The 2nd easiest fix would be to raise a stack error and have a low stack depth
(maybe just 1) for the result of v (in u^:v). Its cool that there is a v and
not just n option.
In terms of imagination,
v can be an explicit function (less hard to return a gerund) that queries state
as well as x and y to form a gerund. It can be as simple a need as to set part
of the gerund in an if statement, which has to be done through a ^:v approach.
> Similarly in the forms u^:[v0]`v1`v2, ([x] v1 y) may not produce a
gerund.
This one I think has no controversy. in u^:v, v has all the information needed
to produce a final gerund. If v needs/wants to only create part of a gerund it
can be called as:
u^:(v0`v1 , v) or
u^:('`v1`v2'&v)
and so the gerund component producing a gerund can be forbiden without
preventing "sophistication" of computing the initial gerund.
----- Original Message -----
From: chris burke <[email protected]>
To: Source forum <[email protected]>
Sent: Wednesday, May 11, 2016 10:02 PM
Subject: [Jsource] In u^:v, v may not return a gerund (proposal)
---------- Forwarded message ----------
From: Henry Rich <[email protected]>
Date: 8 May 2016 at 12:17
Subject: In u^:v, v may not return a gerund (proposal)
To: [email protected]
I am working on an old problem:
f^:] 0&]`]
crashes, because executing ] creates a gerund, which executes to produce
the same gerund, etc.
The easiest fix is to decree that in u^:v, ([x]v y) may not produce a
gerund. Similarly in the forms u^:[v0]`v1`v2, ([x] v1 y) may not produce a
gerund.
Anybody got a problem with that? I can't think of any case where creating
a gerund makes any sense at all.
Henry
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