I'm new to Lisp and I'm working through some tutorials. Below is the given code for differentiating polynomials. I want to streamline it so (d) and (simplify) are one function / can be done in one step. I was thinking something (defun differentiate (poly x) (simplify (d (poly x)))) but then it thinks poly is a function, which doesn't work.
Maybe this doesn't work because the input for (d) has to be quoted? i.e. (d '(+ x y) 'x)
Sorry for the large chunk of code but I thought it would be best to put it all in. The relevant functions are at the bottom.
;;
;; Constructors for polynomials
;;
(defun make-constant (num)
num)
(defun make-variable (sym)
sym)
(defun make-negation (poly)
(list '- poly))
(defun make-sum (poly1 poly2)
(list '+ poly1 poly2))
(defun make-difference (poly1 poly2)
(list '- poly1 poly2))
(defun make-product (poly1 poly2)
(list '* poly1 poly2))
(defun make-power (poly num)
(list '** poly num))
;;
;; Recognizers for polynomials
;;
(defun constant-p (poly)
(numberp poly))
(defun variable-p (poly)
(symbolp poly))
(defun negation-p (poly)
(and (listp poly) (eq (first poly) '-) (null (rest (rest poly)))))
(defun sum-p (poly)
(and (listp poly) (eq (first poly) '+)))
(defun difference-p (poly)
(and (listp poly) (eq (first poly) '-) (not (null (rest (rest poly))))))
(defun product-p (poly)
(and (listp poly) (eq (first poly) '*)))
(defun power-p (poly)
(and (listp poly) (eq (first poly) '**)))
;;
;; Selectors for polynomials
;;
(defun constant-numeric (const)
const)
(defun variable-symbol (var)
var)
(defun negation-arg (neg)
(second neg))
(defun sum-arg1 (sum)
(second sum))
(defun sum-arg2 (sum)
(third sum))
(defun difference-arg1 (diff)
(second diff))
(defun difference-arg2 (diff)
(third diff))
(defun product-arg1 (prod)
(second prod))
(defun product-arg2 (prod)
(third prod))
(defun power-base (pow)
(second pow))
(defun power-exponent (pow)
(third pow))
;;
;; Unevaluated derivative
;;
(defun make-derivative (poly x)
(list 'd poly x))
(defun derivative-p (poly)
(and (listp poly) (eq (first poly) 'd)))
;;
;; Differentiation function
;;
(defun d (poly x)
(cond
((constant-p poly) 0)
((variable-p poly)
(if (equal poly x)
1
(make-derivative poly x)))
((negation-p poly)
(make-negation (d (negation-arg poly) x)))
((sum-p poly)
(make-sum (d (sum-arg1 poly) x)
(d (sum-arg2 poly) x)))
((difference-p poly)
(make-difference (d (difference-arg1 poly) x)
(d (difference-arg2 poly) x)))
((product-p poly)
(make-sum (make-product (product-arg1 poly)
(d (product-arg2 poly) x))
(make-product (product-arg2 poly)
(d (product-arg1 poly) x))))
((power-p poly)
(make-product (make-product (power-exponent poly)
(make-power (power-base poly)
(1- (power-exponent poly))))
(d (power-base poly) x)))))
;;
;; Simplification function
;;
(defun simplify (poly)
"Simplify polynomial POLY."
(cond
((constant-p poly) poly)
((variable-p poly) poly)
((negation-p poly)
(let ((arg (simplify (negation-arg poly))))
(make-simplified-negation arg)))
((sum-p poly)
(let ((arg1 (simplify (sum-arg1 poly)))
(arg2 (simplify (sum-arg2 poly))))
(make-simplified-sum arg1 arg2)))
((product-p poly)
(let ((arg1 (simplify (product-arg1 poly)))
(arg2 (simplify (product-arg2 poly))))
(make-simplified-product arg1 arg2)))
((difference-p poly)
(let ((arg1 (simplify (difference-arg1 poly)))
(arg2 (simplify (difference-arg2 poly))))
(make-simplified-difference arg1 arg2)))
((power-p poly)
(let ((base (simplify (power-base poly)))
(exponent (simplify (power-exponent poly))))
(make-simplified-power base exponent)))
((derivative-p poly) poly)))
(defun make-simplified-negation (arg)
"Given simplified polynomial ARG, construct a simplified negation of ARG."
