Thanks Jürgen. Yes, with new ⍬ constant ( value, not function ); it is consist and no confusion.
On Wed, Dec 25, 2019 at 7:50 AM Dr. Jürgen Sauermann < mail@jürgen-sauermann.de> wrote: > Hi, > > after some more thinking and considering Elias' comment I have come to the > conclusion that it is cleaner to parse ⍬ as a constant than as a niladic > function. > > In particular since otherwise ⍬ (the empty numeric vector) and ' ' (the > empty > text vector) would bind differently and therefore an unexpected > inconsistency > between ' ' and ⍬ would occur. > > The new parsing is effective as of *SVN 1212*. > > The change only makes a difference in the following situation: > > A. There is an APL literal that contains ⍬, and > B. That literal is the left argument of a dyadic function, and > C. The literal is not enclosed in parentheses. > > For example: > > > *1 ⍬ 3 + 4 * > If ⍬ is parsed as a niladic function (the old scheme) one gets: > > > * 4⎕CR 1 ⍬ 3 + 4 ⍝ 1 ⍬ 3 + 4 ←→ 1 ⍬ (3 + 4) *┏→━━━━━━┓ > ┃1 ┏⊖┓ 7┃ > ┃ ┃0┃ ┃ > ┃ ┗━┛ ┃ > ┗∊━━━━━━┛ > > > If ⍬ is parsed as a constant (which binds stronger to other constants) then > the new scheme gives: > > * 4⎕CR 1 ⍬ 3 + 4 ⍝ 1 ⍬ 3 + 4 ←→ (1 ⍬ 3) + 4* > ┏→━━━━━━┓ > ┃5 ┏⊖┓ 7┃ > ┃ ┃0┃ ┃ > ┃ ┗━┛ ┃ > ┗∊━━━━━━┛ > > > Best Regards, > Jürgen > > > > On 12/24/19 4:36 PM, Dr. Jürgen Sauermann wrote: > > 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 > > On Tue, 24 Dec 2019, 14:48 Dr. Jürgen Sauermann, <mail@jürgen-sauermann.de> > wrote: > >> 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 >> >> On Mon, Dec 23, 2019 at 7:20 PM David Tran <email55...@gmail.com> wrote: >> >>> 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 >>> >>> >>> >>> >> > >
