Re: Lexical binding and macros.

Re: Lexical binding and macros.

[Barry Margolin]

In article <mailman.40.1292017369.4804.help-gnu-emacs@gnu.org>,
 Alin Soare <alinsoar@voila.fr> wrote:

> Are the macros and closures equivalent ?

No.  Emacs Lisp has macros, it doesn't have closures.  But in dialects 
of Lisp that have both, they're completely different.

> The set of macros is greater than the set of  closures?
> 
> The set of closures is greater than the set of macros?

These questions don't even make sense.

-- 
Barry Margolin, barmar@alum.mit.edu
Arlington, MA
*** PLEASE post questions in newsgroups, not directly to me ***
*** PLEASE don't copy me on replies, I'll read them in the group ***

Re: Lexical binding and macros.

[LanX]

On 12 Dez., 05:10, Barry Margolin <bar...@alum.mit.edu> wrote:
> In article <mailman.40.1292017369.4804.help-gnu-em...@gnu.org>,
>  Alin Soare <alins...@voila.fr> wrote:
>
> > Are the macros and closures equivalent ?
>
> No.  Emacs Lisp has macros, it doesn't have closures.  But in dialects
> of Lisp that have both, they're completely different.


ehm ... isn't "lexical-let" from the common lisp extension supposed to
support closures?

Re: Lexical binding and macros.

[Pascal J. Bourguignon]

LanX <lanx.perl@googlemail.com> writes:

> On 12 Dez., 05:10, Barry Margolin <bar...@alum.mit.edu> wrote:
>> In article <mailman.40.1292017369.4804.help-gnu-em...@gnu.org>,
>>  Alin Soare <alins...@voila.fr> wrote:
>>
>> > Are the macros and closures equivalent ?
>>
>> No.  Emacs Lisp has macros, it doesn't have closures.  But in dialects
>> of Lisp that have both, they're completely different.
>
>
> ehm ... isn't "lexical-let" from the common lisp extension supposed to
> support closures?

They're not really closures.   Closures are equivalent to objects, so
you can simulate them by creating a kind of object, and this is what
lexical-let does.  But given emacs lisp VM, it cannot do it quite
efficiently.

-- 
__Pascal Bourguignon__                     http://www.informatimago.com/
A bad day in () is better than a good day in {}.

Re: Lexical binding and macros.

[Stefan Monnier]

>> ehm ... isn't "lexical-let" from the common lisp extension supposed to
>> support closures?
> They're not really closures.

Actually, they are closures.  Admittedly, they're not as efficient as
one might like, but other than that, they work very well, thank you.


        Stefan

Re: Lexical binding and macros.

[Elena]

On Dec 15, 4:24 am, Stefan Monnier <monn...@iro.umontreal.ca> wrote:
> >> ehm ... isn't "lexical-let" from the common lisp extension supposed to
> >> support closures?
> > They're not really closures.
>
> Actually, they are closures.  Admittedly, they're not as efficient as
> one might like, but other than that, they work very well, thank you.
>
>         Stefan

Do they still leak memory?  From http://c2.com/cgi/wiki?EmacsLisp :

"Note that variables bound with lexical-let are never released, even
if they are never used. Try

 (loop for i from 1 to 100000 collect (lexical-let ((x i)) '()))

and watch it eat memory. So making infinity (ZeroOneInfinity) lexical
variables is out of the question except for very small values of
infinity."

Re: Lexical binding and macros.

[Pascal J. Bourguignon]

Elena <egarrulo@gmail.com> writes:

> On Dec 15, 4:24 am, Stefan Monnier <monn...@iro.umontreal.ca> wrote:
>> >> ehm ... isn't "lexical-let" from the common lisp extension supposed to
>> >> support closures?
>> > They're not really closures.
>>
>> Actually, they are closures.  Admittedly, they're not as efficient as
>> one might like, but other than that, they work very well, thank you.
>>
>>         Stefan
>
> Do they still leak memory?  From http://c2.com/cgi/wiki?EmacsLisp :
>
> "Note that variables bound with lexical-let are never released, even
> if they are never used. Try
>
>  (loop for i from 1 to 100000 collect (lexical-let ((x i)) '()))
>
> and watch it eat memory.

Just wait till the garbage collector kicks in.
The variables are named by un-interned symbols.  The should be garbage
collected as soon as the 'closures' using them are collected.  In your
example, that's immediately, since no closure is created.

-- 
__Pascal Bourguignon__                     http://www.informatimago.com/
A bad day in () is better than a good day in {}.

Re: Lexical binding and macros.

[Stefan Monnier]

> Do they still leak memory?  From http://c2.com/cgi/wiki?EmacsLisp :

> "Note that variables bound with lexical-let are never released, even
> if they are never used. Try

>  (loop for i from 1 to 100000 collect (lexical-let ((x i)) '()))

> and watch it eat memory. So making infinity (ZeroOneInfinity) lexical
> variables is out of the question except for very small values of
> infinity."

Additionally to what Pascal already explained, I'll add that,
lexical-let, like `loop' are relatively heavy macros, so you definitely
don't want to run them interpreted (where the macro is re-expanded each
time).  If you byte-compile the code above, the memory use will
be identical to what you get with

  (loop for i from 1 to 100000 collect (let ((x i)) '()))

  
        Stefan

Re: Lexical binding and macros.