(cond
((and (constant-p arg) (zerop arg)) arg)
((negation-p arg) (negation-arg arg))
(t (make-negation arg))))
(defun make-simplified-sum (arg1 arg2)
"Given simplified polynomials ARG1 and ARG2, construct a simplified sum of ARG1 and ARG2."
(cond
((and (constant-p arg1) (zerop arg1)) arg2)
((and (constant-p arg2) (zerop arg2)) arg1)
((negation-p arg1) (make-simplified-difference
arg2 (negation-arg arg1)))
((negation-p arg2) (make-simplified-difference
arg1 (negation-arg arg2)))
(t (make-sum arg1 arg2))))
(defun make-simplified-difference (arg1 arg2)
"Given simplified polynomials ARG1 and ARG2, construct a simplified difference of ARG1 and ARG2."
(cond
((and (constant-p arg2) (zerop arg2)) arg1)
((and (constant-p arg1) (zerop arg1)) (make-simplified-negation arg2))
((negation-p arg2) (make-simplified-sum
arg1 (negation-arg arg2)))
(t (make-difference arg1 arg2))))
(defun make-simplified-product (arg1 arg2)
"Given simplified polynomials ARG1 and ARG2, construct a simplified product of ARG1 and ARG2."
(cond
((and (constant-p arg1) (zerop arg1)) (make-constant 0))
((and (constant-p arg2) (zerop arg2)) (make-constant 0))
((and (constant-p arg1) (= arg1 1)) arg2)
((and (constant-p arg2) (= arg2 1)) arg1)
((and (constant-p arg1) (= arg1 -1)) (make-simplified-negation arg2))
((and (constant-p arg2) (= arg2 -1)) (make-simplified-negation arg1))
(t (make-product arg1 arg2))))
(defun make-simplified-power (base exponent)
"Given simplified polynomials BASE and EXPONENT, construct a simplified power with base BASE and exponent EXPONENT."
(cond
((and (constant-p exponent) (= exponent 1)) base)
((and (constant-p exponent) (zerop exponent)) (make-constant 1))
(t (make-power base exponent))))
The inputs to poly don't "need to be quoted". Instead of
(d '(+ x y) 'x)
you could write any of the following:
(d (list '+ 'x 'y) 'x)
(d (make-sum 'x 'y) 'x)
(let ((ex 'x) (why 'y))
(d (list '+ ex why) ex))
In your attempt in the question, you're trying to call d with the result of (poly x) which tries to call a function a named poly (actually, the function binding of the symbol poly, or some other possible things, but that's probably more in-depth than we need to get right now) with the value of the variable x:
(defun differentiate (poly x)
;; call poly with the value of x to produce a value y, and
;; then call d with y to produce a value z, and then call
;; simplify with z.
(simplify (d (poly x))))
That won't work, of course, because there's no function named poly, and even if there was, d expects two arguments, not one. Instead, the first thing you should do is call d with two arguments, viz., the values of the variables poly and x, and then call simplify with the result of that:
(defun differentiate (poly x)
(simplify (d poly x)))
Related
I have problem with macros in my lisp interpreter writtein in JavaScript. the problem is in this code:
(define log (. console "log"))
(define (alist->object alist)
"(alist->object alist)
Function convert alist pairs to JavaScript object."
(if (pair? alist)
((. alist "toObject"))))
(define (klist->alist klist)
"(klist->alist klist)
Function convert klist in form (:foo 10 :bar 20) into alist
in form ((foo . 10) (bar . 20))."
