From mboxrd@z Thu Jan  1 00:00:00 1970
From: Oleh Krehel <ohwoeowho@gmail.com>
Subject: Re: Why does evaluating a piece of Elisp code seemingly not expand	a
 macro?
Date: Fri, 15 Jan 2016 11:57:17 +0100
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References: <87a8o7duj6.fsf@mbork.pl>
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In-Reply-To: <87a8o7duj6.fsf@mbork.pl> (Marcin Borkowski's message of "Fri, 15
	Jan 2016 11:08:45 +0100")
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To: Marcin Borkowski <mbork@mbork.pl>
Cc: Org-Mode mailing list <emacs-orgmode@gnu.org>

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