[David Kastrup]

Stefan Monnier <monnier@iro.umontreal.ca> writes:

>> Do they still leak memory?  From http://c2.com/cgi/wiki?EmacsLisp :
>
>> "Note that variables bound with lexical-let are never released, even
>> if they are never used. Try
>
>>  (loop for i from 1 to 100000 collect (lexical-let ((x i)) '()))
>
>> and watch it eat memory. So making infinity (ZeroOneInfinity) lexical
>> variables is out of the question except for very small values of
>> infinity."
>
> Additionally to what Pascal already explained, I'll add that,
> lexical-let, like `loop' are relatively heavy macros, so you definitely
> don't want to run them interpreted (where the macro is re-expanded each
> time).

Why would they be reexpanded each time?  They are macros.  Their
expansion is done once and merely evalled each time.

Or do I misunderstand something here?

-- 
David Kastrup

Re: Lexical binding and macros.

[Pascal J. Bourguignon]

David Kastrup <dak@gnu.org> writes:

> Stefan Monnier <monnier@iro.umontreal.ca> writes:
>
>>> Do they still leak memory?  From http://c2.com/cgi/wiki?EmacsLisp :
>>
>>> "Note that variables bound with lexical-let are never released, even
>>> if they are never used. Try
>>
>>>  (loop for i from 1 to 100000 collect (lexical-let ((x i)) '()))
>>
>>> and watch it eat memory. So making infinity (ZeroOneInfinity) lexical
>>> variables is out of the question except for very small values of
>>> infinity."
>>
>> Additionally to what Pascal already explained, I'll add that,
>> lexical-let, like `loop' are relatively heavy macros, so you definitely
>> don't want to run them interpreted (where the macro is re-expanded each
>> time).
>
> Why would they be reexpanded each time?  They are macros.  Their
> expansion is done once and merely evalled each time.
>
> Or do I misunderstand something here?


In the case of Common Lisp, implementations providing an interpreter are
allowed to macro expand everytime.  Most implementations do cache the
expansion however.  (A lot of CL implementations just compile everything
automatically anyways).

For example, clisp performs the macroexpansions when defining the
function, and one time again when compiling it:


CL-USER> (defmacro m (x)
           (format t "(m ~S) is expanded" x)
           `(list ',x ',x))
M
CL-USER> (defun f ()
           (m 1))
(m 1) is expanded
F
CL-USER> (list (f) (f) (f))

((1 1) (1 1) (1 1))
CL-USER> (compile 'f)
(m 1) is expanded
F
NIL
NIL
CL-USER> 


In the case of emacs 23.2.1,

    (defmacro m (x)
      (message "(m %S) is expanded" x)
      `(list ',x ',x))

    (defun f ()
      (m 1))

    (list (f) (f) (f)) --> ((1 1) (1 1) (1 1))

produces in *Messages*:

    (m 1) is expanded [3 times]

So it looks like the emacs lisp interpreter doesn't not cache the macro
expansions.

emacs lisp expands it once more when compiling the function too:
(byte-compile 'f) --> #[nil "\300\211D\207" [1] 2]
*Messages*: (m 1) is expanded

-- 
__Pascal Bourguignon__                     http://www.informatimago.com/
A bad day in () is better than a good day in {}.

Re: Lexical binding and macros.

[Elena]

On Dec 15, 5:16 pm, David Kastrup <d...@gnu.org> wrote:
> Stefan Monnier <monn...@iro.umontreal.ca> writes:
> >> Do they still leak memory?  Fromhttp://c2.com/cgi/wiki?EmacsLisp:
>
> >> "Note that variables bound with lexical-let are never released, even
> >> if they are never used. Try
>
> >>  (loop for i from 1 to 100000 collect (lexical-let ((x i)) '()))
>
> >> and watch it eat memory. So making infinity (ZeroOneInfinity) lexical
> >> variables is out of the question except for very small values of
> >> infinity."
>
> > Additionally to what Pascal already explained, I'll add that,
> > lexical-let, like `loop' are relatively heavy macros, so you definitely
> > don't want to run them interpreted (where the macro is re-expanded each
> > time).
>
> Why would they be reexpanded each time?  They are macros.  Their
> expansion is done once and merely evalled each time.

Macros expansion can change if such macros depend on some other macros
which have been redefined after last expansion.

Re: Lexical binding and macros.

[Stefan Monnier]

>> Additionally to what Pascal already explained, I'll add that,
>> lexical-let, like `loop' are relatively heavy macros, so you
>> definitely don't want to run them interpreted (where the macro is
>> re-expanded each time).
> Why would they be reexpanded each time?

Because it's easier to implement it that way in a pure interpreter.

> They are macros.  Their expansion is done once and merely evalled
> each time.
> Or do I misunderstand something here?

Their expansion *can be* done once (which is what happens when you
byte-compile), but the interpreter doesn't bother with
such optimizations.


        Stefan

Lexical binding and macros.

[Alin Soare]

Are the macros and closures equivalent ?

The set of macros is greater than the set of  closures?

The set of closures is greater than the set of macros?


Alin