Re: pointer notation

[km <km@mrna.tn.nic.in>]

Hi all,
i should have posted the snippet before itself to be clear, which i am doing now.

It is from a double-linked list snippet as follows : 
----------------CODE START---------------------------------------
#include<stdio.h>
#include<stdlib.h>
struct node
{
int content;
struct node *next;
struct node *previous;
};
/* prototype declaration */
void add(struct node **s,int c);
/*main program*/
int main()
{
struct node *p;
p = NULL;
add(&p,21);
return 0;
}

/* function */
void add(struct node **s,int c)
{
	struct node *r,*q=*s;

if(*s==NULL)
{
	*s = (struct node *)malloc(sizeof(struct node));
	(*s)->previous = NULL; 		/* (*s) ??? whats that ? */
	(*s)->next = NULL;
	(*s)->content = c
}
else
{
/*bla bla*/
}
}
----------------CODE END-------------------------------------
In the above code we can see the statement called "(*s)->previous = NULL;". I didnt get what exactly it means to be and how different is *s different from (*s) in this context. can it be written in any other form alternatively ? 
	thanks for the feedback,
regards,
KM


-----------------------------------------------------------------
On Tue, Oct 28, 2003 at 10:00:01AM -0200, Luciano Moreira - igLnx wrote:
> Try to think that p is a MACRO, like as (arbitrary macro):
> "#define p  pBaseList+5*(x->element)+1"
> 
> When you do:
> "p->next"
> it ll be:
> "pBaseList+5*(x->element)+1->next"
> 
> the code above seems to be unsafe, so, if you do:
> "(p)->next"
> or:
> "#define p  (pBaseList+5*(x->element)+1)"
> it ll free or code of  "precedence bugs".
> 
> Luciano
> 
> km wrote, On 28/10/2003 07:42:
> 
> >Hi all,
> >how different is the notation 
> >(p)->next (what do the parenthesis around the pointer signify ? )
> >different from  
> >p ( while p is a pointer)
> >
> >regards,
> >KM
> >
> >-
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> >
> > 
> >
> 
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