Re: [9fans] thread

[Digby Tarvin <digbyt@acm.org>]

Hi,

That is a fascinating little subtlety which I hadn't appreciated
before.

I was aware that the array type was special in that the application of
the '&' operator to it resulted in a change of type but not of value.

But if I understand correctly, there is a converse situation with
the type '(*)[n]' (ie address of an array) where the application
of the unary '*' operator also produces a change of type but not value.

Or said another way, the address of an array of ints is a different type
to an array of ints, but for any given array the two values are identical:

main()
{
        int     a[1];
        int     (*la)[1];

        a[0] = 10;
        la = &a;
        printf("  a = %x\n", a);	/* array of ints */
        printf(" &a = %x\n", &a);	/* address of array of ints */
        printf(" la = %x\n", la);	/* address of array of ints */
        printf("*la = %x\n", *la);	/* array of ints */
}

Which produces (gcc on BSD - sorry, Plan9 is not at hand)
	  a = efbfd91c
	 &a = efbfd91c
	 la = efbfd91c
	*la = efbfd91c

Subtle, but at least it is symmetric.

Regards,
DigbyT

dmr@plan9.bell-labs.com:
> So far as I can determine, the Plan 9 C compiler is
> conformant here.
>
> If you have, say, an
>
> 	int A[10];
>
> then just A is the address of A and has type int *.
> &A is a pointer to this array of 10 integers; it will
> have the same value, as a pointer, but a different type:
>
>  int (*)[10]
>
	[snip]
>
> 	Dennis
--
Digby R. S. Tarvin                                              digbyt@acm.org
http://www.cthulhu.dircon.co.uk