Gentlemen,
thanks for your patience. I believe that I have now found a
solution that may please both of you. Attached is my testcase
file, please have a look.
I still believe that left values using A⊃V
or A/B are ambiguous
as to whether their result shall be treated like a variable or
like
part of variable (see my previous comment on V← vs. V[]←).
The fact that other APL interpreters decide that in one way or
the other seems not really satisfactory but somewhat
arbitrary.
I would therefore propose to avoid such constructs for the
sake
of portability.
Best Regards,
Jürgen
On 3/9/23 6:56 PM, Hans-Peter Sorge
wrote:
Hi Jürgen,
thank you for considering it.
Your work is admirable.
Best Regards
Hans-Peter
Am 09.03.23 um 17:49 schrieb Dr.
Jürgen Sauermann:
Hi Hans-Peter,
you are probably right. This one looks rather tricky,
therefore
I may need some time. But I am working on it.
Best Regards,
Jürgen
On 3/8/23 9:05 PM, Hans-Peter
Sorge wrote:
Hello Jürgen,
sorry that I'm so insistent
in
APL2 Programming: Language Reference
it is stated:
Selective Specification: Compress can be used for
selective specification:
M←3 2ρι6
M
1 2
3 4
5 6
(1 0/M)←'ABC'
M
A 2
B 4
C 6
Using our example V
V←1 'bc' M
V
1 bc 1 2
3 4
5 6
The selective specification
(1 0/3⊃V)←'ABC'
V
1 bc ABC
does not yield the same content for 3⊃V as for modified
M.
Best Regards
Hans-Peter
Am 08.03.23 um 17:44 schrieb
Dr. Jürgen Sauermann:
Hi,
I fixed a discrepancy between (3⊃V)← and (1/3⊃V)←. SVN
1657.
We know have:
V←1 'bc' (3 3⍴⍳9) ◊
8 ⎕CR V
┌→─────────────┐
│1 ┌→─┐ ┌→────┐│
│ │bc│ ↓1 2 3││
│ └──┘ │4 5 6││
│ │7 8 9││
│ └─────┘│
└ϵ─────────────┘
(3⊃V)←1 ◊ 8 ⎕CR V
┌→───────┐
│1 ┌→─┐ 1│
│ │bc│ │
│ └──┘ │
└ϵ───────┘
(1/3⊃V)←1 ◊ 8 ⎕CR V
┌→───────┐
│1 ┌→─┐ 1│
│ │bc│ │
│ └──┘ │
└ϵ───────┘
However, I don't quite get why, as suggested below,
the results of
(3⊃V)← and of (1/3⊃V)←
should differ?
For me 1/X is the same as X
(except for a length 1 axis added by 1/ when
X is
scalar) and therefore (3⊃V)←1 and
(1/3⊃V)←1 should
IMHO yield the same.
Best Regards,
Jürgen
On 3/6/23 9:16 PM,
Hans-Peter Sorge wrote:
Hello Jürgen,
I agree with your case 1/2 Statement.
The examples I was showing is actually "off by 1".
I was referring to
(1/3⊃V)←1
having
a←1
b←'ABC'
c←3 3⍴⍳9
V←a b c
(3⊃V)
1
2 3 SVN 1657
4 5 6
7 8 9
As expected with case 1:
(3⊃V)←1
V
1 ABC 1
V←a b c
Not
expected:
(1/3⊃V)←1
V
1 ABC 1
Expected:
(1/3⊃V)←1
V
1
ABC 1 1 1
1 1 1
1 1 1
as with
(1/c)←1
c
1 1 1
1 1 1
1 1 1
And that's Dyalog too.
Please restore compatibility:-)
Best Regards
Hans-Peter
Am 06.03.23 um 16:10
schrieb Dr. Jürgen Sauermann:
Gentlemen,
thanks for the discussion, fixed in SVN 1655.
Hans-Peter, I am sorry that this change creates an
incompatibility in your code.
My thinking for the old solution was this:
V←0 0 0 ◊ V←1 ◊ V ∩ case 1.
1
V←0 0 0 ◊ V[]←1 ◊ V ⍝ case
2.
1 1 1
This applies to GNU APL, APL2, and Dyalog. The
question is then if (A⊃V) in
(A⊃B)←X should behave like case 1 or like case 2
above. The case (A⊃B)←X
with nested (A⊃B)is described neither in the "IBM
APL2 Language Reference"
nor in the "ISO 13751" standard, leaving some room
for interpretation.
However, both APL2 and Dyalog agree on case 1 and
therefore I changed
GNU APL to behave the same.
