type of a variable as a mixin-able string?

[Timothee Cour via Digitalmars-d-learn]

Suppose I have:

import std.range;
auto a=iota(complex_expr_returning_3());

I'd like to have a function/trait/template/compiler magic that takes
variable a and generates a string that can be mixed in to represent the
type of a. The difficulty is that typeid(a).to!string doesn't work for
Voldermort types:
typeid(a).to!string produces "std.range.iota!(int, int).iota.Result", which
is not very useful as it can't be mixed in a program, being a voldemort
type, eg:
mixin("std.range.iota!(int, int).iota.Result a;"); //compile error

However, the compiler could help reduce it to something as simple as
possible, eg:
"typeof(iota(int.init))"
here this would work:

mixin("typeof(iota(int.init)) a;"); //works

Is it possible to do that in a generic way, such that it hides as much as
possible the details of the expression (here, "complex_expr_returning_3()"
should be simplified to int.init)

Or is there another trick I could use to instantiate a variable with same
type as a?


NOTE:
typeof(a) a2;
works even though
http://www.drdobbs.com/cpp/voldemort-types-in-d/232901591?pgno=2 says it
won't work, "Sorry, that won't work, the compiler will not allow a
Voldemort Type to be instantiated outside of its scope "
(although that's doesn't help with my original problem)
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