Say I have a macro like this:
(define-syntax (choose stx)
(define data (syntax->datum stx))
(define args (cadr data))
(define body (cddr data))
(define output
`(apply (case (car ,args)
,(map (lambda (choice)
`((,(car choice)) ,(cadr choice)))
body)
(else (displayln "error")))
(cdr ,args)))
(println output)
#'(void))
If I use this on something like this (there could be more options):
(choose args
("run" runsomething)
("del" delsomethingelse))
It transforms it to
(apply
(case (car args)
((("run") runsomething)
(("del") delsomethingelse))
(else (displayln "error")))
(cdr args))
Which is not valid code, because the map gave it extra parentheses. Instead I want it to give me this:
(apply
(case (car args)
(("run") runsomething)
(("del") delsomethingelse)
(else (displayln "error")))
(cdr args))
How could I do something like this?
Use unquote-splicing (aka ,#) to get rid of the list surrounding map.
Example:
(define xs '(a b c))
`(1 2 ,xs 3 4) ; => '(1 2 (a b c) 3 4)
`(1 2 ,#xs 3 4) ; => '(1 2 a b c 3 4)
However I notice that you use syntax->datum on the input stx
of the syntax transformer. That removes lexical information, which
could end up causing problems. It recommend using either syntax-case
or syntax-parse, which use pattern matching to pick out the elements
of the input syntax and templates to generate the output.
(define-syntax (choose stx)
(syntax-case stx ()
[(_choose args
(datum fun-expr)
...)
#'(apply (case (car args)
[(datum) fun-expr]
...)
(cdr args))]))
(define (run-it . xs) (list 'ran-it xs))
(define (del-it . xs) (list 'delt-it xs))
(choose (list "run" 1 2 3)
("run" run-it)
("del" del-it))
Output: '(ran-it (1 2 3))
Related
I am new in Lisp and i need some help.
I need to simplify next expressions:
from (+ (+ A B) C) to (+ A B C)
and from (- (- A B) C) to (- A B C).
If you could help me with one of them I'll understand how i need to do this to the next one.
Thanks a lot.
Assuming you have an input that matches this pattern, (+ e1 ... en), you want to recursively simplify all e1 to en, which gives you s1, ..., sn, and then extract all the si that start with a + to move their arguments one level up, to the simplified expression you are building.
An expression e matches the above pattern if (and (consp e) (eq '+ (car e))).
Then, all the ei are just given by the list that is (cdr e).
Consider the (+) case, how could you simplify it?
To apply a function f to a list of values, call (mapcar #'f list).
To split a list into two lists, based on a predicate p, you might use a loop:
(let ((sat nil) (unsat nil))
(dolist (x list (values sat unsat))
(if (funcall predicate x)
(push x sat)
(push x unsat))))
There is a purely functional way to write this, can you figure it out?
Here is a trivial simplifier written in Racket, with an implementation of a rather mindless simplifier for +. Note that this is not intended as anything serious: it's just what I typed in when I was thinking about this question.
This uses Racket's pattern matching, probably in a naïve way, to do some of the work.
(define/match (simplify expression)
;; simplifier driver
(((cons op args))
;; An operator with some arguments
;; Note that this assumes that the arguments to operators are always
;; expressions to simplify, so the recursive level can be here
(simplify-op op (map simplify args)))
((expr)
;; anything else
expr))
(define op-table (make-hash))
(define-syntax-rule (define-op-simplifier (op args) form ...)
;; Define a simplifier for op with arguments args
(hash-set! op-table 'op (λ (args) form ...)))
(define (simplify-op op args)
;; Note the slightly arcane fallback: you need to wrap it in a thunk
;; so hash-ref does not try to call it.
((hash-ref op-table op (thunk (λ (args) (cons op args)))) args))
(define-op-simplifier (+ exprs)
;; Simplify (+ ...) by flattening + in its arguments
(let loop ([ftail exprs]
[results '()])
(if (null? ftail)
`(+ ,#(reverse results))
(loop (rest ftail)
(match (first ftail)
[(cons '+ addends)
(append (reverse addends) results)]
[expr (cons expr results)])))))
It is possible to be more aggressive than this. For instance we can coalesce runs of literal numbers, so we can simplify (+ 1 2 3 a 4) to
(+ 6 a 4) (note it is not safe in general to further simplify this to (+ 10 a) unless all arithmetic is exact). Here is a function which does this coalescing for for + and *:
(define (coalesce-literal-numbers f elts)
;; coalesce runs of literal numbers for an operator f.
