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From: manuel@cui.UUCP
Newsgroups: net.lang.lisp
Subject: programming trick
Message-ID: <136@cui.UUCP>
Date: Fri, 28-Feb-86 17:43:28 EST
Article-I.D.: cui.136
Posted: Fri Feb 28 17:43:28 1986
Date-Received: Sun, 2-Mar-86 07:39:51 EST
Reply-To: manuel@cui.UUCP (James Stewart)
Organization: University of Geneva/Switzerland
Lines: 49

I've often run into the problem of wanting to apply a function to
every element of a list, but been stuck because the function takes
more than one argument.  As a simple example:

  (defun F (x y)
    (+ x y))

  (setq L '(2 4 6 8))

How do we add 10 to each element of L?

  (mapcar 'F L '10)                  ; Error: bad list argument to map

  (mapcar 'F L '(10))                ; (12)

  (mapcar 'F L '(10 10 10 10))       ; (12 14 16 18)

  (mapcar '(lambda (x) (F x 10)) L)  ; (12 14 16 18)

The first example doesn't work for obvious reasons.  The second returns
a list whose length is the minimum of the argument lists which were given
to mapcar.  The third works correctly, but is cumbersome because we first
had to know that the length of L was 4.  The fourth is the best solution,
but obscures the code because the number 10 has been hidden inside the
lambda (this gets much worse for more complicated expressions).  It would
be better to have something with the form of the first or second examples.
The following trick can be used to do this:

  (defun repeated (x)              ; returns an 'infinite' list of x's
    ((lambda (list-x)
       (rplacd list-x list-x))
     (list x)))

  (mapcar 'F L (repeated '10))     ; (12 14 16 18)

The 'repeated' function returns a CONS cell whose CDR points to the cell
itself.  This results in an (apparant) infinite list of x's.  Using this
as an argument list to mapcar is clearer and doesn't depend on the lengths 
of the other argument lists.  In general, it also costs less in space and
in time.

Hope this helps ...


James Stewart
University of Geneva

UUCP:   seismo!mcvax!cernvax!cui!manuel
BITNET: stewart@cgeuge52