this post was submitted on 10 Oct 2023
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Lisp

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A counter in racket-scheme:

#lang typed/racket

(define my-counter!
    (let ([t 0])
        (lambda ()
	        (set! t (+ 1 t))
	    t);lambda
	);let
);define 
(print (my-counter!))
(print (my-counter!))

A counter in sbcl-lisp:

load "~/quicklisp/setup.lisp")

(declaim (optimize (speed 3) (safety 3)))

(let ((c 0))
    (defun my-counter! ()
        (lambda ()
            (setf c (+ 1 c))
	    c); lambda
	 ) ;defun
) ;let

(defun main ()
(print (funcall (my-counter!)))
(print (funcall (my-counter!)))
)

(sb-ext:save-lisp-and-die "test.exe" :toplevel #'main :executable t)

Could someone elaborate why i need "funcall" in lisp and not in scheme ? And why the different placing of let ?

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[–] funk443@alien.top 1 points 11 months ago

In Common Lisp, as opposed to Scheme, it is not possible that the car of the compound form to be evaluated is an arbitrary form. If it is not a symbol, it must be a lambda expression, which looks like (lambda lambda-list form*).

This