"Ali Cehreli" <acehreli@yahoo.com> wrote in message
news:pan.2004.07.21.15.41.08.653476.18901@yahoo.co m...[color=blue]
> On Wed, 21 Jul 2004 15:23:30 -0700, Howard wrote:
>[color=green]
> > Given that an array is not a type in the first place,[/color]
>
> An array of a specific size is a type.
>[color=green]
> > I don't see how
> > you can create a reference to it. You can't even create a pointer to an
> > array (just a pointer to its first element).[/color]
>
> These are incorrect. You can have references and pointers to arrays
> as well (not only to their first elements). The following program
> demonstrates this:
>
> #include <iostream>
>
> size_t const arraySize = 10;
>
> typedef int Array[arraySize];
>
> // Function taking reference-to-array
> void fill(Array & array, int value)
> {
> for (size_t i = 0; i != arraySize; ++i)
> {
> array[i] = value;
> }
> }
>
> // Function taking const-reference-to-array
> void printMiddle(Array const & array)
> {
> std::cout << array[arraySize / 2] << '\n';
> }
>
> int main()
> {
> Array one = { 0 };
> fill(one, 1);
> Array two = { 0 };
> fill(two, 2);
>
> // Pointer to array (not to the first element!)
> Array * p = 0;
>
> p = &one;
>
> // Dereferencing pointer-to-array
> printMiddle(*p);
>
> p = &two;
> printMiddle(*p);
> }
>
> Ali[/color]
Well, now you've *created* a type (Array), which *can* be pointed to or
referenced. And the only reason you can do it at all is because you use a
compile-time constant to declare the array as a fixed size. That allows the
compiler to know how much memory the new type requires. This won't work for
a variable-length array passed as a parameter, because there is no type, and
you can't declare it as a type (because it's variable-length). (So, maybe I
should have specified that more clearly originally, but I thought it was
obvious in the context of the conversation my post was in.)
-Howard
