It is a canon of J that an empty list has no contents. Thus (# S:0
(0$a:)) is the same as (# S:0 '') .
# S:0 ''
0
Surely this is right: the argument has level 0, so apply # to the
argument in its entirety.
As a demonstration of the canon:
L. 0$a:
0
Henry Rich
On 12/27/2022 8:46 PM, Igor Zhuravlov wrote:
On Mon, Dec 26, 2022 at 3:04 PM Raul Miller <[email protected]> wrote:
I do not understand the comment on this line:
# S: 0 (0 {. a:) NB. output isn't an empty list
0
I mean the result of this sentence execution should be an empty list:
$ # S: 0 (0 {. a:)
0
but it gives a list of length 1:
$ # S: 0 (0 {. a:)
1
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