Hi,
yes. In some sense. Actually
1 2 X 4
is the same as 1 2 (X) 4
because there is a rule that says so. However, the point here is
that
1 2 X 4
can be (very) different from
(1 2 X 4)
The difference does not occur in simple examples, but may in
more complex ones.
1 2
X 4 + 1
is generally different from
(1 2 X 4) + 1
simply because the order of evaluation differs between X and +.
More generally, if
you put the left argument of a dyadic function in parentheses
then that most likely
(though not neccessarily) changes the valuation order (and hence
the result). I know
that this is odd, but so are niladic functions when it comes to
parsing. According to
the IBM APL 2 parentheses rules, the parantheses around (⍬) are
redundant (and
make no difference while the parentheses around (⍳0)
are not redundant. As a
consequence, even though ⍳0 yields essentially ⍬ they are not
entirely equivalent.
Best Regards,
Jürgen Sauermann
On 12/24/19 3:16 PM, Elias Mårtenson
wrote:
I think such a change would make sense. A niladic
function is in some sense a shortcut to writing a specific
value, so it should probably be treated as such.
It a little odd that:
1 2 X 4
is different from:
1 2 (X) 4
Regards,
Elias
Hi David,
not sure either if this is a bug. The APL standard has
quite a specification gap
when it comes to binding strength. In GNU APL, the binding
between values
(aka strand notation) is stronger than the binding between
values and functions.
For example:
V←⍳0
∇Z←F
Z←⍳0
∇
Both V and F result in the same value ⍬:
V≡F
1
However, they are parsed differently:
2 V
3⊃a
3
2 F 3⊃a
2 99
In the first case: 2
V 3⊃a the value V binds strongly to 2 and
3 so that value (2 V 3)
becomes the left argument of ⊃. In the second
case: 2 F 3
the value 3 binds
stronger to ⊃ than to F so that first is
evaluated 3⊃a and then the result of ⊃
is
tied to the F and 2.
In GNU APL ⍬ is a niladic function, so it
behaves like F above.
This behaviour could be changed, but I hesitate to
do that since it might break
existing code. The impact of such a change would
not only affect ⍬ but all niladic
functions.
Best Regards,
Jürgen Sauermann
On 12/24/19 2:49 AM, David Tran wrote:
Ooops, missing something on my examples,
correction: ( missing ⊂ on ⍳3 )
a ← 'abc'(⊂⍳3)99
the 3 examples are the same:
2(⍳0)3⊃a
2⍬3⊃a
(2⍬3)⊃a
Hi,
Not sure this is a bug or not, for me, (⍳0) ≡
⍬, so it seems that both can be replaced each
other; consider below example:
a←'abc'(⍳3)99
2(⍳0)3⊃a ⍝ ≡ 3
2⍬3⊃a ⍝ ≡ 2 ⍬ 99
(2⍬3)⊃a ⍝ ≡ 3
Doesn't the second example should return 3 as
first example, without the need parentheses as
third example?
(btw. my version is build from SVN around Oct )
Thanks,
Dave
|