From mboxrd@z Thu Jan  1 00:00:00 1970
From: Mike Pastore <mike@oobak.org>
Subject: Re: pointer notation
Date: Tue, 28 Oct 2003 18:40:50 +0000
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References: <20031028094205.GA4228@mrna.tn.nic.in>
	 <3F9E5A41.40204@ig.com.br>  <20031028174605.GA5517@mrna.tn.nic.in>
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To: km <km@mrna.tn.nic.in>
Cc: linux-c-programming@vger.kernel.org

Quite simply, s is a pointer to a pointer to a struct. You want to
dereference it once (with a prefix "*") before dereferencing it a second
time (with an infix "->"). Maybe the following pseudo-equivalent code
will help you understand:

    struct node **s, *t; /* introduce temporary "t" */
    t = *s; /* dereference s, store result in t */
    t->previous = NULL; /* set member of t (*s) to NULL */

The infix "->" binds tighter (has a higher precedence) than the prefix
"*". The incorrect statement "*s->previous = NULL;" is therefore
equivalent to "*(s->previous) = NULL;" and will attempt to get the
"previous" member of s and then dereference the result. But s has no
previous member! It is a pointer to a pointer to a struct, NOT a pointer
to a struct!

The use of parenthesis in "(*s)->previous = NULL;" establishes a clear
order of operations (dereference and THEN get member), which a couple
folks have already mentioned.

--
Mike Pastore
mike@oobak.org

On Tue, 2003-10-28 at 17:46, km wrote:
> 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
> > >
> > >-
> > >To unsubscribe from this list: send the line "unsubscribe 
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> > >
> > >
> > > 
> > >
> > 
> > -
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> 
> 
> 
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