[code]
void foo(int[int] foobar) {
  foobar[1] = 1;
}
void main() {
  int[int] bar;
  foo(bar);
  printf("%d", bar.length);
}
[/code]

If assoc. array 'bar' is not initialized, then:
If 'foobar' is 'in', program outputs '0', i.e. all changes in array are discarded.
If 'foobar' is 'out' or 'inout', program outputs '1'.

If 'bar' is initialized (it can be empty or not), with 'in', 'inout', 'out' program output is '1'.

It's a similar issue as with dynamic arrays. Inserting elements into one involves reallocating the storage for it, and since 'in' uses value semantics, it's working on a copy, not the original.

"Sark7" <sark7@mail333.com> wrote in message news:opr7dyu6bkut8jae@news.digitalmars.com...
> [code]
> void foo(int[int] foobar) {
>    foobar[1] = 1;
> }
> void main() {
>    int[int] bar;
>    foo(bar);
>    printf("%d", bar.length);
> }
> [/code]
>
> If assoc. array 'bar' is not initialized, then:
> If 'foobar' is 'in', program outputs '0', i.e. all changes in array are
> discarded.
> If 'foobar' is 'out' or 'inout', program outputs '1'.
>
> If 'bar' is initialized (it can be empty or not), with 'in', 'inout',
> 'out' program output is '1'.