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From: jrv@siemens.UUCP
Newsgroups: net.lang.c
Subject: Re: Re: Pointers to arrays in C
Message-ID: <32700004@siemens.UUCP>
Date: Mon, 7-Apr-86 10:36:00 EST
Article-I.D.: siemens.32700004
Posted: Mon Apr  7 10:36:00 1986
Date-Received: Wed, 9-Apr-86 23:34:26 EST
References: <2389@brl-smoke.UUCP>
Lines: 57
Nf-ID: #R:brl-smoke:-238900:siemens:32700004:000:1482
Nf-From: siemens!jrv    Apr  7 10:36:00 1986




>It *is* a pointer to an array.  Try running this program:
>
>    main()
>    {
>	char (*x)[10];
>	printf("%d\n%d\n%d\n", sizeof(*x), sizeof(x), sizeof((*x)[2]));
>    }
>
>It prints (on a VAX running ULTRIX):
>
>10
>4
>1
>
>So, *x is an object of size 10 (the array), x is an object of size 4 (a
>pointer), and (*x)[2] is a single byte (a char).  C's strange type
>syntax strikes again.
>
>Paul Dietz
>dietz@slb-doll.csnet
>/* End of text from siemens:net.lang.c */


When I learned C an instructor showed me a trick to use to keep straight
in one's mind what a particular data construction was. To find out what
data type something is go back to the original declaration of it cover over
what you have in the expression and what is left is the data type.
For example using the program above.

sizeof(*x)
	char (*x)[10];
cover        ^^^^        and you get an array of 10 characters


sizeof(x)
	char (*x)[10];
cover          ^         and you get a pointer to an array of 10 characters


sizeof((*x)[2])
	char (*x)[10];
cover        ^^^^^^^^	and you get a character


So by this "convention" C's syntax is consistent and the sizeof() operator
worked as I would expect. Admittedly this scheme works easily on simple
data types (hey, it was an intro class :-)) and there are some expressions
which I have seen in net.lang.c on which I did not even attempt it.


Jim Vallino
Siemens Research and Technology Lab.
Princeton, NJ
{allegra,ihnp4,seismo,philabs}!princeton!siemens!jrv