> 0$a: is one of those places where the implementation
doesn't match the Dictionary:
datatype > 0$a:
boxed
datatype (>"0) 0$a:
boolean
$ > 0$a:
0
$ (>"0) 0$a:
0 0
I think it is important for $ (0$0){i. 10 to be
0 , $ 0{i. 10
and similarly for $ (0$a:){i. 10 to be
0 , $ a:{i. 10
so I think the implementation of { is correct.
The problem, in my opinion, is in defining that two
nouns may match when they are not equivalent. That means
that reasoning such as
I mean (0$0)-:0$<0 so I kind
> of want ((0$0){y) -: (0$0<){y .
which would be sensible if 'match' meant 'equivalent', is
just not valid.
; 0$a: used to produce 0$a:, IIRC. But that was wrong,
because ; removes a level of boxing and it should do that even
if there are no items. So the boolean result is better.
Henry Rich
> -----Original Message-----
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Dan Bron
> Sent: Wednesday, September 27, 2006 5:55 PM
> To: Programming forum
> Subject: [Jprogramming] Dyad { with empty RHA
>
> Is the following intentional and correct?
>
> $ (q =. 0$ 0 ) { i. 10
> 0
> $ (r =. 0$ < 0 ) { i. 10
> 0 10
>
> That is, when 0-:#x then ( ($x) $ y) -:&$ x{y except
> when 32-:3!:0 x , in which case ( ($x) $ ,:y) -:&$ x{y ?
>
> I mean, it makes some sense, because when an empty x is
> "open" (i.e. q ), then it contains 0 integers, which select
> 0 items of y .
>
> However when x is "boxed" (i.e. r ), the boxes that
> aren't there are arbitrary, so to account for all the shapes
> that could have been selected, copies of all of y are provided.
>
> Another interpretation might be that since 32 -: 3!:0 >^:n
> r for any n , it could be that { believes r is trebly
> boxed, and therefore be EXCLUDING zero items of y (i.e.
> selecting all of y ) and reshaping THAT with the (empty)
> shape of x .
>
> But I'm not sure I like it. I mean (0$0)-:0$<0 so I kind
> of want ((0$0){y) -: (0$0<){y . This isn't like -. q
> -:&:{. r because no fills come in to play. Or maybe they do
> under the covers?
>
> Anyway, this came up with a function I believe should be a
> tautology, but isn't:
>
> ravel =: {~ (<@#: i.@:(*/) )@:$
> check =: , -: ravel NB. Theoretical
> tautology
> check i. 10
> 1
> check i. 10 10
> 1
> check i. 0
> 0
>
> Oh, and I thought I understood:
>
> datatype > 0$a: NB. "boxed" is one way of saying
> "the boxes could have held ANY datatype"
> boxed
>
> But then why:
>
> datatype ; 0$a: NB. yeah, I know a:-:<$0 but that
> doesn't make any promises about 0$a:
> boolean
>
> ?
>
> -Dan
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