Extracting known keywords from make-keyword-procedure

[Sorawee Porncharoenwase]

In Python, we can extract known keywords easily:

def f(a, *args, b=2, c=3, **kwargs):
 return (a, args, b, c, kwargs)

print(f(42, 43, c=44, d=45)) #=> (42, (43,), 2, 44, {'d': 45})

Doing so in Racket is not as easy. 

#lang racket/base

(provide lambda/kw define/kw)

(require racket/list

         (for-syntax racket/base))

(define-syntax-rule (lambda/kw (kws kw-args . rst) body ...)

  (let ([f (λ (kws kw-args . rst) body ...)])

    (define-values (ignored used-kws) (procedure-keywords f))

    (make-keyword-procedure

     (λ (actual/kws actual/kw-args . actual/rest)

       (define-values (used-kws+kw-args left-kws+kw-args)

         (partition (λ (e) (member (car e) used-kws))

                    (map cons actual/kws actual/kw-args)))

       (keyword-apply f

                      (map car used-kws+kw-args)

                      (map cdr used-kws+kw-args)

                      (list* (map car left-kws+kw-args)

                             (map cdr left-kws+kw-args)

                             actual/rest))))))

(define-syntax-rule (define/kw (f . rst) body ...)

  (define f (lambda/kw rst body ...)))

(module+ test

  (require rackunit)

  (define/kw (f kws kw-args #:b b #:c [c 16] #:d [d 32] a . rst)

    (list kws kw-args a b c d rst))

  (check-equal? (f 1 2 #:b 4 #:c 8 #:e 64) '((#:e) (64) 1 4 8 32 (2))))

While I am happy with the above macro, I wonder if this could be a part of `make-keyword-procedure`, so that we can simply use:

  (define g (make-keyword-procedure

             (λ (kws kw-args #:b b #:c [c 16] #:d [d 32] a . rst)

               (list kws kw-args a b c d rst))))

  (check-equal? (g 1 2 #:b 4 #:c 8 #:e 64) '((#:e) (64) 1 4 8 32 (2)))