Right, but both forks (1&* + ^.) and (] + ^.) have rank _ instead of rank
0. The adverb ("0) could act on the current result when the rank is not 0
to conform with the Dictionary, for example, producing,
(1 + %)d._1
(1&* + ^.)"0
(instead of (1&* + ^.)) as J is apparently doing for,
(1 + %)d.0
(1 + %)"0
(1 + %)"0 d.0
(1 + %)"0
On Fri, Oct 6, 2017 at 1:38 AM, Arie Groeneveld <[email protected]>
wrote:
>
> (1 + %) d._1 NB. Not OK
>> 1&* + ^.
>>
>
> It's the same as (]+^.)
>
> (1&* -: ]) i.10
> 1
>
>
>
>
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