* Why does evaluating a piece of Elisp code seemingly not expand a macro?
@ 2016-01-15 10:08 Marcin Borkowski
2016-01-15 10:57 ` Oleh Krehel
2016-01-15 21:10 ` Samuel W. Flint
0 siblings, 2 replies; 9+ messages in thread
From: Marcin Borkowski @ 2016-01-15 10:08 UTC (permalink / raw)
To: Org-Mode mailing list
This piece of code:
#+BEGIN_SRC elisp :results value verbatim :exports both
(defmacro forty-two ()
(* 6 7))
(defun print-answer ()
(message "The answer is %s." (forty-two)))
(symbol-function 'print-answer)
#+END_SRC
yields this:
#+RESULTS:
: (lambda nil (message "The answer is %s." (forty-two)))
and not that:
: (lambda nil (message "The answer is %s." 42))
Why?
--
Marcin Borkowski
http://octd.wmi.amu.edu.pl/en/Marcin_Borkowski
Faculty of Mathematics and Computer Science
Adam Mickiewicz University
^ permalink raw reply [flat|nested] 9+ messages in thread
* Re: Why does evaluating a piece of Elisp code seemingly not expand a macro?
2016-01-15 10:08 Why does evaluating a piece of Elisp code seemingly not expand a macro? Marcin Borkowski
@ 2016-01-15 10:57 ` Oleh Krehel
2016-01-17 22:56 ` Marcin Borkowski
2016-01-15 21:10 ` Samuel W. Flint
1 sibling, 1 reply; 9+ messages in thread
From: Oleh Krehel @ 2016-01-15 10:57 UTC (permalink / raw)
To: Marcin Borkowski; +Cc: Org-Mode mailing list
Marcin Borkowski <mbork@mbork.pl> writes:
> Why?
Macro-expand the defun to get:
(defalias 'print-answer
#'(lambda nil
(message
"The answer is %s."
(forty-two))))
`lambda' is a macro that /quotes/ its body. Therefore, the body of
`defun' is not evaluated or expanded when it's defined.
You probably wanted something like this instead:
(macroexpand-all
'(lambda nil
(message
"The answer is %s."
(forty-two))))
;; =>
;; (function
;; (lambda nil
;; (message
;; "The answer is %s."
;; 42)))
Which could be wrapped in a new macro:
(defmacro defun-1 (name arglist &optional docstring &rest body)
(unless (stringp docstring)
(setq body
(if body
(cons docstring body)
docstring))
(setq docstring nil))
(list 'defun name arglist docstring (macroexpand-all body)))
The above seems to work, at least superficially:
(symbol-function
(defun-1 print-answer ()
(message "The answer is %s." (forty-two))))
;; =>
;; (lambda nil
;; (message
;; "The answer is %s."
;; 42))
By the way, it might be more appropriate to ask similar questions on
help-gnu-emacs@gnu.org.
Oleh
^ permalink raw reply [flat|nested] 9+ messages in thread
* Re: Why does evaluating a piece of Elisp code seemingly not expand a macro?
2016-01-15 10:08 Why does evaluating a piece of Elisp code seemingly not expand a macro? Marcin Borkowski
2016-01-15 10:57 ` Oleh Krehel
@ 2016-01-15 21:10 ` Samuel W. Flint
2016-01-15 22:06 ` Nick Dokos
2016-01-15 22:24 ` Marcin Borkowski
1 sibling, 2 replies; 9+ messages in thread
From: Samuel W. Flint @ 2016-01-15 21:10 UTC (permalink / raw)
To: Marcin Borkowski; +Cc: Org-Mode mailing list
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>>>>> Marcin Borkowski writes:
MB> This piece of code: #+BEGIN_SRC elisp :results value verbatim
MB> :exports both (defmacro forty-two () (* 6 7))
That is not a macro. That's a function. The return value of a macro
(the result of the last expression in the implicit progn) needs to be a
(quasi-)quoted expression.
This macro simply evaluates to 42. This should be a function.
If you want a macro, you could have:
#+BEGIN_SRC: emacs-lisp
(defmacro forty-two ()
'(* 6 7))
#+END_SRC
For what you want, you could have it be:
#+BEGIN_SRC: emacs-lisp
(defmacro forty-two ()
`,(* 6 7))
#+END_SRC
[...]
HTH,
Sam
--
Samuel W. Flint
4096R/266596F4
(9477 D23E 389E 40C5 2F10 DE19 68E5 318E 2665 96F4)
(λs.s s) λs.s s
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^ permalink raw reply [flat|nested] 9+ messages in thread
* Re: Why does evaluating a piece of Elisp code seemingly not expand a macro?
2016-01-15 21:10 ` Samuel W. Flint
@ 2016-01-15 22:06 ` Nick Dokos
2016-01-15 22:24 ` Marcin Borkowski
1 sibling, 0 replies; 9+ messages in thread
From: Nick Dokos @ 2016-01-15 22:06 UTC (permalink / raw)
To: emacs-orgmode
swflint@flintfam.org (Samuel W. Flint) writes:
>>>>>> Marcin Borkowski writes:
>
> MB> This piece of code: #+BEGIN_SRC elisp :results value verbatim
> MB> :exports both (defmacro forty-two () (* 6 7))
>
> That is not a macro. That's a function. The return value of a macro
> (the result of the last expression in the implicit progn) needs to be a
> (quasi-)quoted expression.
