Raul,
I cannot parse ident and I cannot get trace to parse it either.
Is there a monadic part a:1 and a dyadic part (u b.1)?
Or is there an adverb 1 :(u b.1) within an adverb?
Neither of these seem plausible because the a: is not compatible.
On 10/22/06, Miller, Raul D <[EMAIL PROTECTED]> wrote:
Pascal Jasmin wrotete:
> I would suggest that the distinction between (0 n $ y1) and
> 0 0 $ y1 never provides any value and always "gets in the way".
If you are only building up lists using , that might be true.
However, if you are manipulating collections using other operations,
such as {. then this is not at all true.
For what it's worth, here's an identity adjective:
ident=:1 :'a:1 :(u b.1)'
For example:
+ident 99
0
+ident i. 9 9
0 0 0 0 0 0 0 0 0
--
Raul
--
(B=) <-----my sig
Brian Schott
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