;; This relies on the fact that (f) returns a good identity for f
;; (so in particular it returns an exact number). Thisis true for Racket
;; and CL and I think any Lisp worth its salt.
;;
;; Note that it's important here that (eqv? 1 1.0) is false.
;;;
(define id (f))
(let loop ([tail elts]
[accum id]
[results '()])
(cond [(null? tail)
(if (not (eqv? accum id))
(reverse (cons accum results))
(reverse results))]
[(number? (first tail))
(loop (rest tail)
(f accum (first tail))
results)]
[(eqv? accum id)
(loop (rest tail)
accum
(cons (first tail) results))]
[else
(loop (rest tail)
id
(list* (first tail) accum results))])))
And here is a modified simplifier for + which uses this. As well as coalescing it notices that (+ x) can be simplified to x.
(define-op-simplifier (+ exprs)
;; Simplify (+ ...) by flattening + in its arguments
(let loop ([ftail exprs]
[results '()])
(if (null? ftail)
(let ([coalesced (coalesce-literal-numbers + (reverse results))])
(match coalesced
[(list something)
something]
[exprs
`(+ ,#exprs)]))
(loop (rest ftail)
(match (first ftail)
[(cons '+ addends)
(append (reverse addends) results)]
[expr (cons expr results)])))))
Here is an example of using this enhanced simplifier:
> (simplify 'a)
'a
> (simplify 1)
1
> (simplify '(+ 1 a))
'(+ 1 a)
> (simplify '(+ a (+ b c)))
'(+ a b c)
> (simplify '(+ 1 (+ 3 c) 4))
'(+ 4 c 4)
> (simplify '(+ 1 2 3))
6
For yet more value you can notice that the simplifier for * is really the same, and change things to this:
(define (simplify-arith-op op fn exprs)
(let loop ([ftail exprs]
[results '()])
(if (null? ftail)
(let ([coalesced (coalesce-literal-numbers fn (reverse results))])
(match coalesced
[(list something)
something]
['()
(fn)]
[exprs
`(,op ,#exprs)]))
(loop (rest ftail)
(match (first ftail)
[(cons the-op addends)
#:when (eqv? the-op op)
(append (reverse addends) results)]
[expr (cons expr results)])))))
(define-op-simplifier (+ exprs)
(simplify-arith-op '+ + exprs))
(define-op-simplifier (* exprs)
(simplify-arith-op '* * exprs))
And now
(simplify '(+ a (* 1 2 (+ 4 5)) (* 3 4) 6 (* b)))
'(+ a 36 b)
Which is reasonably neat.
You can go further than this, For instance when coalescing numbers for an operator you can simply elide sequences of the identity for that operator: (* 1 1 a 1 1 b) can be simplified to (* a b), not (* 1 a 1 b). It may seem silly to do that: who would ever write such an expression, but they can quite easily occur when simplifying complicated expressions.
There is a gist of an elaborated version of this code. It may still be buggy.
When I compile the following code, SBCL complains that g!-unit-value and g!-unit are undefined. I'm not sure how to debug this. As far as I can tell, flatten is failing.
When flatten reaches the unquoted part of defunits, it seems like the entire part is being treated as an atom. Does that sound correct?
