On Wednesday, 13 April 2016 at 11:36:07 UTC, Mithun Hunsur wrote:
Yeah, that also works; you have to define a symbol (if you don't have one you can already use) in order to get to it, so it's a little wasteful. Still useful to know, though!
No, it's not necessary. You should be able to walk the chain of __traits(parent) by starting with a local symbol, e.g. a lambda predicate.
Have a look at std.traits.moduleName which does most of the work.
