std.Algebraic alias this

[Radu via Digitalmars-d-learn]

There is a weird rule on how compiler treats alias this for the N 
and S types bellow.

As you can see, somehow S losses it's type and ends up a plain 
string withing the Algebraic type. I would expect that all types 
should be the treated the same, why a string alias will be 
different then a bool or a double?

int main(string[] argv)
{
	import std.variant	: Algebraic;

	struct N { double val; alias val this; }
	struct S { string val; alias val this; }
	alias T = Algebraic!(N, S);
	pragma(msg, T.AllowedTypes); // prints (N, string)??

	T t = N(1.0);
	assert(t.get!N == 1.0); // works
	// t = 0.4; // fails, OK, variant.d(586): Error: static assert  
"Cannot store a double in a VariantN!(8u, N, S). Valid types are 
(N, string)"
	// t = S("foo"); // this fails, why? Error: static assert  
"Cannot store a S in a VariantN!(8u, N, S). Valid types are (N, 
string)"
	t = "bar"; // this works... why?
	// assert(t.get!S == "bar"); //this fails Variant: attempting to 
use incompatible types immutable(char)[] and main.main.S at 
std\variant.d(1475)
	assert(t.get!string == "bar"); // works, why?
	
	return 0;
}