The following uses code from the book Let over Lambda:
Paul Graham Utilities
(defun symb (&rest args)
(values (intern (apply #'mkstr args))))
(defun mkstr (&rest args)
(with-output-to-string (s)
(dolist (a args) (princ a s))))
(defun group (source n)
(if (zerop n) (error "zero length"))
(labels ((rec (source acc)
(let ((rest (nthcdr n source)))
(if (consp rest)
(rec rest (cons (subseq source 0 n) acc))
(nreverse (cons source acc))))))
(if source (rec source nil) nil)))
(defun flatten (x)
(labels ((rec (x acc)
(cond ((null x) acc)
((atom x) (cons x acc))
(t (rec (car x) (rec (cdr x) acc))))))
(rec x nil)))
Let Over Lambda Utilities - Chapter 3
(defmacro defmacro/g! (name args &rest body)
(let ((g!-symbols (remove-duplicates
(remove-if-not #'g!-symbol-p
(flatten body)))))
`(defmacro ,name ,args
(let ,(mapcar
(lambda (g!-symbol)
`(,g!-symbol (gensym ,(subseq
(symbol-name g!-symbol)
2))))
g!-symbols)
,#body))))
(defun g!-symbol-p (symbol-to-test)
(and (symbolp symbol-to-test)
(> (length (symbol-name symbol-to-test)) 2)
(string= (symbol-name symbol-to-test)
"G!"
:start1 0
:end1 2)))
(defmacro defmacro! (name args &rest body)
(let* ((o!-symbols (remove-if-not #'o!-symbol-p args))
(g!-symbols (mapcar #'o!-symbol-to-g!-symbol o!-symbols)))
`(defmacro/g! ,name ,args
`(let ,(mapcar #'list (list ,#g!-symbols) (list ,#o!-symbols))
,(progn ,#body)))))
(defun o!-symbol-p (symbol-to-test)
(and (symbolp symbol-to-test)
(> (length (symbol-name symbol-to-test)) 2)
(string= (symbol-name symbol-to-test)
"O!"
:start1 0
:end1 2)))
(defun o!-symbol-to-g!-symbol (o!-symbol)
(symb "G!" (subseq (symbol-name o!-symbol) 2)))
Let Over Lambda - Chapter 5
(defun defunits-chaining (u units prev)
(if (member u prev)
(error "~{ ~a~^ depends on~}"
(cons u prev)))
(let ((spec (find u units :key #'car)))
(if (null spec)
(error "Unknown unit ~a" u)
(let ((chain (second spec)))
(if (listp chain)
(* (car chain)
(defunits-chaining
(second chain)
units
(cons u prev)))
chain)))))
(defmacro! defunits (quantity base-unit &rest units)
`(defmacro ,(symb 'unit-of- quantity)
(,g!-unit-value ,g!-unit)
`(* ,,g!-unit-value
,(case ,g!-unit
((,base-unit) 1)
,#(mapcar (lambda (x)
`((,(car x))
,(defunits-chaining
(car x)
(cons
`(,base-unit 1)
(group units 2))
nil)))
(group units 2))))))
This is kind of tricky:
Problem: you assume that backquote/comma expressions are plain lists.
You need to ask yourself this question:
What is the representation of a backquote/comma expression?
Is it a list?
Actually the full representation is unspecified. See here: CLHS: Section 2.4.6.1 Notes about Backquote
We are using SBCL. See this:
* (setf *print-pretty* nil)
NIL
* '`(a ,b)
(SB-INT:QUASIQUOTE (A #S(SB-IMPL::COMMA :EXPR B :KIND 0)))
So a comma expression is represented by a structure of type SB-IMPL::COMMA. The SBCL developers thought that this representation helps when such backquote lists need to be printed by the pretty printer.
Since your flatten treats structures as atoms, it won't look inside...
But this is the specific representation of SBCL. Clozure CL does something else and LispWorks again does something else.
Clozure CL:
? '`(a ,b)
(LIST* 'A (LIST B))
LispWorks:
CL-USER 87 > '`(a ,b)
(SYSTEM::BQ-LIST (QUOTE A) B)
Debugging
Since you found out that somehow flatten was involved, the next debugging steps are:
First: trace the function flatten and see with which data it is called and what it returns.
Since we are not sure what the data actually is, one can INSPECT it.
A debugging example using SBCL:
* (defun flatten (x)
(inspect x)
(labels ((rec (x acc)
(cond ((null x) acc)
((atom x) (cons x acc))
(t (rec (car x) (rec (cdr x) acc))))))
(rec x nil)))
STYLE-WARNING: redefining COMMON-LISP-USER::FLATTEN in DEFUN
FLATTEN
Above calls INSPECT on the argument data. In Common Lisp, the Inspector usually is something where one can interactively inspect data structures.
