On Thu, 27 Oct 2011 13:53:02 -0400, Steve Teale
<[email protected]> wrote:
On Thu, 27 Oct 2011 13:09:43 -0400, Steven Schveighoffer wrote:
On Thu, 27 Oct 2011 12:58:57 -0400, Steve Teale
<[email protected]> wrote:
On Thu, 27 Oct 2011 16:16:26 +0200, Alex Rønne Petersen wrote:
On 27-10-2011 15:50, Steve Teale wrote:
Surely Variant.init should do the trick?
Dmitry,
Are you talking about the new std.variant - I don't see 'init'
currently.
Steve
He is probably referring to the 'init' property on the *type*, i.e.
int.init and so on.
- Alex
Same catch 22. In order to have an init property, the Variant would
have to have been set to some type, at which point hasValue() would say
yes.
Steve
import std.variant;
void main()
{
Variant v = 1;
assert(v.hasValue());
v = v.init;
assert(!v.hasValue());
}
-Steve
Exactly, and v has reverted to being typeless:
Sorry to jump in mid-stream, but you seemed to be suggesting Variant
didn't have an init without an assigned type.
Why wouldn't you just wrap variant if you want to introduce a nullable
variant type? Why does Variant have to be muddied with requirements for
database API? Database null is an entirely different animal from D's null.
-Steve
