extern char *foo vs. extern char foo[]

[r91400 at memqa.uucp]

In article <1990Jun13.030910.11593 at sharkey.cc.umich.edu%sevihc>, reso%sevihc at sharkey.cc.umich.edu (Dennis Reso) writes:
> I found it much easier to deal with pointers and "arrays" when I
> stopped looking at "[]" as meaning "array", and treated it like
> the postfix "offset from pointer" operator. Thus the notation
> "int a[5][10]" rather than "int a[5,10]", where "[10]" operates
> on the expression "a[5]" which yeilds a pointer. Think of it as
> just another shorthand notation:
> 
>                 "a[5][10]"      == "*(*(a+5) + 10)"
>             as  "record->field" == "(*record).field"

I thought that, in this case, a[5] does not yield a pointer, but rather
the fifth element in the array.  Is that not why people prefer sometimes
to create their arrays dynamically, so that a[5] WOULD return a pointer?

...something like
	int *a[6],b[120];
	a[0]=&b[0]; a[1]=&b[20]; a[2]=&b[40]; 
	a[3]=&b[60]; a[4]=&b[80]; a[5]=&b[100];
Indeed, I wouldn't do it quite this way (it may not even be syntactically
correct), but it is an illustration of what I mean.  This construction
does support the syntax a[5][10], and a[5] does return a pointer, and
it saves on a multiplication.  Simply declaring int a[6][20] would not
do this.

So, while the illustration he gave is insightful, C doesn't do things
that way unless you tell it to.

Michael C. Grant
Motorola Inc., Austin, Texas
(Motorola does not care what I think.)