initialize slots without a modifier?

I apologize if this hits the sisc-users list twice; it hasn't shown up for
about 7 hours so I'm retrying...


In the SSS manual chapter 7, there are examples of making slots with
accessors and modifiers:

(define-generics :x :y :y!)
(define-class (<foo>))
(define-class (<bar>))
(define-class (<baz> <foo> <bar>)
  (x :x)
  (y :y :y!))

Then later, the procedure for initialization ("constructor") is explained:

(define-generics :x :x!)
(define-class (<baz>)
  (x :x :x!))
(define-method (initialize (<baz> b) (<number> x)) (:x! b x))
(define baz (make <baz> 2))
(:x baz)  ;=> 2

So what I don't understand is how you would ever modify your slots in
initialize if you didn't create a modifier?  Like, in the first
example, x only has an accessor.  How do I set it?  Is it constant, or
unbound?

I would imagine that the slot descriptor in the define-class specifies
what slots an object has and then their visibility; i.e. in the
following

(define-generics :y :z :z!)
(define-class (<foobar>)
  (x)
  (y :y)
  (z :z :z!))

x is for internal use only, y can be accessed but not modified, z is
read-write (in general, for clients).

Am I off-base here?  What's the right way to think about this and what
am I missing?

Thanks!

P.S.  It might be nice to update the documentation with more
real-world examples, i.e. instead of <foo> and <bar> make like a
simple <queue> or <stack> class that people can relate to better.

Oh, and the section in Chapter 8 about s2j interaction, where there's
a table of "how to do __X__ in Java in SISC" is great...this might not
be a bad idea for the oo chapter too, even though I'm sure SISC's oo
system is a heckuva lot more powerful than Java's.

Just a thought.

-- 
Mike J. Bell on gmail

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