Overloading a struct constructor?

Question

My question is the same as the one asked here, but the answer given there doesn\'t actually work.

I have a bunch of structs inheriting from a parent struct A

Answer 1

Looking further into the new struct keywords that soegaard mentioned, I think I've come up with a solution that's a bit cleaner and, more importantly, much easier to abstract into a macro (those modules and requires were giving me a really hard time). I thought I'd share it for future reference. Again, this requires one of the nightly Racket builds. I'm using 6.5.0.7.

(Note: the use of case-lambda over defining keyword arguments here is just my preference.)

#lang racket

(require (for-syntax syntax/transformer))

(struct horse (color)
  #:name Horse
  #:constructor-name Horse
  #:transparent)

(define make-horse
  (case-lambda
    [() (Horse "black")]
    [(color) (Horse color)]))

(define-match-expander horse
  ; match expander
  (λ (pat)
    (syntax-case pat ()
      [(_ more ...) #'(Horse more ...)]))

  ; constructor
  ; edit: changing this to use make-variable-like-transformer,
  ;  as suggested in the comments.
  #;(syntax-id-rules ()
      [(_ args ...) (make-horse args ...)]
      [horse Horse])
  (make-variable-like-transformer #'make-horse))

Sample usage:

> (define black-beauty (horse))
> black-beauty
(horse "black")

> (define ginger (horse "red"))
> ginger
(horse "red")

> (match black-beauty
    [(horse color) color])
"black"

> (match ginger
    [(horse color) color])
"red"

> (horse-color black-beauty)
"black"

> (horse-color ginger)
"red"

In short: instantiation, matching, and accessing fields seem to work as you would expect if you'd just used struct. The struct id can also be used as an identifier without any syntax issues. I don't really think that part has much practical use, but I thought it was nice.