Path: utzoo!attcan!uunet!seismo!sundc!pitstop!sun!quintus!ok
From: ok@quintus.uucp (Richard A. O'Keefe)
Newsgroups: comp.lang.fortran
Subject: Re: RE:  An array by any other name. . .
Message-ID: <415@quintus.UUCP>
Date: 16 Sep 88 10:16:00 GMT
References: <994@amelia.nas.nasa.gov> <3641@lanl.gov>
Sender: news@quintus.UUCP
Reply-To: ok@quintus.UUCP (Richard A. O'Keefe)
Organization: Quintus Computer Systems, Inc.
Lines: 29

In article <3641@lanl.gov> jlg@lanl.gov (Jim Giles) writes:
>Furthermore, in C x[i][j] is _supposed_ to be semantically equivalent
>to *(*(x+j)+i).  Statically allocated 2-d arrays aren't implemented
>this way!  Many versions of dynamically allocated 2-d arrays are.

x[i][j] *is* equivalent to *(*(x+i)+j).
Suppose we have
	double x[20][30];
then	x[i], or equivalently *(x+i) would be of type
	array 30 of double.
This expression is immediately converted to a pointer to its first
element (which thus is of type "pointer to double"), and
	x[i][j] = *(( (double*)& *(x+i) )+j)
I just checked with a UNIX C compiler, and given the declaration above,
x[i][j] and *(*(x+i)+j) yielded the same compiled code.

>In order to get a dynamically allocated to behave semantically like
>a statically allocated one I have to use a _different_ declaration
>syntax.

Nope.	double (*y)[20][30];
	y = (double (*)[20][30]) malloc(sizeof y);
"cdecl" explains this as
	cast malloc(sizeof y) into pointer to array 20 of array 30 of double
Now use it as
	(*y)[i][j]

The only difference is that instead of writing "x" you write "(*y)", and
you do that consistently in the declaration and the uses.