(let iter ((klist klist) (result '()))
(if (null? klist)
result
(if (and (pair? klist) (pair? (cdr klist)) (key? (car klist)))
(begin
(log ":::" (cadr klist))
(log "data" (. (cadr klist) "data"))
(iter (cddr klist) (cons (cons (key->string (car klist)) (cadr klist)) result)))))))
(define (make-empty-object)
(alist->object '()))
(define empty-object (make-empty-object))
(define klist->object (pipe klist->alist alist->object))
;; main function that give problems
(define (make-tags expr)
(log "make-tags" expr)
`(h ,(key->string (car expr))
,(klist->object (cadr expr))
,(if (not (null? (cddr expr)))
(if (and (pair? (caddr expr)) (let ((s (caaddr expr))) (and (symbol? s) (eq? s 'list))))
`(list->array (list ,#(map make-tags (cdaddr expr))))
(caddr expr)))))
(define-macro (with-tags expr)
(make-tags expr))
I call this macro using this code:
(define (view state actions)
(with-tags (:div ()
(list (:h1 () (value (cdr (assoc 'count (. state "counter")))))
(:button (:onclick (lambda () (--> actions (down 1)))) "-")
(:button (:onclick (lambda () (--> actions (up 1)))) "+")))))
which should expand to almost the same code:
(define (view state actions)
(h "div" (make-empty-object)
(list->array (list
(h "h1" (make-empty-object) (value (cdr (assoc 'count (. state "counter")))))
(h "button" (klist->object `(:onclick ,(lambda () (--> actions (down 1))))) "-")
(h "button" (klist->object `(:onclick ,(lambda () (--> actions (up 1))))) "+")))))
This function works. I have problem with expanded code using my macro that call the main function, don't know how LIPS should behave when it find:
(:onclick (lambda () (--> actions (down 1))))
inside code and you try to process it like this:
,(klist->object (cadr expr))
Right now my lisp works that lambda is marked as data (have data flag set to true this is a hack to prevent of recursive evaluation of some code from macros) and klist->object function get lambda code as list, instead of function.
How this should work in Scheme or Common Lisp? Should klist->object get function object (lambda get evaluated) or list structure with lambda as first symbol? If second then how I sould write my function and macro to evaluate lambda should I use eval (kind of hack to me).
Sorry don't know how to test this, with more bug free LISP.
EDIT:
I've tried to apply the hint from #jkiiski in guile (because in my lisp it was not working)
;; -*- sheme -*-
(define nil '())
(define (key? symbol)
"(key? symbol)
Function check if symbol is key symbol, have colon as first character."
(and (symbol? symbol) (eq? ":" (substring (symbol->string symbol) 0 1))))
(define (key->string symbol)
"(key->string symbol)
If symbol is key it convert that to string - remove colon."
(if (key? symbol)
(substring (symbol->string symbol) 1)))
(define (pair-map fn seq-list)
"(seq-map fn list)
Function call fn argument for pairs in a list and return combined list with
values returned from function fn. It work like the map but take two items from list"
(let iter ((seq-list seq-list) (result '()))
(if (null? seq-list)
result
(if (and (pair? seq-list) (pair? (cdr seq-list)))
(let* ((first (car seq-list))
(second (cadr seq-list))
(value (fn first second)))
(if (null? value)
(iter (cddr seq-list) result)
(iter (cddr seq-list) (cons value result))))))))
(define (klist->alist klist)
"(klist->alist klist)
Function convert klist in form (:foo 10 :bar 20) into alist
in form ((foo . 10) (bar . 20))."
(pair-map (lambda (first second)
(if (key? first)
(cons (key->string first) second))) klist))
(define (h props . rest)
(display props)
(display rest)
(cons (cons 'props props) (cons (cons 'rest rest) nil)))
(define (make-tags expr)
`(h ,(key->string (car expr))
(klist->alist (list ,#(cadr expr)))
,(if (not (null? (cddr expr)))
(if (and (pair? (caddr expr)) (let ((s (caaddr expr))) (and (symbol? s) (eq? s 'list))))
`(list->array (list ,#(map make-tags (cdaddr expr))))
(caddr expr)))))
(define-macro (with-tags expr)
(make-tags expr))
(define state '((count . 10)))
(define xxx (with-tags (:div ()
(list (:h1 () (cdr (assoc 'count state)))
(:button (:onclick (lambda () (display "down"))) "-")
(:button (:onclick (lambda () (display "up"))) "+")))))
but got error:
ERROR: Unbound variable: :onclick
I've found solution for my lisp, Here is code:
(define (pair-map fn seq-list)
"(seq-map fn list)
Function call fn argument for pairs in a list and return combined list with
values returned from function fn. It work like the map but take two items from list"
(let iter ((seq-list seq-list) (result '()))
(if (null? seq-list)
result
(if (and (pair? seq-list) (pair? (cdr seq-list)))
(let* ((first (car seq-list))
(second (cadr seq-list))
(value (fn first second)))
(if (null? value)
(iter (cddr seq-list) result)
(iter (cddr seq-list) (cons value result))))))))
(define (make-tags expr)
(log "make-tags" expr)
`(h ,(key->string (car expr))
(alist->object (quasiquote
;; create alist with unquote for values and keys as strings
,#(pair-map (lambda (car cdr)
(cons (cons (key->string car) (list 'unquote cdr))))
(cadr expr))))
,(if (not (null? (cddr expr)))
(if (and (pair? (caddr expr)) (let ((s (caaddr expr))) (and (symbol? s) (eq? s 'list))))
`(list->array (list ,#(map make-tags (cdaddr expr))))
(caddr expr)))))
So in my code I'm writing some kind of meta macro I'm writing quasiquote as list that will get evaluated the same as if I use in my original code:
(klist->object `(:onclick ,(lambda () (--> actions (down 1)))))
I'm using alist->object and new function pair-map, so I can unquote the value and convert key symbol to string.
is this how it should be implemented in scheme? not sure If I need to fix my lisp or macros are working correctly there.
I'm trying to write let over lambda defmacro/g! in guile scheme. I have this:
(use-modules (srfi srfi-1))
(define (flatten x)
(let rec ((x x) (acc '()))
(cond ((null? x) acc)
((not (pair? x)) (cons x acc))
(else
(rec (car x)
(rec (cdr x) acc))))))
(define (g!-symbol? s)
(and (symbol? s)
(let ((symbol-string (symbol->string s)))
(and (> (string-length symbol-string) 2)
(equal? (string-downcase (substring symbol-string 0 2)) "g!")))))
(define-macro (define-macro/g! name-args . body)
(let ((syms (delete-duplicates
(filter g!-symbol? (flatten body)))))
`(define-macro ,name-args
(let ,(map
(lambda (s)
`(,s (gensym ,(substring (symbol->string s) 2))))
syms)
,#body))))
but when I try to macro expand define-macro/g! using this:
(use-modules (language tree-il))
(tree-il->scheme (macroexpand '(define-macro/g! (foo . body) `(let ((g!car ,(car body))) g!car))))
I've got this:
$15 = (if #f #f)
why I've got this result? How can I expand define-macro/g!?
I need to use this code:
(define macro '(define-macro/g! (foo . body) `(let ((g!car ,(car body))) g!car)))
(tree-il->scheme (macroexpand macro 'c '(compile load eval)))
i have a simple boolean expression presented as a lisp list like this :
'(OR 0 (AND A1 A2))
the previous list is the presentation of ((A1 AND A2)OR 0).
anyway , i am writing a function to simplify this expression ..
for example :
calling the function "reduce" like this :
(reduce '(OR 0 (AND A1 A2)))
would yield
(AND A1 A2)
i firstly tried to create base rules , so i defined
the following identities:
(AND 1 S) == S,
(OR 0 S ) == S,
(AND 0 S) == 0,
( OR 1 S) == 1,
(NOT O) == 1,
(NOT 1) == 0.*
i was thinking of defining 6 functions , one for each rule , and then
call them one by one in a wrapper , i am new to lisp so i don't have a clue to implement that , i made it in java once , but i dont know how to handle such problem using the syntax of lisp so please help me ..