As an example we are calling flatten with a backquote expression:
* (flatten '`(a ,b))
The object is a proper list of length 2.
0. 0: SB-INT:QUASIQUOTE
1. 1: (A ,B)
We are in the interactive Inspector. The commands now available:
> help
help for INSPECT:
Q, E - Quit the inspector.
<integer> - Inspect the numbered slot.
R - Redisplay current inspected object.
U - Move upward/backward to previous inspected object.
?, H, Help - Show this help.
<other> - Evaluate the input as an expression.
Within the inspector, the special variable SB-EXT:*INSPECTED* is bound
to the current inspected object, so that it can be referred to in
evaluated expressions.
So the command 1 walks into the data structure, here a list.
> 1
The object is a proper list of length 2.
0. 0: A
1. 1: ,B
Walk in further:
> 1
The object is a STRUCTURE-OBJECT of type SB-IMPL::COMMA.
0. EXPR: B
1. KIND: 0
Here the Inspector tells us that the object is a structure of a certain type. That's what we wanted to know.
We now leave the Inspector using the command q and the flatten function continues and returns a value:
> q
(SB-INT:QUASIQUOTE A ,B)
For anyone else who is trying to get defmacro! to work on SBCL, a temporary solution to this problem is to grope inside the unquote structure during the flatten procedure recursively flatten its contents:
(defun flatten (x)
(labels ((flatten-recursively (x flattening-list)
(cond ((null x) flattening-list)
((eq (type-of x) 'SB-IMPL::COMMA) (flatten-recursively (sb-impl::comma-expr x) flattening-list))
((atom x) (cons x flattening-list))
(t (flatten-recursively (car x) (flatten-recursively (cdr x) flattening-list))))))
(flatten-recursively x nil)))
But this is horribly platform dependant. If I find a better way, I'll post it.
In case anyone's still interested in this one, here are my three cents. My objection to the above modification of flatten is that it might be more naturally useful as it were originally, while the problem with representations of unquote is rather endemic to defmacro/g!. I came up with a not-too-pretty modification of defmacro/g! using features to decide what to do. Namely, when dealing with non-SBCL implementations (#-sbcl) we proceed as before, while in the case of SBCL (#+sbcl) we dig into the sb-impl::comma structure, use its expr attribute when necessary and use equalp in remove-duplicates, as we are now dealing with structures, not symbols. Here's the code:
(defmacro defmacro/g! (name args &rest body)
(let ((syms (remove-duplicates
(remove-if-not #-sbcl #'g!-symbol-p
#+sbcl #'(lambda (s)
(and (sb-impl::comma-p s)
(g!-symbol-p (sb-impl::comma-expr s))))
(flatten body))
:test #-sbcl #'eql #+sbcl #'equalp)))
`(defmacro ,name ,args
(let ,(mapcar
(lambda (s)
`(#-sbcl ,s #+sbcl ,(sb-impl::comma-expr s)
(gensym ,(subseq
#-sbcl
(symbol-name s)
#+sbcl
(symbol-name (sb-impl::comma-expr s))
2))))
syms)
,#body))))
It works with SBCL. I have yet to test it thoroughly on other implementations.
I have a series of expressions to convert from postfix to prefix and I thought that I would try to write a program to do it for me in DrRacket. I am getting stuck with some of the more complex ones such as (10 (1 2 3 +) ^).
I have the very simple case down for (1 2 \*) → (\* 1 2). I have set these expressions up as a list and I know that you have to use cdr/car and recursion to do it but that is where I get stuck.
My inputs will be something along the lines of '(1 2 +).
I have for simple things such as '(1 2 +):
(define ans '())
(define (post-pre lst)
(set! ans (list (last lst) (first lst) (second lst))))
For the more complex stuff I have this (which fails to work correctly):
(define ans '())
(define (post-pre-comp lst)
(cond [(pair? (car lst)) (post-pre-comp (car lst))]
[(pair? (cdr lst)) (post-pre-comp (cdr lst))]
[else (set! ans (list (last lst) (first lst) (second lst)))]))
Obviously I am getting tripped up because (cdr lst) will return a pair most of the time. I'm guessing my structure of the else statement is wrong and I need it to be cons instead of list, but I'm not sure how to get that to work properly in this case.
