t stays in
memory for prolonged time, parallel to program running its thing).
Did I get it right?
On 20 April 2013 23:41, Tonino Jankov wrote:
> I mean, I think that *in both cases* the original sequence *at one point
> in time* must be, entirely realized, in memory.
>
> And if there i
, in its entirety present in memory, it means that
memory can handle the whole collection.
Maybe my questions sound a bit dubious, but anyway, I'm a bit sold out on
this lisp, so I want to get it right.
On 20 April 2013 23:33, Tonino Jankov wrote:
> Marko, you say "There is no doublin
Marko, you say "There is no doubling: *t* and *d* share the same underlying
lazy sequence and will refer to the same objects. The trouble is only that
you force the evaluation of *(count d)* while *(count t)* still waits to be
evaluated, so *t* must definitely stay bound to the head of the shared
s