>
Not so.
> This macro simply evaluates to 42. This should be a function.
>
Maybe it should be a function, but it *is* a macro:
--8<---------------cut here---------------start------------->8---
(defmacro forty-two () (* 6 7))
==> forty-two
(symbol-function 'forty-two)
==> (macro lambda nil (* 6 7))
--8<---------------cut here---------------end--------------->8---
> If you want a macro, you could have:
>
> #+BEGIN_SRC: emacs-lisp
> (defmacro forty-two ()
> '(* 6 7))
> #+END_SRC
>
That's a different macro:
--8<---------------cut here---------------start------------->8---
(defmacro forty-two () '(* 6 7))
==> forty-two
(symbol-function 'forty-two)
==> (macro lambda nil (quote (* 6 7)))
--8<---------------cut here---------------end--------------->8---
--
Nick
^ permalink raw reply [flat|nested] 9+ messages in thread
* Re: Why does evaluating a piece of Elisp code seemingly not expand a macro?
2016-01-15 21:10 ` Samuel W. Flint
2016-01-15 22:06 ` Nick Dokos
@ 2016-01-15 22:24 ` Marcin Borkowski
1 sibling, 0 replies; 9+ messages in thread
From: Marcin Borkowski @ 2016-01-15 22:24 UTC (permalink / raw)
To: Samuel W. Flint; +Cc: Org-Mode mailing list
On 2016-01-15, at 22:10, Samuel W. Flint <swflint@flintfam.org> wrote:
>>>>>> Marcin Borkowski writes:
>
> MB> This piece of code: #+BEGIN_SRC elisp :results value verbatim
> MB> :exports both (defmacro forty-two () (* 6 7))
>
> That is not a macro. That's a function. The return value of a macro
> (the result of the last expression in the implicit progn) needs to be a
> (quasi-)quoted expression.
IIUC, a macro is a function - a function returning a Lisp form.
> This macro simply evaluates to 42. This should be a function.
Yes, I wanted it to evaluate to 42. I expected the constant 42 in the
code. (And this is what happens if I evaluate these forms outside an
Org block.)
> If you want a macro, you could have:
>
> #+BEGIN_SRC: emacs-lisp
> (defmacro forty-two ()
> '(* 6 7))
> #+END_SRC
>
> For what you want, you could have it be:
>
> #+BEGIN_SRC: emacs-lisp
> (defmacro forty-two ()
> `,(* 6 7))
> #+END_SRC
But this is _not_ what I want! What I want is to understand the
difference between just C-M-x'ing these forms and evaluating them in
Org-mode.
> Sam
Regards,
--
Marcin Borkowski
http://octd.wmi.amu.edu.pl/en/Marcin_Borkowski
Faculty of Mathematics and Computer Science
Adam Mickiewicz University
^ permalink raw reply [flat|nested] 9+ messages in thread
* Re: Why does evaluating a piece of Elisp code seemingly not expand a macro?
2016-01-15 10:57 ` Oleh Krehel
@ 2016-01-17 22:56 ` Marcin Borkowski
2016-01-18 13:54 ` Stefan Monnier
0 siblings, 1 reply; 9+ messages in thread
From: Marcin Borkowski @ 2016-01-17 22:56 UTC (permalink / raw)
To: Oleh Krehel; +Cc: Help Gnu Emacs mailing list, Org-Mode mailing list
On 2016-01-15, at 11:57, Oleh Krehel <ohwoeowho@gmail.com> wrote:
> Marcin Borkowski <mbork@mbork.pl> writes:
>
>> Why?
>
> Macro-expand the defun to get:
>
> (defalias 'print-answer
> #'(lambda nil
> (message
> "The answer is %s."
> (forty-two))))
>
> `lambda' is a macro that /quotes/ its body. Therefore, the body of
> `defun' is not evaluated or expanded when it's defined.
Interesting.
1. Why is lambda sharp-quoted? I remember reading (in Artur's blog)
that it shouldn't be.
2. I always thought that macros get expanded on compilation (or defining
the function). If I evaluate all forms I've written about outside Org
(using C-M-x, for instance), the `forty-two' macro seems to get
expanded.
In the manual (info "(elisp)Expansion"), I could find this:
--8<---------------cut here---------------start------------->8---
Note that Emacs tries to expand macros when loading an uncompiled
Lisp file. This is not always possible, but if it is, it speeds up
subsequent execution. *Note How Programs Do Loading::.
--8<---------------cut here---------------end--------------->8---
Does it mean that C-M-x is different than loading? Or C-x C-e, for that
matter? Is this covered by the manual? (If not, it might need
correcting.)
> You probably wanted something like this instead:
>
> (macroexpand-all
> '(lambda nil
> (message
> "The answer is %s."
> (forty-two))))
> ;; =>
> ;; (function
> ;; (lambda nil
> ;; (message
> ;; "The answer is %s."
