â [lisp] Må¼ æè¿ç¥ã£ãã®ã ããLispã®åæã«ã¯Så¼ã®ä»ã«Må¼ã¨ããè¨æ³ãèãããã¦ãããããã http://en.wikipedia.org/wiki/M-expression ãã³ã³ãã¥ã¼ã¿ããã°ã©ã ã®ä¸çã«ãã㦠Må¼ (meta-expressions) ã¯ã Lispè¨èªã®ä¸é¨ã¨ãã¦ãSå¼ã®äººéãèªããã¨ã®åºæ¥ãå½¢æ ã¨ãªããã¨ãç®çã«èããããã Må¼ã¯Lispã®åæã®è«æã«ããã¦çè«çãªè¨èªã¨ãã¦ä½¿ããã¦ããããå®éã«å®è£ ããããã¨ã¯ãªãã£ãã ã Så¼ã¯äººéãèªããããããªãã®ããï½ã¨ããã³ããããªãã¾ããã å®è¡åã«Så¼ã«å¤æãããããmeta-expressionãã¨åã¥ãããã模æ§ã [1, 2, 3] (quote (1 2 3)) or '(1 2 3) car[X] (car X) car[append[[1,2,3], [4,5,6]]] (car
$Id: ycombinator.html,v 1.6 2002/06/27 23:37:39 aamine Exp $ [ruby-list:35058] ã«åºæ¿ãåã㦠Y combinator ã解èªãã¦ã¿ãã ãããªããèªãããããªã以ä¸ã®åèãã¼ã¸ãèªãã ã»ããããã åèã«ãã (ã¨ãããã»ã¨ãã©ãã®ã¾ã¾ãª) ãã¼ã¸ (è±èª) http://www.ececs.uc.edu/~franco/C511/html/Scheme/ycomb.html åæ© å帰é¢æ°ã¯å帰ããã¨ãã«èªåèªèº«ãååã§å¼ã¶ã®ãæ®éã§ããã ããããªãã¨ããã¦ååã使ãããé¢æ°ãã®ãã®ãå¼ã¶ããã« ããããã æ±ããã ã¾ãåç´ãª fact (éä¹) ã以ä¸ã«ç¤ºããè¨èªã¯ Scheme ã§ããã (define fact (lambda (n) (if (zero? n) 1 (* n (fact (- n
Leon Bambrick / éæ¨é 訳 2006å¹´9æ24æ¥ æ¥æ ããã«Lispã®çå®ãæããã«ãããããã¤ãã®ä»£æ¿ãæ示ãããã Lispãå¦ã¶ãã¨ã¯ããªãã®äººçãå¤ããã ããªãã®è³ã¯ããã大ãããªãããããªã«å¤§ãããªããã®ã ã¨ã¯æããªãã£ãã»ã©ã«ãªãã ããã ããªãã¯èªåã®ã¢ããªã±ã¼ã·ã§ã³ããã¹ã¦ãã»ãã®ä¸æ¡ãã®ã³ã¼ãã§æ¸ãæããã ããã 社ä¼ã¯ããªããé¿ããããã«ãªããããªãã社ä¼ãé¿ããããã«ãªãã ããªãã¯èªåã®ã¾ããã®ç©ãã¾ããã®äººãã¹ã¦ã«ä¸æºãæããããã«ãªãã Lispã¯é常ã«ã·ã³ãã«ã§ãããã»ãã®æ°åã§å¦ã¶ãã¨ãã§ãããç§ã¯ãã£ããã¹ãå¾ ã£ã¦ããéã«å¦ãã ã Lispã¯é常ã«ã·ã³ãã«ã§ãããã©ããªè¨èªã使ããã¨æ°ãã¼ã¸ã®ã³ã¼ãã§å®è£ ã§ãããããããããªãã¨ããããã¨ã¯ãªãã ãããã²ã¨ãã³Lispãå¦ãã ãªããããªãã¯Lisp以å¤ã®è¨èªã§ä½ããæ¸ããã¨ã¯æããªããªã
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LISPUSERLISPMEMOLisp is like a ball of mud - you can throw anything you want into it, and it's still Lisp. -- Anonymous Ruby æ¹é¢ã§ã¿ããããã¿ã«è§¦çºããã¾ããï¼Lisp ã¯ããã°ã©ããã«ãªè¨èªãª ã®ã§ãªã¼ãã¼ãããã°ã©ããã«ã§ãï¼ããã§ãªã¼ãã¼ãããã£ã¦ let ã defun ãªã©ã®æ§æãæ¬å¼§ãããªã㦠end ã«ã§ããã¦ã¿ã¾ããããï¼ 100 è¡ç¨åº¦ãç®æ¨ã«ãã©ã¤ãã¾ããï¼ä½¿ã£ãã©ã¤ãã©ãªã¯ CL-Yacc ã®ã¿ã§ãï¼ ã§ãããã£ãã®ããã¡ãï¼ #@suck-lisp defun fib (n) if (< n 0) (error "oops") elif (= n 0) 0 elif (= n 1) 1 else let x <- (fib (- n 1))
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If you have a little Scheme under your belt, just plunge into the challenge right now: We want to find three integers, x y and z such that both are in the range [2,9] and that they are the lengths of the sides of a right triangle (x^2 = y^2 + z^2). The challenge is to write two procedures, in-range and fail such that the solution to the problem looks like this: (let ((x (in-range 2 9)) (y (in-rang
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