I think the manual page on insert (
http://www.jsoftware.com/help/dictionary/d420.htm ) sufficiently
explains this behavior:
"0 makes the function operate on scalars and, as per the Vocabulary on
Insert (/) :
" If y has no items (that is, 0=#y), the result of u/y is the neutral
or identity element of the function u . A neutral of a function u is a
value e such that x u e ↔ x or e u x ↔ x, for every x in the domain
(or some significant sub-domain such as boolean) of u . This
definition of insertion over an argument having zero items extends
partitioning identities of the form u/y ↔ (u/k{.y) u (u/k}.y) to the
cases k e. 0,#y ."
I do think there is something not right in the vocabulary (excuse the
blasphemy) with the requirement of 0=#y, since # equals 1 for scalars.
Using $$ instead would make it correct though.
Hope this helps
Jan-Pieter
2014-09-10 17:24 GMT+02:00 Joe Bogner <joebog...@gmail.com>:
> On Wed, Sep 10, 2014 at 9:57 AM, Linda Alvord <lindaalv...@verizon.net>
> wrote:
>
>>
>> Here's the example that is puzzling me.
>>
>> *i.5
>> 0 1 1 1 1
>> */"0 i.5
>> 0 1 2 3 4
>>
>>
> It looks like /"0 yields the right side without evaluating the left side
>
> asdfasdf/"0 i.5
>
> 0 1 2 3 4
>
>
> This might be helpful in understanding:
> http://jsoftware.2058.n7.nabble.com/Adverb-and-conjunction-parsing-rules-td2021.html#a2022
> ... I don't totally get it myself so I can't explain it any further than
> sharing what I found.
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