lisp

How to convince Lisp SBCL to do inline fixnum arithmetic?

由 ℡╲_俬逩灬. 提交于 2021-02-08 05:01:36
问题 I've found some techniques in other SO answers, but apparently I've been unable to convince SBCL to do inline fixnum arithmetic: (declaim (optimize (speed 2) (safety 1))) (declaim (ftype (function (fixnum fixnum) double-float) fixnumtest) (inline fixnumtest)) (defun fixnumtest (i j) (declare (type fixnum i j)) (let* ((n (the fixnum (+ i j))) (n+1 (the fixnum (1+ n)))) (declare (type fixnum n n+1)) (/ 1.0d0 (the fixnum (* n n+1)) ) ) ) (defun main () (format t "~11,9F~%" (fixnumtest 2 3)) )

Common Lisp学习笔记

由 老子叫甜甜 提交于 2021-02-07 22:47:47
全局变量以*号包裹 ------------------------------------------------------------- 列表：(list 元素1 元素2 元素3 ...) 属性表：(list 符号1 元素1 符号2 元素2 ...) 从属性表取值：(getf 属性表 符号) 遍历列表：(dolist 元素 列表)，用PHP表述就相当于foreach(列表 as 元素) 压入元素：(push 元素 变量) ------------------------------------------------------------- 定义函数：(defun 函数名 (参数表) 函数体) 定义变量：(defvar 变量名 变量值) 变量赋值：(setf 变量名 变量值) ------------------------------------------------------------- format格式化参数的含义 # ~{ ... ~}：拆分一个属性表的键值对 # ~a：占位符 # ~t：制表符 # ~%：换行符 来源： oschina 链接： https://my.oschina.net/u/103341/blog/121425

Practical use of fold/reduce in functional languages

由 喜欢而已 提交于 2021-02-07 12:16:31
问题 Fold (aka reduce ) is considered a very important higher order function. Map can be expressed in terms of fold (see here). But it sounds more academical than practical to me. A typical use could be to get the sum, or product, or maximum of numbers, but these functions usually accept any number of arguments. So why write (fold + 0 '(2 3 5)) when (+ 2 3 5) works fine. My question is, in what situation is it easiest or most natural to use fold ? 回答1: The point of fold is that it's more abstract.

Practical use of fold/reduce in functional languages

由 无人久伴 提交于 2021-02-07 12:15:39
问题 Fold (aka reduce ) is considered a very important higher order function. Map can be expressed in terms of fold (see here). But it sounds more academical than practical to me. A typical use could be to get the sum, or product, or maximum of numbers, but these functions usually accept any number of arguments. So why write (fold + 0 '(2 3 5)) when (+ 2 3 5) works fine. My question is, in what situation is it easiest or most natural to use fold ? 回答1: The point of fold is that it's more abstract.

(compose) in Common Lisp

由 巧了我就是萌 提交于 2021-02-07 06:17:46
问题 We find this function builder to realize composition in P.Graham's "ANSI Common Lisp" (page 110). The arguments are n>0 quoted function names. I don't understand it completely, so I'll quote the code here and specify my questions underneath it: (defun compose (&rest fns) (destructuring-bind (fn1 . rest) (reverse fns) #'(lambda (&rest args) (reduce #'(lambda (v f) (funcall f v)) rest :initial-value (apply fn1 args))))) The argument list to compose is reversed and unpacked, its (now first)

(compose) in Common Lisp

由 好久不见. 提交于 2021-02-07 06:16:43
问题 We find this function builder to realize composition in P.Graham's "ANSI Common Lisp" (page 110). The arguments are n>0 quoted function names. I don't understand it completely, so I'll quote the code here and specify my questions underneath it: (defun compose (&rest fns) (destructuring-bind (fn1 . rest) (reverse fns) #'(lambda (&rest args) (reduce #'(lambda (v f) (funcall f v)) rest :initial-value (apply fn1 args))))) The argument list to compose is reversed and unpacked, its (now first)

Why sharp quote lambda expressions?

由 巧了我就是萌 提交于 2021-02-06 09:58:55
问题 It is a technique used frequently in On Lisp , which is on Common Lisp: > (mapcar #'(lambda (x) (+ x 10)) '(1 2 3)) (11 12 13) Why is sharp-quote needed or even possible? lambda expressions return function objects, and sharp quoting returns function objects from names. I also have heard contradictory information on whether lambda expressions are names - in particular On Lisp contradicts the standard, but his code seems to work which also contradicts the standard. In elisp it seems it's not

Why sharp quote lambda expressions?

由 不打扰是莪最后的温柔 提交于 2021-02-06 09:58:25
问题 It is a technique used frequently in On Lisp , which is on Common Lisp: > (mapcar #'(lambda (x) (+ x 10)) '(1 2 3)) (11 12 13) Why is sharp-quote needed or even possible? lambda expressions return function objects, and sharp quoting returns function objects from names. I also have heard contradictory information on whether lambda expressions are names - in particular On Lisp contradicts the standard, but his code seems to work which also contradicts the standard. In elisp it seems it's not

【SICP练习】5 练习1.9

由 ぃ、小莉子 提交于 2021-02-06 07:59:42
 以下是第一个加起两个正整数的方法，其中 inc 将参数加 1 ， dec 将参数减 1 。 (define (+ a b) (if(= a 0) b (inc (+ (dec a) b)))) 用代换模型展示 (+ 4 5) 如下： (+ 4 5) (inc (+ 3 5)) (inc (inc (+ 2 5))) (inc (inc (inc (+ 1 5)))) (inc (inc (inc (inc (+ 0 5))))) (inc (inc (inc (inc 5)))) (inc (inc (inc 6))) (inc (inc 7)) (inc 8) 9 如上所示，在代换模型展示中包含了伸展和收缩两个阶段，并且伸展阶段所需的额外存储量和计算所需的步数都正比于参数 a 。因此这是一个线性递归过程。 以下是另一个加起两个正整数的方法。 (define (+ a b) (if(= a 0) b (+ (dec a) (inc b)))) 同样用代换模型展示 (+ 4 5) 如下： (+ 4 5) (+ 3 6) (+ 2 7) (+ 1 8) (+ 0 9) 9 在这个过程中并没有任何增长或者收缩，而其计算过程可用固定数目的状态变量（ a ）描述。这是一个线性迭代过程。 版权声明：本文为 NoMasp柯于旺 原创文章，未经许可严禁转载！欢迎访问我的博客：http:/

【SICP练习】2 练习1.6

由 こ雲淡風輕ζ 提交于 2021-02-06 07:47:19
 练习 1.6 这道题通过由一个新版本的 if 来引出，主要讨论的还是应用序和正则序的问题。我看到“将 if 提供为一种特殊形式”时还满头雾水，并不太清楚什么特殊形式。当再返回看 if 的语法时才发现，这在第 12 页 if 的一般表达式下面一段。如果 <predicate> 得到真值，解释器就去求值 <consequent> 并返回其值。注意，在此处已经返回其值了，并没有进行后续运算。 而通过 cond 写出来的常规过程的 if ，在解释器采用应用序求值的情况下，如果第一次运算 good-enough? 时为真，则直接返回了 guess 。 原文中的求平方根的程序： (define (new-if predicate then-clauseelse-clause) (cond(predicate then-clause) (elseelse-clause))) (define (sqrt-iter guess x) (new-if(good-enough? guess x) guess (sqrt-iter (improve guess x) x))) (define(sqrt-iter guess x) (if(good-enough? guess x) guess (sqrt-iter (improve guess x) x))) 于是博主进行了如下测试： (sqrt
订阅 lisp