Path: utzoo!utgpu!water!watmath!clyde!rutgers!sri-unix!quintus!ok
From: ok@quintus.UUCP (Richard A. O'Keefe)
Newsgroups: comp.lang.c
Subject: Re: Arrays and structures
Message-ID: <552@cresswell.quintus.UUCP>
Date: 20 Jan 88 05:53:10 GMT
References: <22151@linus.UUCP>
Organization: Quintus Computer Systems, Mountain View, CA
Lines: 43

In article <22151@linus.UUCP>, jfjr@mitre-bedford.ARPA (Jerome Freedman) writes:
> Suppose I have the following
> typedef struct
>   { 
>     /* some irrelevant field declarations */
>    char string[80];
>    int  vector [4];
>   } structure_type;
> 
> Now, tell me about the string and vector fields of second structure.
> Since they are arrays then the fields are actually pointers.

Wrong.  Since they are arrays, the fields are actually ARRAYS.
When you *MENTION* an array-valued field, as in
	structure_type Foobaz;
	.... Foobaz.string ...
you get a pointer, just like if you mention ANY array you get a pointer.

If you declare
	structure_type destination, source;
	...
	destination = source;
then the contents of the two fields (80*sizeof (char) bytes of string[]
and 4*sizeof (int) bytes of vector[]) are moved.  No pointers are moved,
copied or changed, because there aren't any pointers.

> Is the situation different if I change the type declaration to
> ...
>    char *string;
>    int  *vector;

Yes.  If you declare arrays, you get arrays.  If you declare pointers
you get pointers.

The easiest way to answer questions like this is to type in a tiny
example and ask your C compiler.
	cc -S foobaz.c			-- UNIX
	/* look at foobaz.s */

	cc foobaz/list/machine_code	-- VMS
	/* look at foobaz.lst */

On other systems, you may have to ask a debugger.