Best Regards,
Jürgen
On 3/4/23 8:25 PM,
Hans-Peter Sorge wrote:
Hi,
Works as expected
⊃'Sue' 'Maria' 'Annalisa'
is an array 3 by 8.
⊂⊃'Susan' 'Mary' 'Annalisa'
is an element (⊂) of a 3 by 8 array (⊃'Susan'
'Mary' 'Annalisa' ).
Finally each element in ⊃'Sue' 'Maria' 'Annalisa'
gets assigned an array of ⊃'Susan' 'Mary'
'Annalisa'
Greetings
Hans-Peter
Am 04.03.23 um 16:53 schrieb Mr. Sunday:
Hi,
I have an issue with reassigning an element of a
nested array. Here is an example.
14535:15a:~% apl --version
BUILDTAG:
---------
Project: GNU APL
Version / SVN: 1.8 / SVN: 1651M
Build Date: 2023-03-02 00:25:07 UTC
Build OS: Darwin 21.6.0 x86_64
config.status: default ./configure options
Archive SVN: 1621
var←0 0 0 ⋄ (1⊃var)←5 4 ⋄ (2⊃var)←3 4⍴⍳12 ⋄
(3⊃var)←⊃'Sue' 'Maria' 'Annalisa' ⋄ var ⋄
(3⊃var)←⊂⊃'Susan' 'Mary' 'Annalisa' ⋄ var
┌→────────────────────────────┐
│┌→──┐ ┌→─────────┐ ┌→───────┐│
││5 4│ ↓1 2 3 4│ ↓Sue ││
│└───┘ │5 6 7 8│ │Maria ││
│ │9 10 11 12│ │Annalisa││
│ └──────────┘ └────────┘│
└ϵ────────────────────────────┘
┌→───────────────────────────────────────────────────────────────────────────────────────────────────────────┐
│┌→──┐ ┌→─────────┐
┌→──────────────────────────────────────────────────────────────────────────────────────┐│
││5 4│ ↓1 2 3 4│ ↓┌→───────┐ ┌→───────┐
┌→───────┐ ┌→───────┐ ┌→───────┐ ┌→───────┐
┌→───────┐ ┌→───────┐││
│└───┘ │5 6 7 8│ │↓Susan │ ↓Susan │
↓Susan │ ↓Susan │ ↓Susan │ ↓Susan │
↓Susan │ ↓Susan │││
│ │9 10 11 12│ ││Mary │ │Mary │
│Mary │ │Mary │ │Mary │ │Mary │
│Mary │ │Mary │││
│ └──────────┘ ││Annalisa│ │Annalisa│
│Annalisa│ │Annalisa│ │Annalisa│ │Annalisa│
│Annalisa│ │Annalisa│││
│ │└────────┘ └────────┘
└────────┘ └────────┘ └────────┘ └────────┘
└────────┘ └────────┘││
│
│
││
│ │┌→───────┐ ┌→───────┐
┌→───────┐ ┌→��──────┐ ┌→───────┐ ┌→───────┐
┌→───────┐ ┌→───────┐││
│ │↓Susan │ ↓Susan │
↓Susan │ ↓Susan │ ↓Susan │ ↓Susan │
↓Susan │ ↓Susan │││
│ ││Mary │ │Mary │
│Mary │ │Mary │ │Mary │ │Mary │
│Mary │ │Mary │││
│ ││Annalisa│ │Annalisa│
│Annalisa│ │Annalisa│ │Annalisa│ │Annalisa│
│Annalisa│ │Annalisa│││
│ │└────────┘ └────────┘
└────────┘ └────────┘ └────────┘ └────────┘
└────────┘ └────────┘││
│
│
││
│ │┌→───────┐ ┌→───────┐
┌→───────┐ ┌→───────┐ ┌→───────┐ ┌→───────┐
┌→───────┐ ┌→───────┐││
│ │↓Susan │ ↓Susan │
↓Susan │ ↓Susan │ ↓Susan │ ↓Susan │
↓Susan │ ↓Susan │││
│ ││Mary │ │Mary │
│Mary │ │Mary │ │Mary │ │Mary │
│Mary │ │Mary │││
│ ││Annalisa│ │Annalisa│
│Annalisa│ │Annalisa│ │Annalisa│ │Annalisa│
│Annalisa│ │Annalisa│││
│ │└────────┘ └────────┘
└────────┘ └────────┘ └────────┘ └────────┘
└────────┘ └────────┘││
│
└ϵ──────────────────────────────────────────────────────────────────────────────────────┘│
└ϵϵ──────────────────────────────────────────────────────────────────────────────────────────────────────────┘
-- Everyday is Sunday.