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Re: return macro
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From: |
Damien Mattei |
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Subject: |
Re: return macro |
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Date: |
Thu, 30 Sep 2021 10:46:06 +0200 |
i do not remember well the technic of this commented code, in this case
note that it is 'break' keyword that cause the current continuation being
called and escaping from the loop to the continuing code...
there is no 'return' here but 'break'.
Damien
On Thu, Sep 30, 2021 at 5:38 AM adriano <randomlooser@riseup.net> wrote:
> Il giorno lun, 28/06/2021 alle 03.15 +0200, Taylan Kammer ha scritto:
> > On 28.06.2021 01:10, Damien Mattei wrote:
> > > hi,
> > >
> > > i wanted to create a macro that is used like a function definition
> > > and
> > > allow return using call/cc:
> > >
> > > (define-syntax def
> > > (syntax-rules (return)
> > > ((_ (name args ...) body body* ...)
> > > (define name (lambda (args ...)
> > > (call/cc (lambda (return) body body* ...)))))
> > > ((_ name expr) (define name expr))))
> > >
> > > unfortunaly i got error:
> > >
> > > scheme@(guile-user)> (def (test x) (cond ((= x 3) 7) ((= x 2)
> > > (return 5))
> > > (else 3)))
> > > ;;; <stdin>:2:42: warning: possibly unbound variable `return'
> > > scheme@(guile-user)> (test 2)
> > > ice-9/boot-9.scm:1685:16: In procedure raise-exception:
> > > Unbound variable: return
> > >
> > >
> > > any idea?
> >
> > Hi Damien,
> >
> > This is because of the "hygiene" rule of Scheme, where the notion of
> > "lexical
> > scope" is taken very seriously: for an identifier to be bound, it
> > must be
> > visible in the lexical (textual) surroundings where it has been
> > bound.
> >
> > So for instance, in the following example code:
> >
> > (def (test x)
> > (cond
> > ((= x 3) (return 7))
> > ((= x 2) (return 5))))
> >
> > We can't see a binding for "return" anywhere in the text, therefore
> > it cannot
> > be bound.
> >
> > This is good "default" behavior because it makes code more flexible
> > and easier
> > to understand.
> >
> > An easy way of overcoming this issue is to let the user explicitly
> > name the
> > return identifier however they like:
> >
> > (define-syntax def
> > (syntax-rules ()
> > ((_ (name ret arg ...) body body* ...)
> > (define (name arg ...)
> > (call/cc (lambda (ret) body body* ...))))))
> >
> > Now you could define:
> >
> > (def (test return x)
> > (cond
> > ((= x 3) (return 7))
> > ((= x 2) (return 5))))
> >
> > Or for instance:
> >
> > (def (test blubba x)
> > (cond
> > ((= x 3) (blubba 7))
> > ((= x 2) (blubba 5))))
> >
> > However, sometimes you're sure that you want to make an implicit
> > binding for
> > an identifier, and for those cases you need to write an "unhygienic"
> > macro
> > which can be achieved with the more complex macro system "syntax-
> > case".
> >
> > Here's how your desired macro could be defined. I will use
> > identifiers like
> > "<foo>" just for easier readability; they don't have any special
> > meaning:
> >
> > (define-syntax def
> > (lambda (stx)
> > (syntax-case stx ()
> > ((_ (<name> <arg> ...) <body> <body>* ...)
> > (let ((ret-id (datum->syntax stx 'return)))
> > #`(define (<name> <arg> ...)
> > (call/cc (lambda (#,ret-id) <body> <body>* ...))))))))
> >
> > There's a few things here to take note of:
> >
> > - Unlike with syntax-rules, the syntax-case is contained in a lambda
> > which
> > takes a single argument: a "syntax object" which is passed to
> > syntax-case
> >
> > - Unlike with syntax-rules, the "body" of the macro (where it begins
> > with a
> > 'let') is not immediately part of the generated code; that 'let' is
> > actually
> > executed during compile-time. The body of the macro must result in
> > an
> > object of the type "syntax object" that represents the generated
> > code.
> >
> > - You see that I define a variable called "ret-id" which I bind to
> > the result
> > of the expression:
> >
> > (datum->syntax stx 'return)
> >
> > which means "create a syntax object in the same lexical environment
> > as stx,
> > and is represented by the symbol 'return'."
> >
> > - The actual code generation begins within the #`(...) which is a
> > shorthand
> > for (quasisyntax (...)) just like '(...) is short for (quote
> > (...)). The
> > result of a quasisyntax expression is a syntax object. Basically,
> > it's the
> > most convenient way of creating a syntax object, but like syntax-
> > rules it's
> > also hygienic by default and you need to insert "unhygienic" syntax
> > objects
> > into it explicitly.
> >
> > - Within the quasisyntax, I use #,ret-id which is short for (unsyntax
> > ret-id)
> > to inject the unhygienic syntax object that holds the symbol
> > 'return' into
> > the generated code.
> >
> > For someone used to macros in the Common Lisp or Elisp style, this
> > may seem
> > over-complicated. It's the cost of the "hygienic by default"
> > behavior.
> >
> > By the way I assume that you're just toying around with the language
> > to learn.
> > If you were thinking of using a 'def' macro like this in real code, I
> > would
> > discourage it because there's already a built-in mechanism that
> > allows the
> > programmer something very similar, called 'let/ec':
> >
> > (import (ice-9 control))
> >
> > (define (test x)
> > (let/ec return
> > (cond
> > ((= x 3) (return 7))
> > ((= x 2) (return 5)))))
>
>
>
> What does this return (defined with let/ec) do ?
>
> In the orevious versions I could see the call to call/cc so I could
> (somewhat) figure out the "jump" imlpied by calling return
>
> But in this last case, where is the return behaviour defined ?
>
>
>
>