and now corrected!
(defmacro anaphoric-recur [parm-binds expr & parms]
"An anaphoric recursive function that takes a vector of blind
bindable vars, an expression that can handle the bindable vars.
and the parameters. 'self' is used to call the function
recursively."
`(letfn [(~'self ~parm-binds
(do ~expr))]
(~'self ~...@parms)))
> (anaphoric-recur [x] (if (= x 0) 1 (* 2 (self (dec x)))) 5)
32
On Jun 30, 2:13 pm, Tim Robinson <tim.blacks...@gmail.com> wrote:
> Thanks for all the replies. Sorry if my responses are delayed. I'm
> still on newb moderation mode, so I find my response can take 2 - 10
> hours to be published.
>
> > note that z is free in that expression, which looks suspicious.
>
> yup, it was supposed to be x. It was a copy paste error. I normally
> would evaluate the function, but in this case I was expecting errors.
>
> As for acc or other options, I was hoping to create a recursive fn
> that could generically accept an anonymous function as an argument,
> which was what I was
> thinking.
>
> Using a form like:
>
> ((recursive-fn #(do-stuff (self (dec %))) (recur self)) 5)
>
> (obviously not the way - just the notion)
>
> kind of thing.
>
> When I am home tonight I can play aorund with the examples provided.
>
> Thanks,
> Tim
>
> On Jun 30, 8:59 am, Dominic Cooney <dominic.coo...@gmail.com> wrote:
>
>
>
> > If you name the Y combinator, then you can write recursive anonymous
> > functions:
>
> > <http://rosettacode.org/wiki/Y_combinator#Clojure>
>
> > You can't use Clojure's recur as written because it isn't in tail
> > position--the result of the function isn't just the value of the recur
> > expression (in the first instance it is (* 2 (recur ...)) however I
> > note that z is free in that expression, which looks suspicious.
>
> > If your goal is to define a function that computes 2^n with
> > tail-recursion, then consider threading an "accumulator" parameter
> > through the recursive function; something like this:
>
> > ((fun [acc x]
> > (if (= x 0)
> > acc
> > (recur (* 2 acc) (dec x))) 1 5)
>
> > Of course, you might consider a helper wrapper function that supplies
> > the 1 "default" accumulator.
>
> > Many functional languages optimize tail calls to jmp instructions;
> > that's hard to do efficiently in general on the JVM because it doesn't
> > support tail calls. I expect Clojure has designed recur to work the
> > way it does because some algorithms depend on the tail call
> > optimization, but the restrictions of recur make it possible to
> > compile efficiently as a jmp. So it's a nice compromise.
>
> > <http://clojure.org/special_forms#SpecialForms--(recurexprs*)>
>
> > HTH,
>
> > Dominic
>
> > On Wed, Jun 30, 2010 at 2:44 PM, Tim Robinson <tim.blacks...@gmail.com>
> > wrote:
> > > So I am reading On Lisp + some blogs while learning Clojure (I know,
> > > scary stuff :)
>
> > > Anyway, I've been playing around to see if I can get an anonymous
> > > recursive function to work, but alas I am still a n00b and not even
> > > sure what Clojure's approach to this would be.
>
> > > How would I do this in Clojure?:
>
> > > My first attempt:
>
> > > ((fn [x]
> > > (if (= x 0)
> > > 1
> > > (* 2 (recur (dec z))))) 5)
>
> > > Then my second:
>
> > > ((fn [x]
> > > (let [z (if (= x 0)
> > > 1
> > > (* 2 x))]
> > > (recur (dec z)))) 5)
>
> > > Ideally one could do:
>
> > > ((recursive-fn #( if (= % 0) 1 (* 2 %)) (recur it)) 5)
>
> > > Obviously none work, and I believe I understand why. I just don't
> > > understand how to actually do this or if for some reason Clojure
> > > avoids this for some reason.
>
> > > I am not actually trying to accomplish a specific task, so the example
> > > was made to be simple not meaningful. I'm just playing around to learn
> > > recursive functions and Clojure style.
>
> > > Thanks,
> > > Tim
>
> > > --
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