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