Were you thinking of something like this?
(define (pp sxp)
(cond
((null? sxp) sxp)
((list? sxp) (let-values (((args op) (split-at-right sxp 1)))
(cons (car op) (map pp args))))
(else sxp)))
then
> (pp '(1 2 *))
'(* 1 2)
> (pp '(10 (1 2 3 +) ^))
'(^ 10 (+ 1 2 3))
Try something like this:
(define (postfix->prefix expr)
(cond
[(and (list? expr) (not (null? expr)))
(define op (last expr))
(define args (drop-right expr 1))
(cons op (map postfix->prefix args))]
[else expr]))
This operates on the structure recursively by using map to call itself on the arguments to each call.
I wonder if it's possible to write a macro in Racket that would translate every form of shape (c(a|d)+r xs), where c(a|d)+r is a regular expression matching car, cdr, caar, cadr, ... etc, into
the corresponding composition of first and rest.
For example, this macro should take (caadr '(1 2 3 4 5)) and transform that to (first (first (rest '(1 2 3 4 5)))).
Something like this in Shen (Mark Tarver's new programming language): https://groups.google.com/group/qilang/browse_thread/thread/131eda1cf60d9094?hl=en
It is very possible to do exactly that in Racket, and in a much shorter way than done above. There are two (not-really) tricks involved:
Using Racket's #%top macro makes it possible to create such bindings-out-of-thin-air. This macro is getting used implicitly around any variable reference that is unbound ("top" because these things are references to toplevel variables).
Macros become much simpler if you make them do the necessary minimum, and leave the rest to a function.
Here's the complete code with comments and tests (the actual code is tiny, ~10 lines).
#lang racket
;; we're going to define our own #%top, so make the real one available
(require (only-in racket [#%top real-top]))
;; in case you want to use this thing as a library for other code
(provide #%top)
;; non-trick#1: doing the real work in a function is almost trivial
(define (c...r path)
(apply compose (map (λ(x) (case x [(#\a) car] [(#\d) cdr])) path)))
;; non-trick#2: define our own #%top, which expands to the above in
;; case of a `c[ad]*r', or to the real `#%top' otherwise.
(define-syntax (#%top stx)
(syntax-case stx ()
[(_ . id)
(let ([m (regexp-match #rx"^c([ad]*)r$"
(symbol->string (syntax-e #'id)))])
(if m
#`(c...r '#,(string->list (cadr m)))
#'(real-top . id)))]))
;; Tests, to see that it works:
(caadadr '(1 (2 (3 4)) 5 6))
(let ([f caadadr]) (f '(1 (2 (3 4)) 5 6))) ; works even as a value
(cr 'bleh)
(cadr '(1 2 3)) ; uses the actual `cadr' since it's bound,
;; (cadr '(1)) ; to see this, note this error message
;; (caddddr '(1)) ; versus the error in this case
(let ([cr list]) (cr 'bleh)) ; lexical scope is still respected
You can certainly write something that takes in a quoted s-expression and outputs the translation as a quoted s-expression.
Start with simply translating well-formed lists like '(#\c #\a #\d #\r) into your first/rest s-expressions.
Now build the solution with symbol?, symbol->string, regexp-match #rx"^c(a|d)+r$", string->list, and map
Traverse the input. If it is a symbol, check the regexp (return as-is if it fails), convert to list, and use your starting translator. Recurse on the nested expressions.
