[retroforth] Re: RetroForth 8.0
- From: Charles Childers <charles.childers@xxxxxxxxx>
- To: retroforth@xxxxxxxxxxxxx
- Date: Sat, 15 Jan 2005 18:44:59 -0500
> I dont like this at the moment. Could you explain better what happens after
> "does (foo)" ? Is it possible to say:
>
> : (foo) 10 + ;
> : foo create literal, does (foo) s" I created a (foo) doing thing!" cr ;
Yes, though you probably want to use ." or a type after the s". In
this case, you'd get:
10 foo a
I created a (foo) doing thing!
3 foo b
I created a (foo) doing thing!
a
20
b
13
> If so: how do you implement it? Is not it more complicated now, since the
> return address on stack does not play a role?
: does 32 parse find m: literal F: compile M: ;; ;
Overall it's less complicated since it doesn't have to modify already
compiled code. does> had to change the "call dovar" that create
compiles to "call code_after_does" and patch a bit of code after does>
to get the address of the data on the stack.
> "Factoring" is fine. Wasting dictionary with trash is another thing. "does>"
> is the default OO-part of FORTH and making a new "class" of words would need
> to implement two words for implementation now.
loc:
: (foo) 10 + . ;
: foo create literal, does (foo) ;
' foo
; loc alias foo
Now only "foo" will be visible in the dictionary. The loc: ;loc
pairings make local factoring possible, so you can factor more heavily
without actually polluting the overall dictionary.
> Why should I have to factor out for the following things:
>
> : char-out create 1, does> c@ emit ;
> 32 char-out space
> 10 char-out cr
> 9 char-out tab
You could do this:
: char-out create literal, does emit ;
32 char-out space
10 char-out cr
9 char-out tab
--
Charles
Other related posts:
