Re: cl-loop - do you understand it well?

[Michael Heerdegen <michael_heerdegen@web.de>]

Emanuel Berg <incal@dataswamp.org> writes:

> >   (cl-block nil
> >     (let* ((i 0)
> >            (funs nil))
> >       (while (<= i 1)
> >         (setq funs (nconc funs (list (lambda nil i))))
> >         (setq i (+ i 1)))
> >       (apply #'+ (mapcar #'funcall funs))))
> >
> No, where does the 4 come from?

There are two closures in FUNS, and both share the same lexical variable
i.  When the closures are `funcall'ed i is bound to 2, which is the
value i had when the loop is exited.  2*2=4.

The goal of this exercise was to demonstrate that closures share
modifiable environments instead of copies of values present at closure
creation time.

Michael.