On Tuesday, 16 August 2016 at 19:23:51 UTC, Engine Machine wrote:
On Tuesday, 16 August 2016 at 17:39:14 UTC, Lodovico Giaretta
wrote:
On Monday, 15 August 2016 at 19:31:14 UTC, Engine Machine
wrote:
[...]
I don't know if this is exactly what you want:
=====================================
import std.traits: hasMember;
struct Q(T)
if (hasMember!(T, "x"))
{
T s;
@property auto x() { return s.x; }
}
auto makeQ(T)(auto ref T val)
{
return Q!T(val);
}
auto s = myTypeWithFieldX();
auto q = makeQ(s);
assert(q.x == s.x);
=====================================
`Q` can store any type with an `x` field and gives access to
it. The auxiliary function `makeQ` acts as a constructor for
`Q` with template parameter deduction.
I realize now that D simply cannot do what I ask directly
because It's type parameters are part of the type. It can't be
told that in some cases they are not. Any use of the
parameters, as I am using them, will result in this issue.
I'm now trying a different method which builds the based types
using partial oop and partial CT code. The CT code is only for
performance and convenience anyways.
You are right in that it is not possible to strip part of the
type informations.
The only problem is Q is still defined by T. Your makeQ then
requires the type implicitly, which I don't necessarily have.
You've simply added complexity to the problem but the issue is
still there.
How is it possible that you don't have T? Your variable will
surely have a type! Can you provide more background? Because
maybe I'm not completely getting your point here...