Given the complexity of your solution, here's my implementation that's a lot shorter and more readable:
(defun reduc (exp)
(if (atom exp)
exp
(flet ((helper (op args n) ; and and or is nearly the same code so we factor it out
(let ((newargs (remove n args)) (cn (- 1 n)))
(cond
((null newargs) n)
((some (lambda (e) (eql cn e)) newargs) cn)
((null (cdr newargs)) (car newargs))
((cons op newargs))))))
(let ((op (car exp)) (args (mapcar #'reduc (cdr exp))))
(ecase op
((not) (if (= 1 (length args))
(let ((arg1 (car args)))
(if (and (numberp arg1) (<= 0 arg1 1)) (- 1 arg1) exp))
(error "'not' must have exactly one parameter")))
((and) (helper op args 1))
((or) (helper op args 0)))))))
Testing:
? (reduc '(OR 0 (AND A1 A2)))
(AND A1 A2)
? (reduc '(OR 0 (AND A1 1 A2)))
(AND A1 A2)
? (reduc '(or ERROR (not 0)))
1
? (reduc '(AND ERROR (not 0)))
ERROR
? (reduc '(OR 0 (AND A1 0)))
0
? (reduc '(OR 0 (AND A1 1)))
A1
i finally came up with this solution .
(defun simplify (EXPR)
(simplify-expr NIL EXPR))
(defun simplify-expr (EXPR1 EXPR2)
(cond
((or (atom EXPR2) (equal EXPR1 EXPR2)) EXPR2)
(T (simplify-expr EXPR2 (simplify-boolean-expr EXPR2)))))
(defun simplify-boolean-expr (EXPR)
(cond
((and (equal (first EXPR) `and) (>= (length EXPR) 3))
(simplify-and-expr (rest EXPR)))
((and (equal (first EXPR) `or) (>= (length EXPR) 3))
(simplify-or-expr (rest EXPR)))
((and (equal (first EXPR) `not) (= (length EXPR) 2))
(simplify-not-expr (rest EXPR)))
(T
(error "~S is not a valid circuit descriptor expression or has an unknown operator." EXPR))))
(defun simplify-and-expr (EXPR)
(let ((SIMPLIFIED_EXPR (remove `T (remove-duplicates EXPR))))
(cond
((null SIMPLIFIED_EXPR) `T)
((member `NIL SIMPLIFIED_EXPR) `NIL)
((null (second SIMPLIFIED_EXPR)) (first SIMPLIFIED_EXPR))
(T (cons `and (simplify-operand SIMPLIFIED_EXPR))))))
(defun simplify-or-expr (EXPR)
(let ((SIMPLIFIED_EXPR (remove `NIL (remove-duplicates EXPR))))
(cond
((null SIMPLIFIED_EXPR) `NIL)
((member `T SIMPLIFIED_EXPR) `T)
((null (second SIMPLIFIED_EXPR)) (first SIMPLIFIED_EXPR))
(T (cons `or (simplify-operand SIMPLIFIED_EXPR))))))
(defun simplify-not-expr (EXPR)
(cond
((equal (first EXPR) `NIL) `T)
((equal (first EXPR) `T) `NIL)
((and (listp (first EXPR)) (equal (first (first EXPR)) `not))
(first (rest (first EXPR))))
(T (cons `not (simplify-operand EXPR)))))
(defun simplify-operand (OPERAND_LIST)
(cond
((null OPERAND_LIST) NIL)
((atom (first OPERAND_LIST))
(cons (first OPERAND_LIST) (simplify-operand (rest OPERAND_LIST))))
(T
(cons (simplify-expr NIL (first OPERAND_LIST)) (simplify-operand (rest OPERAND_LIST))))))
it takes (nil , T) for (0 , 1) and reduces any boolean expression , i tried it and it works fine .
