shared const int i;
static if (is(typeof(i) T == shared U, U))
{
//Prints "shared(const(int))"
pragma(msg, U);
}
This seems like subtly wrong behaviour to me. If T == shared U,
for some U, then shouldn't U be unshared? If T is
shared(const(int)), and T is the same as the type U with the
'shared' qualifier applied to it, then U should be of type
const(int), not shared(const(int)).
I'm bringing this up partially because it seems wrong to me, and
partially because we currently don't have a good why of "shaving"
the outermost qualifier off a type, and this seemed the natural
way to do it to me (I was surprised when it didn't work).
