CODEWITHSUNDEEP
common-lisp
linjunhalida
linjunhalida

Reputation: 4655

How to get rid of funcall in common lisp

According to this document: http://cl-cookbook.sourceforge.net/functions.html

(defun adder (n)
  (lambda (x) (+ x n)))
(funcall (adder 12) 1)

I have to use funcall to call (adder 12), And it is very ignoring to write funcall over and over, is there any way to write code like it in scheme:

((adder 12) 1)

Upvotes: 2

Views: 356

Answers (2)

user797257
user797257

Reputation:

However, you could use something like this (not sure why would you, but the number of characters typed would be the same as it is in Scheme):

(set-macro-character
 #\[
 #'(lambda (stream char)
     (declare (ignore char))
     (set-syntax-from-char #\] #\;)
     (let ((forms (read-delimited-list #\] stream t)))
       (set-syntax-from-char #\] #\x)
       (append '(funcall) forms))))

(defun adder (n)
  #'(lambda (x) (+ x n)))

(format t "sum: ~s~&" [(adder 12) #x128]) ;; 308

This may give you some problems if you will encounter a variable name with brackets in it. Sure, using it is up to you, consider yourself warned.

Upvotes: 1

Rainer Joswig
Rainer Joswig

Reputation: 139261

No. There is none.

You can also see it as a feature: it makes calls of function objects explicit and improves understandability of source code.

Upvotes: 6

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