Re: Returning variable "references" under lexical binding

[Barry Margolin]

In article <87bo84b85k.fsf@gmail.com>, Sean McAfee <eefacm@gmail.com> 
wrote:

> Barry Margolin <barmar@alum.mit.edu> writes:
> > But it's not the same lexical variable in your example. Each time you 
> > call start-my-timer you're creating a new closure over that variable. 
> > The caller has to save that closure somewhere, so that it can pass it to 
> > cancel-my-timer later. There's no functional difference between that and 
> > saving the timer itself.
> 
> Sure there is.  OK, forget about timers, let's go even simpler:
> 
> ;; -*- lexical-binding: t; -*-
> 
> (defun return-variable-n ()
>   (let ((n 1))
>     (in-five-seconds (lambda () (setq n 2)))
>     SOMETHING))
> 
> (setq foo (return-variable-n))
> 
> I want to return SOMETHING such that if I inspect it immediately, I'll
> get 1, but if I save it and inspect it after five seconds have passed,
> I'll get 2.  SOMETHING can't be just "n", because the function returns
> by value and foo would only ever contain 1.

This is different, because now you have something with state that can 
change, and want to be able to inspect. it.

> 
> Similarly, in my original code, I can't just return "timer", because the
> calling code would only ever see the very first timer created, and would
> not be able to see the new values for timer that the callbacks store
> later.

What "new values"? Your code didn't have any way to change the value of 
timer.  All it did was wrap it in a closure, but that closure doesn't do 
anything other than return the initial value. Did you leave something 
out of the original example?

-- 
Barry Margolin, barmar@alum.mit.edu
Arlington, MA
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