODE' is not a typedefed pointer. As such, actually even if it was a typedef, it is still only a local variable, which is destroyed when your function returns. What you really need to do is:

Code:

NODE *new_node( char *s )
{
    NODE *n = malloc( sizeof *n );
    if( n != NULL )
    {
        n->left = n->right = NULL;
        strcpy( n->question_or_animal, s ); // assuming qoa isn't just a pointer too
    }
    return n;
}

Quzah.