The problem with flet is that the functions bound therein must be defined inline. In other words, there's no way to do this:
(new-flet ((a (lambda (f x)
(funcall f (* x 2))))
(b (function-generator)))
(a #'b 10))
I considered defining such a macro myself, but the problem is that flet seems to be the only way to set local function values. symbol-function always gets the global definition only, and function can't be used with setf. Anyone have an idea how this can be done fairly cleanly, if at all?
You can easily build a trampoline
(defun function-generator (x)
(lambda (y) (* x y)))
(let ((fg (function-generator 42)))
(flet ((a (f x) (funcall f (* x 2)))
(b (x) (funcall fg x)))
(a #'b 10)))
A macro implementation of new-flet with this approach is
(defmacro new-flet (bindings &body body)
(let ((let-bindings (list))
(flet-bindings (list))
(args (gensym)))
(dolist (binding bindings)
(let ((name (gensym)))
(push `(,name ,(second binding))
let-bindings)
(push `(,(first binding) (&rest ,args)
(apply ,name ,args))
flet-bindings)))
`(let ,(nreverse let-bindings)
(flet ,(nreverse flet-bindings)
,#body))))
that expands in your example case as
(macroexpand-1 '(new-flet ((a (lambda (f x) (funcall f (* x 2))))
(b (function-generator)))
(a #'b 10)))
==> (LET ((#:G605 (LAMBDA (F X)
(FUNCALL F (* X 2))))
(#:G606 (FUNCTION-GENERATOR)))
(FLET ((A (&REST #:G604)
(APPLY #:G605 #:G604))
(B (&REST #:G604)
(APPLY #:G606 #:G604)))
(A #'B 10)))
Is
(let* ((a (lambda (f x) (funcall f (* x 2))))
(b (function-generator)))
(funcall a b 10))
a fairly clean solution to your problem?
How about binding the variables with let, so that they're setfable, and then using an flet as the body of the let so that they're funcallable and (function …)-able, too. E.g., where I've given a silly little function instead of (generate-function):
(let ((a (lambda (f x)
(funcall f (* x 2))))
(b (lambda (&rest args)
(print (list* 'print-from-b args)))))
(flet ((a (&rest args)
(apply a args))
(b (&rest args)
(apply b args)))
(a #'b 10)))
We can wrap this up in a macro relatively easily:
(defmacro let/flet (bindings &body body)
(let ((args (gensym (string '#:args-))))
`(let ,bindings
(flet ,(loop :for (name nil) :in bindings
:collect `(,name (&rest ,args) (apply ,name ,args)))
,#body))))
Now
(let/flet ((a (lambda (f x)
(funcall f (* x 2))))
(b (lambda (&rest args)
(print (list* 'print-from-b args)))))
(a #'b 10))
expands into the first block of code. Note that you can also use (a b 10) in the body as well, since the binding of b is the same as the value of #'b. You can use setf on the variable as well:
(let/flet ((a (lambda (x)
(print (list 'from-a x)))))
(a 23)
(setf a (lambda (x)
(print (list 'from-new-a x x))))
(a 23))
prints
(FROM-A 23)
(FROM-NEW-A 23 23)
If anyone's interested in a labels equivalent, here it is:
(defmacro my-labels ((&rest definitions) &rest body)
(let ((gensyms (loop for d in definitions collect (gensym)))
(names (loop for d in definitions collect (car d)))
(fdefs (loop for f in definitions collect (cadr f)))
(args (gensym)))
`(let (,#(loop for g in gensyms collect (list g)))
(labels (,#(loop for g in gensyms for n in names
collect `(,n (&rest ,args) (apply ,g ,args))))
,#(loop for g in gensyms for f in fdefs
collect `(setf ,g ,f))
,#body))))
This is sort of like Scheme's letrec.
