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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[–] corbasai@alien.top 1 points 1 year ago

funcall isn't needed, because in your snippet defun-ed f.

funcall for defvar lambdas in CL.

Scheme is a Lisp-1 -> variables and procedures are defined in one namespace.

In CL (Lisp-2) variables and functions separated.