Thanks for joining in.
The amended question had nothing to do with the earlier example
I'm wondering if there's a quick, standard (and easily understood) idiom
(without an internal helper function) to recur on the variable argument list y
The flatten's not ideal - I may want to retain some sub-list structure
(define x 4)
(define t2
(lambda y
(let ((z (car y)))
(display y)(newline)
(display (flatten y))(newline)
(display z)(newline)
(if (> x 0)
(begin
(set! x (- x 1))
;(t2 (cdr y)))
(t2 y))
(void)))))
(t2 1 2 3 4 5 6)
RETURNS
(1 2 3 4 5 6)
(1 2 3 4 5 6)
1
((1 2 3 4 5 6))
(1 2 3 4 5 6)
(1 2 3 4 5 6)
(((1 2 3 4 5 6)))
(1 2 3 4 5 6)
((1 2 3 4 5 6))
((((1 2 3 4 5 6))))
(1 2 3 4 5 6)
(((1 2 3 4 5 6)))
(((((1 2 3 4 5 6)))))
(1 2 3 4 5 6)
((((1 2 3 4 5 6))))
On Tuesday, September 6, 2016 at 9:04:07 PM UTC-4, Jon Zeppieri wrote:
> On Tue, Sep 6, 2016 at 8:50 PM, Jon Zeppieri <zepp...@gmail.com> wrote:
>
> The `(lambda s ...)` is variable-arity, but when you call `ss` internally
> [...]
>
>
> Sorry, I realized after that your `ss` function essentially *is* a wrapper
> around your `subsets` function -- which is exactly what I was suggesting.
> However, there's no reason for it to duplicate so much of the work of
> `subsets`. Since `subsets` works on any list, your definition of `ss` can
> simply be:
>
>
> (define (ss . xs) (subsets xs))
>
>
> The only purpose of `ss` is to package up its arguments as a list and pass
> that list to `subsets`.
>
>
> - Jon
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