Given the following definition of the LISP eval function - what is required to add the defmacro function? (Or even just evaluate a macro)
(defun null. (x)
(eq x '()))
(defun and. (x y)
(cond (x (cond (y 't) ('t '())))
('t '())))
(defun not. (x)
(cond (x '())
('t 't)))
(defun append. (x y)
(cond ((null. x) y)
('t (cons (car x) (append. (cdr x) y)))))
(defun list. (x y)
(cons x (cons y '())))
(defun pair. (x y)
(cond ((and. (null. x) (null. y)) '())
((and. (not. (atom x)) (not. (atom y)))
(cons (list. (car x) (car y))
(pair. (cdr x) (cdr y))))))
(defun assoc. (x y)
(cond ((eq (caar y) x) (cadar y))
('t (assoc. x (cdr y)))))
(defun eval. (e a)
(cond
((atom e) (assoc. e a))
((atom (car e))
(cond
((eq (car e) 'quote) (cadr e))
((eq (car e) 'atom) (atom (eval. (cadr e) a)))
((eq (car e) 'eq) (eq (eval. (cadr e) a)
(eval. (caddr e) a)))
((eq (car e) 'car) (car (eval. (cadr e) a)))
((eq (car e) 'cdr) (cdr (eval. (cadr e) a)))
((eq (car e) 'cons) (cons (eval. (cadr e) a)
(eval. (caddr e) a)))
((eq (car e) 'cond) (evcon. (cdr e) a))
('t (eval. (cons (assoc. (car e) a)
(cdr e))
a))))
((eq (caar e) 'label)
(eval. (cons (caddar e) (cdr e))
(cons (list. (cadar e) (car e)) a)))
((eq (caar e) 'lambda)
(eval. (caddar e)
(append. (pair. (cadar e) (evlis. (cdr e) a))
a)))))
(defun evcon. (c a)
(cond ((eval. (caar c) a)
(eval. (cadar c) a))
('t (evcon. (cdr c) a))))
(defun evlis. (m a)
(cond ((null. m) '())
('t (cons (eval. (car m) a)
(evlis. (cdr m) a)))))
(eval '(car '(a a)) )
The representation of an anonymous macro is by convention a list of the form (macro lambda ...). Try evaling these in your favorite Lisp interpreter (tested in Emacs):
> (defmacro triple (x) `(+ ,x ,x ,x))
triple
> (symbol-function 'triple)
(macro lambda (x) (\` (+ (\, x) (\, x) (\, x))))
Although things don't work that way in Emacs, the only thing left to do is to give the adequate semantics to such a form. That is, when eval. sees ((macro lambda (x) EXPR) FORM), it must
Replace every occurence of x in FORM with EXPR without evaluating EXPR first (as opposed to what happens in a function call);
eval. the result of above.
You can achieve this by adding a clause to the outermost cond in eval. that deals with the ((macro lambda ...) ...) case. Here is a crude prototype:
((eq (caar e) 'macro)
(cond
((eq (cadar e) 'lambda)
(eval. (eval. (car (cdddar e))
(cons (list. (car (caddar e)) (cadr e)) a))
a))))
This code only works for single-argument macros. Fixing that involves writing an auxiliary function substlis. that works like evlis. but without looping to eval.; that is left as an exercise to the reader :-)
To test, define cadr. as a macro thusly:
(defmacro cadr. (x)
(list. 'car (list. 'cdr x)))
After this you would have
> (symbol-function 'cadr.)
(macro lambda (x) (list. (quote car) (list. (quote cdr) x)))
You can construct a form that applies this (macro lambda ...) to an expression, and eval that construction within a context that contains a definition for list. (because it is not considered primitive by the eval. interpreter). For instance,
(let ((e '((macro lambda (x) (list (quote car) (list (quote cdr) x)))
(cons (quote x) (cons (quote y) nil))))
(bindings `((list ,(symbol-function 'list.)))))
(eval. e bindings))
y
Tada!
This is also quite good:
https://web.archive.org/web/20120702032624/http://jlongster.com/2012/02/18/its-not-about-macros-its-about-read.html
"You can implement a macro system in 30 lines of Lisp. All you need is read, and it's easy."
https://gist.github.com/1712455