On 10/27/07, david alis <[EMAIL PROTECTED]> wrote:
> Is the following the expected behaviour of using M.
...
> A =: 5#5
> B =: 5?5
> foo_with_M =: 3 : 0 M.
> n =:?1e5
> data =:(<>:n{::aa)n}aa
> n;n{::data
> )
...
> foo_with_M "0 A
Note that you use a value 'aa' here, which you do not define in
your message.
That said, note that A is 5#5, so I would expect that
foo_with_M"0 A to produce a list of five identical results.
And, since 5 does not appear in B, I think it's reasonable
that the items in foo_with_M"0 B be different from those
produced using A.
--
Raul
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