Consider the following two pieces of Racket code:
;version A
(define-syntax (b stx)
(syntax-case stx () [(X u) #'(display (syntax->datum #'(X u v)))]))
(b 1)
and
;version B
(define-for-syntax (g stx)
(syntax-case stx () [(X u) #'(display (syntax->datum #'(X u v)))]))
(define-syntax (b s) (g s))
(b 1)
Experimentally, both programs return (b 1 v). Question:
Is "version B" a valid Racket? If so, is it fully equivalent to "version A"?
Yes, both of these are fine, and do exactly the same thing, as you've noticed.
Related
Is it possible to write an R5RS macro that would "flatten" arbitrarily deep syntax tree?
Example:
(flatten-syntax (a (b (c d)) e)) => (a b c d e)
My endgoal is to have another macro that would work like this:
(declare-tree (a (b (c d)) e))
=>
(begin (define a #f) (define b #f) (define c #f) (define d #f) (define e #f))
but it should be easy to define if flatten-syntax is available.
The most challenging part for me here is the syntax-rules-only restriction, but if you have syntax-case solution, please also post it.
My progress on this problem stalled at this point:
(define-syntax flatten-syntax-helper
(syntax-rules ()
((_ buf (x . xs))
(flatten-syntax-helper
(flatten-syntax-helper buf x) xs))
((_ buf ())
buf)
((_ buf x)
(x . buf))))
(define-syntax-rule (my-flatten-syntax T)
(flatten-syntax-helper () T))
In guile, ,expand (my-flatten-syntax (a (b (c d)) e)) results in syntax error "failed to match any pattern in form my-flatten-syntax".
Here's my quick attempt:
#lang racket
(define-syntax reverse-macro
(syntax-rules ()
[(_ () (result ...)) '(result ...)]
[(_ (x xs ...) (result ...)) (reverse-macro (xs ...) (x result ...))]))
(define-syntax flatten-syntax-aux
(syntax-rules ()
[(_ ((xs ...) ys ...) (result ...))
(flatten-syntax-aux (xs ... ys ...) (result ...))]
[(_ (x xs ...) (result ...))
(flatten-syntax-aux (xs ...) (x result ...))]
[(_ () (result ...))
(reverse-macro (result ...) ())]))
(define-syntax-rule (flatten-syntax xs)
(flatten-syntax-aux xs ()))
(flatten-syntax (a (b (c d)) e)) ;=> '(a b c d e)
Indeed, you can create declare-tree by adjusting flatten-syntax a little bit, but it might be surprising to you that defining declare-tree directly is in fact much easier:
#lang racket
(define-syntax declare-tree
(syntax-rules ()
[(_ ((xs ...) ys ...))
(begin (declare-tree (xs ...))
(declare-tree (ys ...)))]
[(_ (x xs ...))
(begin (define x #f)
(declare-tree (xs ...)))]
[(_ ())
(begin)]))
(declare-tree (a (b (c d)) e))
(list a b c d e) ;=> '(#f #f #f #f #f)
This is because declare-tree actually doesn't need to flatten the structure. It can generate nested begin, like:
(begin
(begin (define a #f)
(define b #f))
(begin (define c #f)
(define d #f)))
I'm wondering if there is a way to do recursive expansion of macros?
(define-syntax my-define
(syntax-rules ()
[(my-define (fn v ...) body) #'(define (fn v ...) body)]))
(define-syntax my-let
(syntax-rules ()
[(my-let ([v e] ...) body) #'(let ([v e] ...) body)]))
;(my-define (f1 a) a)
; this returns (define (f1 a) a)
;(my-let ([x 10]) x)
; this returns (let ([x 10]) x)
(my-define (f1 a) (my-let ([x 10]) x))
; but this returns (define (f1 a) (my-let [x 10] x)))
The nested case is somehow not expanded. Am I doing something wrong?
Seems to work fine, when you remove the #' (which you probably put it in to debug):
#lang racket
(define-syntax my-define
(syntax-rules ()
[(my-define (fn v ...) body)
(define (fn v ...) body)]))
(define-syntax my-let
(syntax-rules ()
[(my-let ([v e] ...) body)
(let ([v e] ...) body)]))
(my-define (fact n)
(my-let ([k (- n 1)])
(if (zero? n) 1 (* n (fact k)))))
(fact 5)
I'm trying to write a macro behaving just like racket define, but processing fully expanded racket procedures in some way (just expanding for simplicity in the example below):
(define-syntax (define/expand stx)
(syntax-case stx ()
[(_ (head args ...) body body-rest ...)
(let* ([define/racket (syntax/loc stx (define (head args ...) body body-rest ...))]
