CODEWITHSUNDEEP
cpointers
Human Cyborg Relations
Human Cyborg Relations

Reputation: 1244

Pointer to a pointer, how do I get values?

Lets say I have the following code,

typedef struct WordNode WordNode;
struct WordNode {
  int       freq;   // The frequency of the the word.
  char     *word;   // The word itself.
  WordNode *next;   // The next word node in the list.
};

struct WordSet {
  int       size;   // The number of elements in the set.
  WordNode *head;   // The starting node in the set.
};

After this, I've got some functions that initialize these structs. To prevent too much code in this post, I'm going to avoid posting those functions here.

Now, lets say I have the following,

  WordNode **p = wset->head; // can't change this line

Here, p is basically a pointer pointing to a pointer, correct?

And then, if I do this:

(*p) == NULL

This would return true if p is pointing to NULL, right?

Now, how would I get the word stored in wset->head?

Can I do this?

(*(*p))->word

And if I want p to point to the next WordNode, can I do this?

p = (*(*p))->next

I just want to know if all this is valid syntax so that I know I'm using pointers correctly.

Upvotes: 0

Views: 70

Answers (2)

L. Carlier
L. Carlier

Reputation: 305

Just for the sake of simplicity, whenever I have to deal with double pointer and I need the value I all the time go via an intermediate variable

e.g.

WordNode **ptr2ptr = wset->head;
WordNode *ptr = *ptr2Ptr;
WordNode value = *ptr;

or to get the head:

WordNode head = ptr->head;

Now I'm just answering the question which is, how to access the value of a pointer to pointer. You must be careful that your wset->head contains actually a pointer to pointer which is normally not the case in a linked list the way I understand you are trying to do. This is also not the way you have defined your head member...

Upvotes: 0

Psi
Psi

Reputation: 6783

Not really. (*(*p))->word is a total dereferenciation. So it would either be (*(*p)).word or (*p)->word.

You can imagine it that way, that the ->-Operator takes away one reference for you.

obj->field is the same as (*obj).field

Upvotes: 1

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