[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]
Re: [SOLVED with `eval']: Why I cannot use this variable in macro call f
From: |
tomas |
Subject: |
Re: [SOLVED with `eval']: Why I cannot use this variable in macro call from function? |
Date: |
Wed, 9 Jun 2021 18:41:12 +0200 |
User-agent: |
Mutt/1.5.21 (2010-09-15) |
On Wed, Jun 09, 2021 at 05:39:25PM +0300, Jean Louis wrote:
> ;; -*- lexical-binding: t; -*-
> * tomas@tuxteam.de <tomas@tuxteam.de> [2021-06-09 14:35]:
> > On Wed, Jun 09, 2021 at 01:56:45PM +0300, Jean Louis wrote:
> > > * tomas@tuxteam.de <tomas@tuxteam.de> [2021-06-09 11:54]:
> > > > On Wed, Jun 09, 2021 at 11:22:38AM +0300, Jean Louis wrote:
> > > > > * tomas@tuxteam.de <tomas@tuxteam.de> [2021-06-09 10:40]:
> > > > > > You snipped the (for me) interesting part: did you notice how
> > > > > > `eval' jumps over the local declaration?
> > > > >
> > > > > Do you mean variables within `let'?
> > > >
> > > > Yes, it doesn't see them :)
> > >
> > > Maybe in theory it does not see, but in reality it does see it as
> > > `list' is evaluated before `eval', so the interned `rcd-symbol' and
> > > variable `description' they get evaluated before `eval'.
>
> > That sentence doesn't make any sense to me. It does or it doesn't.
>
> Well, I get confused too, you said that it does not see, but it is
> obvious that it does see.
>
> > I propose to you the next experiment:
> >
> > * experiment 3
> >
> > (let ()
> > (let ((x 42))
> > (eval '(progn (setq x 43) (message "in eval: x is %S" x)))
> > (message "inner let: x is %S" x))
> > (message "outer let: x is %S" x))
> >
> > (you might have to switch to the *Messages* buffer to see all three
> > messages).
>
> The outer scope does not see the inner scope.
>
> Then the buffer *Backtrace* jumps up:
>
> Debugger entered--Lisp error: (void-variable x)
> (message "outer let: x is %S" x)
> (let nil (let ((x 42)) (eval '(progn (setq x 43) (message "in eval: x is
> %S" x))) (message "inner let: x is %S" x)) (message "outer let: x is %S" x))
> eval((let nil (let ((x 42)) (eval '(progn (setq x 43) (message "in eval: x
> is %S" x))) (message "inner let: x is %S" x)) (message "outer let: x is %S"
> x)) nil)
> elisp--eval-last-sexp(nil)
> eval-last-sexp(nil)
> funcall-interactively(eval-last-sexp nil)
> call-interactively(eval-last-sexp nil nil)
> command-execute(eval-last-sexp)
Oh, that is interesting. For me (this is copied off the *Messages* buffer;
I started Emacs anew to make sure no global definition of x lingers
around; to make extra sure, I just eval'ed x before: it throws the
expected void-variable x):
in eval: x is 43
inner let: x is 42
outer let: x is 43
"outer let: x is 43"
This is what I actually expect: the setq whithin the eval does set x's
global (or buffer-local, if there is one) value. It doesn't touch the
lexical value, because it doesn't know about it.
> > What are the results? Do they correspond to your expectations? If
> > not, why not?
>
> I did not have any expectations for that piece of code and that
> one does not generate new global variables to return the symbol
> for history, which is what I need, and what is solved with `eval'
> nicely.
>
> Look here:
>
> (let ((x 42))
> (eval (list 'progn (setq x 43) (message "in eval: x is %S" x)))) ⇒ "in
> eval: x is 43"
It seems you have fun generating intentionally obfuscated code. Eval
is a function, so its argument is evaluated. Let's see...
(list ; a function
'progn ; (quote progn) evaluates to the symbol progn
(setq x 43) ; evaluates to 43 SIDE EFFECT: set x to 43 at some
global/buffer-local level
(message "...") ; evaluates to "..." SIDE EFFECT print "..." to *Messages*
What eval "sees" at the end is
(eval '(progn 43 "in eval: x is 43"))
which evaluates to "in eval: x is 43".
On the way to there, two side effects happened. The symbol x was bound to 43 in
some
(global or buffer-local) symbol table (obarray), probably "before" (message ...)
happens (I've assumed so when substituting-in the x there), and something got
printed to *Messages*. Go look there: you'll probably find two copies of
"in eval: x is 43". The one is the side effect, the other the p from the REPL.
> that obviously does work nicely as the list gets evaluated before eval
> receives it. At least I assume it is so, according to learning about
> the LISP in general.
If it works nicely, we're all in agreement.
I'll have to stop here. Other things demand my attention, and I just can't
keep up with your bandwidth.
So happy hacking!
Cheers
- t
signature.asc
Description: Digital signature
- Re: [SOLVED with `eval']: Why I cannot use this variable in macro call from function?, (continued)
- Re: [SOLVED with `eval']: Why I cannot use this variable in macro call from function?, tomas, 2021/06/09
- Re: [SOLVED with `eval']: Why I cannot use this variable in macro call from function?, Jean Louis, 2021/06/09
- Re: [SOLVED with `eval']: Why I cannot use this variable in macro call from function?, tomas, 2021/06/09
- Re: [SOLVED with `eval']: Why I cannot use this variable in macro call from function?, Jean Louis, 2021/06/09
- Re: [SOLVED with `eval']: Why I cannot use this variable in macro call from function?, tomas, 2021/06/09
- Re: [SOLVED with `eval']: Why I cannot use this variable in macro call from function?, Jean Louis, 2021/06/09
- Re: [SOLVED with `eval']: Why I cannot use this variable in macro call from function?, tomas, 2021/06/09
- Re: [SOLVED with `eval']: Why I cannot use this variable in macro call from function?, Jean Louis, 2021/06/09
- Re: [SOLVED with `eval']: Why I cannot use this variable in macro call from function?, tomas, 2021/06/09
- Re: [SOLVED with `eval']: Why I cannot use this variable in macro call from function?, Jean Louis, 2021/06/09
- Re: [SOLVED with `eval']: Why I cannot use this variable in macro call from function?,
tomas <=
- Re: [SOLVED with `eval']: Why I cannot use this variable in macro call from function?, Robert Thorpe, 2021/06/09
- Re: [SOLVED with `eval']: Why I cannot use this variable in macro call from function?, Jean Louis, 2021/06/10
- Re: [SOLVED with `eval']: Why I cannot use this variable in macro call from function?, Robert Thorpe, 2021/06/11
- Re: [SOLVED with `eval']: Why I cannot use this variable in macro call from function?, Jean Louis, 2021/06/11