kragen@pobox.com (Kragen Sitaker) writes:
[...]
> > in fact, (+) is the same as (\x y -> x + y) via eta-reduction.
>
> Yes, and (+) 3 5 is the same as 8 via constant folding and function
> application. But '8' is not a useful answer to the question, "How do
> I add two numbers in OCaml?"
i thought you included the calling part to count the number of tokens it took.
Once again you wanted something adding 2 parameters, not adding 3 and 5 ;pp
>
> > Anyway, you can have the same as your stack-based construct in other
> > languages:
> >
> > Lisp's (funcall '+ 3 5)
>
> That isn't the same; that calls + directly (well, actually, it calls
> it after looking up its function binding in the environment). {add}
> or :noname + ; or [+] are freshly-created lambda-expressions that
> simply call + themselves, but are not themselves identical to +.
i still don't agree, but hey, that's how it goes :)
>
> > but this is pretty dumb... Maybe a better closure example would be better.
> > What about adding the 2 parameters + 1.
>
> These aren't closures, FWIW.
oops. typo
[...]
> > Haskell: (\x y -> x + y + 1) 3 5 12 tokens
> > FORTH, PostScript, dc???
>
>
> PostScript: 3 5 {add 1 add} exec 8 tokens
> dc: 3 5 [+1+] x 8 tokens
> FORTH: 3 5 :noname + 1+ ; execute 7 tokens
wow nice.
haskell could also be
((+) . ((+) 1)) 3 5
but since it's longer than the readable version (14 vs 12 tokens), i can't
fight anymore against stack-languages :)
merd version:
- add 2 numbers:
(+)(3, 5) 8 tokens (where (+) *really* is sugar for "x,y -> x+y")
- add 2 numbers + 1:
(x, y -> x + y + 1)(3, 5) 16 tokens
((1 +) ~ (+))(3, 5) 15 tokens
(all space are optional)
PS: http://merd.net/