[fully-expanded (local-expand define/racket 'top-level (list))])
fully-expanded)]
[(_ id expr) (syntax/loc stx (define id expr))]))
Everything is fine unless recursive definition is met:
(define/expand (sum n)
(if (<= n 0)
0
(+ n (sum (- n 1)))))
Running it raises an error
sum: unbound identifier in module in: sum
pointing the call of the sum (not the definition). Obviously the definition of sum is not captured by local expander. I've tried a straightforward way of fixing it: creating new local definition context and binding head into it:
(define-syntax (define/expand stx)
(syntax-case stx ()
[(_ (head args ...) body body-rest ...)
(let* ([ctx (syntax-local-make-definition-context)] ; <- These two lines added
[_ (syntax-local-bind-syntaxes (list #'head) #f ctx)] ; <--/
[define/racket (syntax/loc stx (define (head args ...) body body-rest ...))]
[fully-expanded (local-expand define/racket 'top-level (list) ctx)])
fully-expanded)]
[(_ id expr) (syntax/loc stx (define id expr))]))
It solves the problem (local expanded successfully expands the procedure into define-values), but creates the other one:
module: out-of-context identifier for definition in: sum
pointing the definition of sum. The reason is probably that expander binds identifiers to one in ctx instead of head in current context.
Intuitively it does not seem to be a rare problem, but I could not find the solution over the network. I thought that I should somehow use local-expand/capture-lifts and syntax-local-lift-expression, but I don't get how to use it properly. Could someone clarify what's going on and/or give a hint how to fix it?
Let's try your first program in a top-level (repl):
#lang racket
(define-syntax (define/expand stx)
(syntax-case stx ()
[(_ (head args ...) body body-rest ...)
(let*
([define/racket (syntax/loc stx (define (head args ...) body body-rest ...))]
[fully-expanded (local-expand define/racket 'top-level (list))])
fully-expanded)]
[(_ id expr)
(syntax/loc stx (define id expr))]))
and then in the repl:
Welcome to DrRacket, version 6.6.0.3--2016-07-28(-/f) [3m].
Language: racket, with debugging [custom]; memory limit: 1024 MB.
> (define/expand (sum n)
(if (<= n 0)
0
(+ n (sum (- n 1)))))
.#<syntax:3:2 (define-values (sum) (lambda ...>
> (sum 5)
15
This shows that your program works at the top-level.
The reason the same approach doesn't work in a module context
is that the #%module-begin uses partial expansion of forms
to detect definitions before expanding expressions.
In other words define/expand must tell #%module-begin that
it expands into a definition of sum but must delay the use
of local-expand until #%module-begin has detected all
bound identifiers at the module level.
This suggests a two step approach:
#lang racket
(define-syntax (delay-expansion stx)
(syntax-case stx ()
[(_delay-expansion more ...)
(let ([fully-expanded (local-expand #'(lambda () more ...) 'module (list))])
(display fully-expanded)
fully-expanded)]))
(define-syntax (define/expand stx)
(syntax-case stx ()
[(_ (head args ...) body body-rest ...)
(syntax/loc stx
(define (head args ...)
((delay-expansion
body body-rest ...))))]
[(_ id expr)
(syntax/loc stx
(define id expr))]))
(define/expand (sum n)
(if (<= n 0)
0
(+ n (sum (- n 1)))))
(sum 5)
See more here: https://groups.google.com/d/msg/racket-users/RB3inP62SVA/54r6pJL0wMYJ
I am working on a language translator in guile scheme, and need to handle the basic case, where you're trying to convert a single word.
(define var 5)
(translate var)
This should return the string var and not the number 5.
How do I do this using R5RS Scheme macros (the define-syntax style)?
Edit:
I'm translating from Scheme to Coffeescript.
(define-syntax translate
(syntax-rules ()
[(_ v) 'v]))
And if you want a string:
(define-syntax translate
(syntax-rules ()
[(_ v) (symbol->string 'v)]))
Hopefully Guile's compiler is smart enough to fold the resulting expression so it essentially becomes a constant string.
With syntax-case and its guard support:
(define-syntax translate
(lambda (stx)
(syntax-case stx ()
[(_ v) (identifier? #'v)
#'(symbol->string 'v)]
[(_ v) (number? (syntax-e #'v))
#'(number->string v)])))
(I've used square brackets for easy comparison with Eli's answer, however, it's not my usual style. ;-))
But if you're using syntax-case, then you can just as well do the conversion at the syntax level instead of producing code that does it at runtime:
(define-syntax translate
(lambda (stx)
(syntax-case stx ()
[(_ v) (identifier? #'v)
(datum->syntax stx (symbol->string (syntax->datum #'v)))]
[(_ v) (number? (syntax-e #'v))
(datum->syntax stx (number->string (syntax->datum #'v)))])))
The main thing here is that the macro code is now plain scheme, for example, you could abstract the common parts into a helper:
(define-syntax translate
(lambda (stx)
(define (rewrap convert x)
(datum->syntax stx (convert (syntax->datum x))))
(syntax-case stx ()
[(_ v) (identifier? #'v) (rewrap symbol->string #'v)]
[(_ v) (number? (syntax-e #'v)) (rewrap number->string #'v)])))
Along the same lines, if this macro is so simple, then there's no real need for syntax-case, other than pulling out the subexpression:
(define-syntax translate
(lambda (stx)
(syntax-case stx ()
[(_ v) (let ([d (syntax->datum #'v)])
(datum->syntax
stx
((cond [(number? d) number->string]
[(symbol? d) symbol->string])
d)))])))
Note, BTW, that there is no magic in syntax-case -- and in the case of this simple pattern, you could just pull out the value yourself:
(define-syntax translate
(lambda (stx)
(let ([d (cadr (syntax->datum #'v))])
(datum->syntax
stx
((cond [(number? d) number->string]
[(symbol? d) symbol->string])
d)))))
There is some boilerplate stuff that syntax-case does that this last version loses:
If you use the macro in an unexpected way like (translate) then this version will throw an error about cadr instead of a more comprehensible syntax error
Similarly, if you use (translate 1 2) then this version will just silently ignore the 2 instead of an error.
And if it's used with something that is neither an identifier nor a number (eg, (translate (+ 1 2))) then this will depend on the unspecified value that cond returns rather than throwing a syntax error.
The other answers are useful enough already, but I thought I'd just point out that it's possible to generalize this technique in a very useful ways: macros to print out expressions and their results for debugging:
(define-syntax log-expr
(syntax-rules ()
((_ expr)
(let ((result expr))
(write (quote expr))
(display " evaluates to ")
(write result)
(newline)
result))))
I wish to expand
(foo x (f n) (f n) (arbitrary) (f n) ...)
into
(begin (x 'f n) (x 'f n) (arbitrary) (x 'f n) ...)
my attempt is:
(define-syntax foo
(syntax-rules ()
((_ l a ...)
(let-syntax ((f (syntax-rules ()
((_ n) (l (quote f) n)))))
(begin a ...)))))
(define (x t1 t2) (cons t1 t2)) ;; for example only
(define (arbitrary) (cons 'a 'b)) ;; for example only
(foo x (f 1) (f 2) (arbitrary) (f 3))
Using a macro stepper I can see that the first stage of the macro expands to
(let-syntax ((f (syntax-rules () ((_ n) (x 'f n)))))
(begin (f 1) (f 2) (arbitrary) (f 3)))
Which, when evaluated in isolation works perfectly, but when executed as a whole I get an error about f being an undefined identifier. I assume this is an issue in scoping, is this type of macro expansion possible?
Yeah, you need to get f from somewhere -- your macro just makes it up, and therefore it is not visible to users of foo. When you do consider that you need to get it from somewhere, the question is where would you get it from? Here's a fixed version of your code that assumes that it is the first thing in the second subform of foo:
(define-syntax foo
(syntax-rules ()
[(_ l (f a) more ...)
(let-syntax ([f (syntax-rules ()
[(_ n) (l 'f n)])])
(list (f a) more ...))]))
(define (x t1 t2) (cons t1 t2))
(define (arbitrary) (cons 'a 'b))
(foo x (f 1) (f 2) (arbitrary) (f 3))
(I also made it expand into a list to see that all forms are transformed.)
However, if you want a global kind of f to be used inside foo, then you really have to do just that: define a global f. Here's a limited way to do that:
;; no body => using `f' is always an error
(define-syntax f (syntax-rules ()))
(define-syntax foo
(syntax-rules ()
[(_ l a ...) (list (foo-helper l a) ...)]))
(define-syntax foo-helper
(syntax-rules (f) ; match on f and transform it
[(_ l (f n)) (l 'f n)]
[(_ l a) a]))
(define (x t1 t2) (cons t1 t2))
(define (arbitrary) (cons 'a 'b))
(foo x (f 1) (f 2) (arbitrary) (f 3))
The main limitation in this is that it will only work if one of the a forms is using f -- but it won't work if it is nested in an expression. For example, this will throw a syntax error:
(foo x (f 1) (f 2) (arbitrary)
(let ([n 3]) (f n)))
You can imagine complicating foo-helper and make it scan its input recursively, but that's a slippery slope you don't want to get into. (You'll need to make special cases for places like inside a quote, in a binding, etc.)
The way to solve that in Racket (and recently in Guile too) is to use a syntax parameter. Think about this as binding f to the same useless macro using define-syntax-parameter, and then use syntax-parameterize to "adjust" its meaning inside a foo to a macro that does the transformation that you want. Here's how this looks like:
;; needed to get syntax parameters
(require racket/stxparam)
;; same useless definition, but as a syntax parameter
(define-syntax-parameter f (syntax-rules ()))
(define-syntax foo
(syntax-rules ()
[(_ l a ...)
;; adjust it inside these forms
(syntax-parameterize ([f (syntax-rules ()
[(_ n) (l 'f n)])])
(list a ...))]))
(define (x t1 t2) (cons t1 t2))
(define (arbitrary) (cons 'a 'b))
(foo x (f 1) (f 2) (arbitrary)
(let ([n 3]) (f n)))