__traits so long and ugly, what about ::?

[Ary Manzana]

I just hate it when you have to write too much to get simple things
done.

Does this type have a member? Current approach:

1. import std.traits;
2. invoke hasMember!(S, "m")

Another approach:

1. I have the type, let's ask it: S::hasMember("m")

Map's thought to code.

My problem is that when you start using D's cool features you end up
with a really hard to understand and obscure code...