On 04/06/2012 12:23 PM, Walter Bright wrote:
On 4/6/2012 2:54 AM, Timon Gehr wrote:
Should add additional information to the type Foo. I don't see any
issues with
it, and not supporting it would be very strange.
How would:
@attr(foo) int x;
int y;
work? Are x and y the same type or not?
Yes, they are.
(But a future extension might leave this choice up to 'foo')
Now, consider:
auto c = b ? x : y;
What type does c have? int or @attr(foo)int ? And that's really just the
beginning. How about:
struct S(T) {
T t;
}
Instantiate it with S!int and S!(@attr(foo)int). Are those the same
instantiation, or different? If the same, does S.t have the attribute or
not?
There is no such thing as an @attr(foo) int, because @attr is not a type
constructor.
And, whatever you choose for the semantics, what is the sensible,
intuitive rule for it?
Custom attributes apply to the symbol declaration. Because any symbol is
declared just once, there is never any ambiguity. There is no conceptual
gap between
@attr(foo) int x;
and
@attr(bar) struct S{...}
Obviously @attr(bar) changes S, but there is only one declaration of S.
