Path: utzoo!utgpu!jarvis.csri.toronto.edu!mailrus!ames!sun-barr!apple!rutgers!att!westmark!mole-end!mat
From: mat@mole-end.UUCP (Mark A Terribile)
Newsgroups: comp.lang.c
Subject: Re: Pointers, Structs, and Arrays
Summary: *.[0], yes, of course ...
Message-ID: <181@mole-end.UUCP>
Date: 11 May 89 11:00:27 GMT
References: <743@mccc.UUCP>
Distribution: usa
Organization: mole-end--private system.  admin: mole-end!newtnews
Lines: 40

In article <743@mccc.UUCP>, pjh@mccc.UUCP (Pete Holsberg) writes:
> I ran across the following code and, although I've figured out what it does,
> I'm not sure how to explain the notation.
 
> struct foo  { char bar[20];  } list[100];

> init()
> {
> 	int i;
> 	for (i=0; i<100; ++i)
> 		*list[i].bar = '\0';
> }
 
> Now I know that the last line points to the first character in the bar array
> In each struct variable, but how?  If bar = &bar[0], how so the '*' and the
> '&' "get together" to make the last line equivalent to
> 		list[i].bar[0]

It's really simple.  Remember first that ``.'' and ``[]'' are what K&R-I call
``primary'' operators; they bind more tightly than anything else.

	list[ i ].bar

is an array of char.  Used as an expression, it is ``converted syntactically''
(it is taken to mean) a pointer-to-char pointing at the first element in the
array.  What array?  The array in the struct.  Ok, then why do we write

	*list[ i ].bar

instead of

	list[ i ].*bar

Because the other operators are primaries, and it is the *entire* expression
from ``list'' to ``bar'' that is the pointer-to-char.  The expression with
the ``*'' in the middle isn't legal C, though it can be legal C++.
-- 

(This man's opinions are his own.)
From mole-end				Mark Terribile