Defining what $: refers to is hard to put into a definition, but I can
tell you what the interpreter does.
During every execution of a verb there is a definition of what $: will
mean if it is encountered. The value of $: is set (and the previous
value pushed onto a stack) in 2 cases:
1. A name is executed. During the execution of the name, $: refers to
the name
2. The parser executes a verb (as [x] v y). $: refers to the executed verb
In
2 +@:* 5
The parser executes only one verb: the anonymous compound +@:* . That's
what $: would be during that execution.
Henry Rich
On 11/29/2018 11:49 AM, David Lambert wrote:
Wonderful example. I might be able to explain to someone else!
On 11/29/18 7:00 AM, [email protected] wrote:
Date: Wed, 28 Nov 2018 17:17:24 -0500
From: Jose Mario Quintana<[email protected]>
To:[email protected]
Subject: Re: [Jprogramming] Recursive verbs
Message-ID:
<CABtFPKu4bFwxO7k3g8vZJ7m_fitPmWC2euh+RJL=ysm07mg...@mail.gmail.com>
Content-Type: text/plain; charset="UTF-8"
David, the trick is to keep in mind: "$: denotes the longest verb that
contains it." This is important because J does not provide a
primitive for
the scope of $: (although one can write a wicked tacit adverb to set the
scope directly to keep everything tacit).
This is what I think is happening,
f =: complete`f_odd`[email protected]
5 f 4
|stack error: f_even
| 5 f 4
Why? Because
5 agenda 4
2
and
5 f_even 4
|stack error: f_even
| 5 f_even 4
Why? Because "the longest verb that contains it." is
f_even
$: (, (_1 2 p. {:))
thus,
2 ($: (, (_1 2 p. {:))) 4
|stack error
| 2 ($:(,(_1 2 p.{:)))4
A similar reasoning applies to,
5 (complete`f_odd`[email protected] f.) 4
|stack error
| x $:(,(_1 2 p.{:))y
Why does the interpreter produce,
complete`f_odd`[email protected] f.
]`(3 : '$: (, (1 2 p. {:)) y' :(4 : 'x $: (, (1 2 p. {:)) y'))`(3 :
'$: (,
(_1 2 p. {:)) y' :(4 : 'x $: (, (_1 2 p. {:)) y'))@.((> * [: >: 2&|) #)
? A simple example can illustrate the rationale. Assume that given the
recursive factorial verb,
fac=. 1:`(* $:@:<:)@.*
one wants to produce a verb which calculates the factorial plus one.
Naturally,
facplusone=. 1 + fac
facplusone 5
121
facplusone f. 5
121
facplusone f.
1 + 3 : '1:`(* $:@:<:)@.* y' :(4 : 'x 1:`(* $:@:<:)@.* y')
(i.e., the interpreter is using an explicit envelope to force a natural
scope)
whereas,
facplusone=. 1 + fac f.
facplusone 5
446
gives the wrong answer (in this case).
I hope it helps
----------------------------------------------------------------------
For information about J forums see http://www.jsoftware.com/forums.htm
---
This email has been checked for viruses by AVG.
https://www.avg.com
----------------------------------------------------------------------
For information about J forums see http://www.jsoftware.com/forums.htm
