Confusion about typedef and pointers in linked lists

Hi everybody!

I was trying to do some linked lists exercises, and I'd like to ask you what is the difference between the following two prototypes?

Code:

void insert(ListNodePtr sPtr, char value);
void insert(ListNodePtr *sPtr, char value);

Is one of this wrong? The meaning is the same or the second one means that sPtr is a pointer to a second pointer?

These were the definitions of the linked list components:

Code:

typedef struct listNode ListNode;
typedef ListNode *ListNodePtr;

struct listNode {
 char data;
 ListNodePtr nextPtr;
};

Thanks,

Diego

For nodes, you'd want to use listNode *sPtr as the parameter. Pointers are many times more flexible, though I am a bit less experienced than I want to be, so I can't really explain exactly why this is better.

Code:

void insert(listNode *sPtr, char value);

and if you want a list that can be connected, you want a pointer to another listnode inside the struct definition.

Code:

struct listNode {

 char data;

 listNode *nextPtr;

};

But having used typedef, ListNodePtr is already a pointer, so, shy should I use ListNodePtr *sPtr instead of ListNodePtr sPtr? I suppose anyway that the firs version is much better (the one you suggested), since in the Deitel and Deitel C how to program, the examples use that version....

Thanks,
Diego

Originally Posted by mrdiegus

I was trying to do some linked lists exercises, and I'd like to ask you what is the difference between the following two prototypes?

Code:

void insert(ListNodePtr sPtr, char value);
void insert(ListNodePtr *sPtr, char value);

Is one of this wrong? The meaning is the same or the second one means that sPtr is a pointer to a second pointer?

There is a big difference.
The first version takes a pointer to ListNode. If you change the pointer inside the function (so it points to somewhere else, i.e. sPtr = xxxx) the change won't be visible to the caller. For example if you have this code:

Code:

ListNodePtr head = NULL;
printf("%p\n", (void *) head);
insert(head, 'a');
printf("%p\n", (void *) head);

"head" will still point to NULL after calling the function no matter how you change "sPtr" within insert().

In your second version, you pass a pointer to pointer to ListNode, thus changes to the pointer to ListNode (*sPtr = xxxx) will be visible to the caller.

You wouldn't say

Code:

void func(int value);
void func(int *value);

are the same, would you? It's the same when you use pointers in function calls. Arguments are always passed by value in C.

Bye, Andreas

Andreas, thanks a lot.

Following the reasoning, are the following equivalent?

Code:

void insert(ListNodePtr *sPtr, char value); // given the typedef written before
void insert(ListNode **sPtr, char value); // replacing typedef with original ones

Thanks again,

Diego

Yes, they are . But there are some reasons for you not to typedef the pointers

- What do you mean by saying some reasons?I wouldn't be convinced and nor should you.
Take a look at my thread here .The problem was that i couldn't see because of typedef that ListPtr was actually a point..My eyes expected to see an asterisk..And so i allocated memory for a pointer, not for where the pointer points to ..

Std10093,

thanks, finally I've understood the matter... I agree with you, the two equivalent prototypes are the evidence, since the latter is much more readable!

Thanks,
Diego