[racket] Making a Racket function "recallable"
Sure, sounds good.
The (let loop ([vs '()]) ...) is a 'named let' (http://docs.racket-lang.org/guide/let.html?q=named%20let#(part._.Named_let)) which as the guide says is equivalent to (letrec ([loop (lambda (vs) ...)]) (loop '()) --- so 'loop' is a recursive function that we're using to iterate until (fib) is >= n, accumulating the values in the vs argument. If you're not familiar with letrec, another definition would be
(define (fib-less-than-n n)
(define fib (mk-fib))
(define (loop vs)
(let ([fib-val (fib)])
(if (>= fib-val n)
(reverse vs)
(loop (cons fib-val vs)))))
(loop '()))
Thinking of loops as just a form of recursion can be a little strange at first. There's more information in the Racket guide here- http://docs.racket-lang.org/guide/Lists__Iteration__and_Recursion.html or another page to check out might be http://mitpress.mit.edu/sicp/full-text/sicp/book/node15.html.
Hope this helps,
- Erik
On Feb 14, 2012, at 1:00 AM, Joe Gilray wrote:
> Thanks Erik. Very nice! Although I miss the recursion.
>
> please help me understand how this works, especially (let loop ([vs '()]). What exactly is going on there?
>
> Thanks again,
> -joe
>
> On Mon, Feb 13, 2012 at 9:51 PM, Erik Silkensen <eriksilkensen at gmail.com> wrote:
> Instead of calling fib-less-than-n recursively, you could add a loop inside, for example something like,
>
> (define (fib-less-than-n n)
> (define fib (mk-fib))
> (let loop ([vs '()])
> (let ([fib-val (fib)])
> (if (>= fib-val n)
> (reverse vs)
> (loop (cons fib-val vs))))))
>
> - Erik
>
> On Feb 13, 2012, at 10:41 PM, Joe Gilray wrote:
>
>> Erik,
>>
>> Intriguing idea and it works "flat", but if I write the following I end up with an infinite loop:
>>
>> (define (fib-less-than-n n)
>> (let ([f1 (mk-fib)])
>> (let ([fib-val (f1)])
>> (if (>= fib-val n) '() (cons fib-val (fib-less-than-n n))))))
>>
>> How would I use your idea in this case?
>>
>> thanks!
>> -joe
>>
>> On Mon, Feb 13, 2012 at 9:18 PM, Erik Silkensen <eriksilkensen at gmail.com> wrote:
>> Another option is you could turn change your fib function to be 'mk-fib' that returns a new instance of the fib function each time it's called:
>>
>> (define (mk-fib)
>> (let ([n0 -1] [n1 1])
>> (lambda ()
>> (let ([next (+ n0 n1)])
>> (set! n0 n1)
>> (set! n1 next))
>> n1)))
>>
>> - Erik
>>
>> On Feb 13, 2012, at 10:10 PM, Danny Yoo wrote:
>>
>> > On Mon, Feb 13, 2012 at 11:52 PM, Joe Gilray <jgilray at gmail.com> wrote:
>> >> Warning: extreme newbie question ahead.
>> >>
>> >> I wrote the following fibonacci function:
>> >>
>> >> ; function that returns the next fibonacci number each time it is called
>> >> ; invoke as (fib)
>> >> (define fib
>> >> (let ([n0 -1] [n1 1])
>> >> (lambda ()
>> >> (let ([next (+ n0 n1)])
>> >> (set! n0 n1)
>> >> (set! n1 next))
>> >> n1)))
>> >
>> >
>> > One thing you can do is turn fib into a "sequence", and then from a
>> > sequence into a stream that knows how to remember its previous values.
>> >
>> > Here's what it might look like:
>> >
>> > ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
>> > #lang racket
>> > (require racket/sequence)
>> >
>> > (define fib
>> > (let ([n0 -1] [n1 1])
>> > (lambda ()
>> > (let ([next (+ n0 n1)])
>> > (set! n0 n1)
>> > (set! n1 next))
>> > n1)))
>> >
>> > (define fib-stream
>> > ;; Here's a sequence of the function:
>> > (let ([fib-sequence (in-producer fib 'donttellmecauseithurts)])
>> >
>> > ;; Let's wrap it and turn it into a stream that remembers...
>> > (sequence->stream fib-sequence)))
>> >
>> >
>> > ;; Ok, we've got a stream. Let's look at its first few elements.
>> > (define (peek-fibs n)
>> > (for ([elt fib-stream]
>> > [i (in-range n)])
>> > (displayln elt)))
>> >
>> > (peek-fibs 10)
>> > (printf "-----\n")
>> > (peek-fibs 20)
>> > ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
>> >
>> > ____________________
>> > Racket Users list:
>> > http://lists.racket-lang.org/users
>>
>>
>
>
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