The example for the produce word is basically your solution already.
\ produce help
click the example
It should look like this:
USING: kernel math prettyprint sequences ;
1337 [ dup 0 > ] [ 2/ dup ] produce nip .
{ 668 334 167 83 41 20 10 5 2 1 0 }
This is basically John's solution, too.
Cheers,
Doug
On Tue, Jun 19, 2012 at 8:09 PM, Marshall Lochbaum <mwlochb...@gmail.com> wrote:
> I figured there would be a word designed for that already... Here's my take
> on this anyway.
>
> I would append the items to a sequence as the algorithm goes along:
>
> { } swap [ dup 1 > ] [ dup 2 /i [ suffix ] dip ] while suffix
>
> accomplishes the task.
>
> Marshall
>
> On Tue, Jun 19, 2012 at 10:50 PM, Graham Telfer <gakouse...@hotmail.com>
> wrote:
>>
>> Doug Coleman <doug.coleman@...> writes:
>>
>> >
>> > In general, you shouldn't want to do this, as all Factor words
>> > (besides the one I'm about to show you) need a fixed number of
>> > parameters at compile-time. Macros expand at compile-time, so that's
>> > one way to get around the restriction.
>> >
>> > The other way it to use a slow word called with-datastack:
>> >
>> > { 1 2 } [ + ] with-datastack
>> >
>> > You can also use smart combinators:
>> >
>> > [ 2 1 - 3 ] sum-outputs
>> >
>> > This works because the quotation infers at compile-time, and
>> > sum-outputs is a "smart" combinator that expands based on the number
>> > of outputs inferred.
>> >
>> > You are trying to do a C varargs function, basically, and just as in
>> > C, most Factor words are not varags.
>> >
>> > If you have more details about what you want to program, we could help
>> > you!
>> >
>> > Doug
>>
>> Hi Doug,
>>
>> I want to divide an integer by 2 until it gets to 1.
>>
>> I have this snippet of code that goes into a loop.
>>
>> dup 1 > [ dup 2 /i ] [ ] if
>>
>> When it gets to 1 I want to push the elements into a sequence.
>>
>>
>>
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>
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