> ;; 42)))
>
> Which could be wrapped in a new macro:
>
> (defmacro defun-1 (name arglist &optional docstring &rest body)
> (unless (stringp docstring)
> (setq body
> (if body
> (cons docstring body)
> docstring))
> (setq docstring nil))
> (list 'defun name arglist docstring (macroexpand-all body)))
>
> The above seems to work, at least superficially:
>
> (symbol-function
> (defun-1 print-answer ()
> (message "The answer is %s." (forty-two))))
> ;; =>
> ;; (lambda nil
> ;; (message
> ;; "The answer is %s."
> ;; 42))
Interesting, I will study this (but not today - it's 23:51 here, I'll
need sleep soon!)
> By the way, it might be more appropriate to ask similar questions on
> help-gnu-emacs@gnu.org.
I posted this reply there, too, though in view of what I wrote above
I still think this is Org-related.
> Oleh
Best,
--
Marcin Borkowski
http://octd.wmi.amu.edu.pl/en/Marcin_Borkowski
Faculty of Mathematics and Computer Science
Adam Mickiewicz University
^ permalink raw reply [flat|nested] 9+ messages in thread
* Re: Why does evaluating a piece of Elisp code seemingly not expand a macro?
2016-01-17 22:56 ` Marcin Borkowski
@ 2016-01-18 13:54 ` Stefan Monnier
2016-01-18 20:03 ` Marcin Borkowski
0 siblings, 1 reply; 9+ messages in thread
From: Stefan Monnier @ 2016-01-18 13:54 UTC (permalink / raw)
To: help-gnu-emacs; +Cc: emacs-orgmode
> Does it mean that C-M-x is different than loading?
Yes.
> Or C-x C-e, for that matter?
As well.
> Is this covered by the manual? (If not, it might need correcting.)
Not really. The basic idea is that macroexpansion can take place
*anytime* (tho, before the code is actually executed). If you care
about when expansion takes place you probably have a bug.
Stefan
^ permalink raw reply [flat|nested] 9+ messages in thread
* Re: Why does evaluating a piece of Elisp code seemingly not expand a macro?
2016-01-18 13:54 ` Stefan Monnier
@ 2016-01-18 20:03 ` Marcin Borkowski
2016-01-18 20:19 ` Stefan Monnier
0 siblings, 1 reply; 9+ messages in thread
From: Marcin Borkowski @ 2016-01-18 20:03 UTC (permalink / raw)
To: Stefan Monnier; +Cc: help-gnu-emacs, emacs-orgmode
On 2016-01-18, at 14:54, Stefan Monnier <monnier@iro.umontreal.ca> wrote:
>> Does it mean that C-M-x is different than loading?
>
> Yes.
>
>> Or C-x C-e, for that matter?
>
> As well.
>
>> Is this covered by the manual? (If not, it might need correcting.)
>
> Not really. The basic idea is that macroexpansion can take place
> *anytime* (tho, before the code is actually executed). If you care
> about when expansion takes place you probably have a bug.
Does that mean that it's possible that a function definition contains
unexpanded macros?
Does that mean that `symbol-function' will expand them?
Does that mean that if I define a macro, then a function using that
macro, and then change the definition of the macro, the behavior of the
function is undefined?
Sorry for so many questions, but I really want to understand this.
(Also, when that happens, I might send a patch for the manual.)
> Stefan
Best,
--
Marcin Borkowski
http://octd.wmi.amu.edu.pl/en/Marcin_Borkowski
Faculty of Mathematics and Computer Science
Adam Mickiewicz University
^ permalink raw reply [flat|nested] 9+ messages in thread
* Re: Why does evaluating a piece of Elisp code seemingly not expand a macro?
2016-01-18 20:03 ` Marcin Borkowski
@ 2016-01-18 20:19 ` Stefan Monnier
0 siblings, 0 replies; 9+ messages in thread
From: Stefan Monnier @ 2016-01-18 20:19 UTC (permalink / raw)
To: Marcin Borkowski; +Cc: help-gnu-emacs, emacs-orgmode
> Does that mean that it's possible that a function definition contains
> unexpanded macros?
Yes.
> Does that mean that `symbol-function' will expand them?
AFAIK it currently never happens there, but if your code relies on this
property it's probably got a bug.
> Does that mean that if I define a macro, then a function using that
> macro, and then change the definition of the macro, the behavior of the
> function is undefined?
Yes.
Stefan
^ permalink raw reply [flat|nested] 9+ messages in thread
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2016-01-15 10:08 Why does evaluating a piece of Elisp code seemingly not expand a macro? Marcin Borkowski
2016-01-15 10:57 ` Oleh Krehel
2016-01-17 22:56 ` Marcin Borkowski
2016-01-18 13:54 ` Stefan Monnier
2016-01-18 20:03 ` Marcin Borkowski
2016-01-18 20:19 ` Stefan Monnier
2016-01-15 21:10 ` Samuel W. Flint
2016-01-15 22:06 ` Nick Dokos
2016-01-15 22:24 ` Marcin Borkowski
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