On Thursday, 14 March 2013 at 21:29:26 UTC, Timon Gehr wrote:
int foo(int delegate() dg){ return dg(); }
void main(){
foo(delegate()=>2); // error, int delegate()pure immutable
// does not convert to int delegate()
}
That make sense. The rules have to be loosened.
Guarantees about const pure functions are lost. This is new.
(Assuming a sound implementation.)
Yes. This is still better than hole in the type system IMO.
It's just the combination of const and inout.
It behaves in the straightforward fashion.
inout is a wildcard standing for mutable, const or immutable.
inout is mutable => const(inout) = const
inout is const => const(inout) = const(const) = const
inout is immutable => const(inout) = const(immutable) =
immutable
Hence const(inout) can be seen as a wildcard standing for const
or immutable (but that's just derived information). Identifying
const(inout) with const, as DMD does it, gratuitously loses
type information.
I see. Type qualifier come at high cost, so I'm not sure.