EDIT: here's some badly written code that can translate source-to-source (assuming the purpose is to read the output)
;; translates a list of characters '(#\c #\a #\d #\r)
;; into first and rest equivalents
;; throw first of rst into call
(define (translate-list lst rst)
(cond [(null? lst) (raise #f)]
[(eq? #\c (first lst)) (translate-list (rest lst) rst)]
[(eq? #\r (first lst)) (first rst)]
[(eq? #\a (first lst)) (cons 'first (cons (translate-list (rest lst) rst) '()))]
[(eq? #\d (first lst)) (cons 'rest (cons (translate-list (rest lst) rst) '()))]
[else (raise #f)]))
;; translate the symbol to first/rest if it matches c(a|d)+r
;; pass through otherwise
(define (maybe-translate sym rst)
(if (regexp-match #rx"^c(a|d)+r$" (symbol->string sym))
(translate-list (string->list (symbol->string sym)) rst)
(cons sym rst)))
;; recursively first-restify a quoted s-expression
(define (translate-expression exp)
(cond [(null? exp) null]
[(symbol? (first exp)) (maybe-translate (first exp) (translate-expression (rest exp)))]
[(pair? (first exp)) (cons (translate-expression (first exp)) (translate-expression (rest exp)))]
[else exp]))
'test-2
(define test-2 '(cadr (1 2 3)))
(maybe-translate (first test-2) (rest test-2))
(translate-expression test-2)
(translate-expression '(car (cdar (list (list 1 2) 3))))
(translate-expression '(translate-list '() '(a b c)))
(translate-expression '(() (1 2)))
As mentioned in the comments, I am curious why you'd want a macro. If the purpose is to translate source into something readable, don't you want to capture the output to replace the original?
Let Over Lambda is a book which uses Common Lisp but it has a chapter in which it defines a macro with-all-cxrs that does what you want.
Here's my implementation (now fixed to use call-site's car and cdr, so you can redefine them and they will work correctly):
(define-syntax (biteme stx)
(define (id->string id)
(symbol->string (syntax->datum id)))
(define (decomp id)
(define match (regexp-match #rx"^c([ad])(.*)r$" (id->string id)))
(define func (case (string-ref (cadr match) 0)
((#\a) 'car)
((#\d) 'cdr)))
(datum->syntax id (list func (string->symbol (format "c~ar" (caddr match))))))
(syntax-case stx ()
((_ (c*r x)) (regexp-match #rx"^c[ad]+r$" (id->string #'c*r))
(with-syntax (((a d) (decomp #'c*r)))
(syntax-case #'d (cr)
(cr #'(a x))
(_ #'(a (biteme (d x)))))))))
Examples:
(biteme (car '(1 2 3 4 5 6 7))) ; => 1
(biteme (cadr '(1 2 3 4 5 6 7))) ; => 2
(biteme (cddddr '(1 2 3 4 5 6 7))) ; => (5 6 7)
(biteme (caddddddr '(1 2 3 4 5 6 7))) ; => 7
(let ((car cdr)
(cdr car))
(biteme (cdaaaaar '(1 2 3 4 5 6 7)))) ; => 6
I'm wondering how to implement mutable versions of cadr, caddr, and the likes in Racket without defining each one separately? ie. not
(define (mcadr exp)
(mcar (mcdr exp)))
It seems that for mutable lists or pairs, Racket only supports mcar and mcdr but not the "expanded" versions. Do I need to know and be good at macros to be able to do this?
Here's a macro solution:
#lang racket/base
(require racket/mpair (for-syntax racket/base))
(define-syntax (define-combinations stx)
(syntax-case stx ()
[(_ n) (integer? (syntax-e #'n))
(let ([n (syntax-e #'n)])
(define options (list (cons "a" #'mcar) (cons "d" #'mcdr)))
(define (add-options r)
(apply append
(map (λ (opt)
(map (λ (l) (cons (string-append (car opt) (car l))
(list (cdr opt) (cdr l))))
r))
options)))
(define combinations
(cdddr
(let loop ([n n] [r '(("" . x))])
(if (zero? n) r (append r (loop (sub1 n) (add-options r)))))))
(define (make-name combo)
(let ([s (string->symbol (string-append "mc" (car combo) "r"))])
(datum->syntax stx s stx)))
(with-syntax ([(body ...) (map cdr combinations)]
[(name ...) (map make-name combinations)])
#'(begin (define (name x) body) ...)))]))
(define-combinations 4)
(mcaddr (mlist 1 2 3 4 5))
You could do:
(define mcaar (compose mcar mcar))
(define mcadr (compose mcar mcdr))
;; ...
(define mcddddr (compose mcdr mcdr mcdr mcdr))
But there is no real getting around the repetition. Even in the Racket source (look in racket/src/list.c), the repetition is there, albeit prettified a little with C macros.