An issue with lazy delegates

[Andrej Mitrovic]

import std.stdio;

void test(T)(lazy T dg)
{
    test2(dg);
}

void test2(T)(lazy T dg)
{
    dg();    // nothing happens
    dg()();  // have to use double-invocation instead
}

void main()
{
    test({ writeln("test"); });
}

Do you think it would be possible for the compiler to avoid wrapping
delegates into another delegate? I'm having this problem with this
sort of template:

import std.conv;
import std.string;

auto onFailThrow(E, T)(lazy T dg)
{
    try
    {
        static if (is(ReturnType!T == void))
            dg();
        else
            return dg();
    }
    catch (Exception ex)
    {
        throw new E(ex.msg);
    }
}

void main()
{
    string x = "x";
    string y = "y";
    onFailThrow!Exception({ to!int(x); });
    onFailThrow!Exception(to!int(y));
}

The first call doesn't do anything because the delegate is wrapped
inside of another delegate. I want this template to be versatile
enough to be used by both lazy expressions and delegate literals, but
I don't know how.

If I write the same template but with "lazy" stripped off I'll have
conflicting templates, but I don't know how I would write constraints
to separate the two. Any ideas?