Path: utzoo!utgpu!water!watmath!clyde!rutgers!sri-spam!ames!pasteur!ucbvax!hplabs!hpcea!hpfcdc!boemker
From: boemker@hpfcdc.HP.COM (Tim Boemker)
Newsgroups: comp.lang.c
Subject: Re: Question about type casting
Message-ID: <5080010@hpfcdc.HP.COM>
Date: 12 Jan 88 23:29:56 GMT
References: <5080009@hpfcdc.HP.COM>
Organization: HP Ft. Collins, Co.
Lines: 37

> In article <5080009@hpfcdc.HP.COM> boemker@hpfcdc.HP.COM (Tim Boemker) writes:
> -What's wrong with this program?
> -main()
> -{
> -	int *i;
> -	((int [1]) i) [0] = 1;
> -}
> 
> How can you turn a pointer into an array?  It makes no sense.
> Just use i[0] = 1; which is the same as *i = 1;.

Consider int A[1].  A is semantically an array, but the expression "A" is a
pointer to the base of that array.  The only differences between a pointer
and array are (1) when the storage is allocated and (2) how index calculations
are made.

Using i[i] instead of *i may be fine if the array in question is one
dimensional, but it doesn't work exactly the same way when the array
is multidimensional.  Consider int *A.  If I know that A points to
a 10x20 array, I might like to address elements as A[i][j].  I can
do the indexing math myself, as in *(A + i * 20 + j), but that's not
what the question is really about.  I want to know why a syntactically
correct and semantically meaningful expression is not allowed.

If you need further provocation, consider this program:

main()
{
  int A[1];

  ( (int [1]) A ) [0] = 0;
}

My C compiler doesn't like that construction, either, presumably for the
same reason that it didn't like the one in the first example.  But please
explain to me why it's illegal to cast an expression into it's natural
type!  Why can't I cast an array as an array?