I don’t quite see what you’re pointing out. You must’ve
misunderstood me, or more probably I have misunderstood
you.

To clarify:
what I meant was that I was surprised because I expected

(-r) -: v”(-r)b.0

and therefor

(-r) -: u@(v”(-r))b.0

and by extension u to be applied to each result
of v applied to each (-r)-cell of the argument.

Now that I know J pretty well, another thing that surprises me
along with this is that all the intrinsic verb ranks are all positive.
This is I guess to be consistent with @ and others’ treatment
of negative rank, but wouldn’t it be logical to have for example

1 _ _1 -: i.b.0   ?

Thanks for your answers.
Louis

> On 08 Aug 2017, at 03:27, Henry Rich <henryhr...@gmail.com> wrote:
> 
> If negative rank passed through compounds,  consider what the result of
> 
> ]@(#"_1) i.2 3
> 
> would be.
> 
> Henry Rich
> 
> On Aug 8, 2017 00:49, "Raul Miller" <rauldmil...@gmail.com> wrote:
> 
>> Negative rank is a convention - it means the rank is relative to the
>> noun rank. I'm not sure what a rank less than 0 would mean otherwise.
>> 
>> Thanks,
>> 
>> --
>> Raul
>> 
>> On Mon, Aug 7, 2017 at 7:15 PM, Louis de Forcrand <ol...@bluewin.ch>
>> wrote:
>>> I see. So negative ranks are sort of placeholders, and are replaced by
>>> positive (effective) ranks internally during evaluation?
>>> Because it could just announce its rank to be negative, and not actually
>>> calculate the effective rank until it is really needed (a lazier
>> effective rank
>>> evaluation if you will):
>>> 
>>>      x u”_1/ y   <->   x u”_1”_1 _ y   NB. evaluate effective rank here
>>> <->   x u”_1”((0>.(#$x)-1) , _) y       NB. and here
>>> <->   x 4 : ‘x u”((0>.(#$x)-1),(0>.(#$y)-1)) y'”((0>.(#$x)-1) , _) y
>>> 
>>> instead of
>>> 
>>>      x u”_1/ y NB. just here
>>> <->   x 4 : ‘x u”((0>.(#$x)-1),(0>.(#$y)-1)) y’/ y
>>> 
>>> What I mean is that I would’ve made v=: u“r with r<0 report rank r, and
>> if an
>>> operator needs to know the rank of v, then just feed it r and let it
>> deal with
>>> calculating the effective rank.
>>> 
>>> Of course their are probably implementation limitations which I do not
>> know of.
>>> 
>>> Thanks for your explanation!
>>> 
>>> Louis
>>> 
>>>> On 07 Aug 2017, at 18:29, Raul Miller <rauldmil...@gmail.com> wrote:
>>>> 
>>>> The rank of +"_1 is infinite because the derived verb has to see the
>>>> full ranks of its arguments to figure out what rank to use for the
>>>> inner verb.
>>>> 
>>>> In other words, -"_1 in -"_1 i.3 3 has an effective rank of 1, but in
>>>> -"_ i.3 it has an effective rank of 0.
>>>> 
>>>> Since it can't know what rank to use until after it sees the nouns,
>>>> its announced rank has to be infinite.
>>>> 
>>>> Thanks,
>>>> 
>>>> --
>>>> Raul
>>>> 
>>>> 
>>>> On Mon, Aug 7, 2017 at 4:40 PM, Louis de Forcrand <ol...@bluewin.ch>
>> wrote:
>>>>>  +"0/~ i.3
>>>>> 0 1 2
>>>>> 1 2 3
>>>>> 2 3 4
>>>>>  +"_1/~ i.3
>>>>> 0 2 4
>>>>>  +"0 b.0
>>>>> 0 0 0
>>>>>  +"_1 b.0
>>>>> _ _ _
>>>>> 
>>>>> I understand that this is dictionary compliant:
>>>>> 
>>>>> "In general, each cell of x is applied to the entire of y . Thus x u/
>> y is equivalent to x u"(lu,_) y where lu is the left rank of u ."
>>>>> 
>>>>>  +"_1 b.0
>>>>> _ _ _
>>>>> 
>>>>> So u"_1/ -: u"_ _ . Wouldn’t it be better though if u"_1/ -: u"_1 _ ,
>>>>> or if (u”_1 b.0) -: _1 _1 _1 (or any other negative rank)?
>>>>> I ran into this while trying to use ,"_1/ , which I can replace by
>>> @{@,&< ,
>>>>> but I still find this strange.
>>>>> 
>>>>> Louis
>>>>> 
>>>>> ----------------------------------------------------------------------
>>>>> For information about J forums see http://www.jsoftware.com/forums.htm
>>>> ----------------------------------------------------------------------
>>>> For information about J forums see http://www.jsoftware.com/forums.htm
>>> 
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