Re: [Bug-apl] Bug in the parser?

[Juergen Sauermann]

Hi Elias, thanks, fixed in SVN 511. /// Jürgen On 11/26/2014 04:38 PM, Elias Mårtenson wrote: Hello Jürgen and thanks for the explanation. Based on what you just explained, I would have assumed the following to work?       m (/⍨) 2 = +/[1] 0 = m ∘.| m←⍳N LENGTH ERROR       m(/⍨)2=+/[1]0=m∘.|m←⍳N But this variation works?       m (/⍨) (2 = +/[1] 0 = m ∘.| m←⍳N) 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 Regards, Elias On 26 November 2014 at 21:45, Juergen Sauermann <juergen.sauerm...@t-online.de> wrote: Hi again, I have analyzed this a bit further. Below are my conclusions. 1. Problem description -------------------------------- Elias' initial question can be reduced to this: Given APL values A and B, say       A←1 2 3 4 5 and       B←1 1 0 1 0 how shall the  following expressions be evaluated:       A /⍨ B       A (/⍨) B       (A) /⍨ B The matter is a bit complicated by the ambiguity of the / token: For historical reasons the token / is ambiguous and could mean the dyadic function compress or the monadic operator reduce. GNU APL resolves this ambiguity at ⎕FX time when possible. For example, in 1 2 3 / B the operator / is  "downgraded" from  "operator reduce" to "function compress" because 1 2 3 is a value and therefore only function compress makes sense. However in A / B it is not known at ⎕FX time whether A is a function or a value. In that case a symbol is assumed to be a function (and hence / an operator) while something in parentheses like (A) is assumed to be a value. That is the reason why A /⍨ B gives a syntax error while (A) /⍨ B does not. IBM APL2 also downgrades / from operator reduce to function compress, but at a later time. The behavior of IBM APL2 in that special case is somewhat sub-optimal because insisting in / being an operator even if it is obviously not leads to the following inconsistency: IBM APL2:       1 (+¨) 1 2       1 (/¨) 1 SYNTAX ERROR GNU APL:       1 (+¨) 1 2       1 (/¨)1  1 2. Alternatives --------------------- I have tested what happens if we would introduce a M M pattern into GNU APL in order to get IBM APL2's behavior. In the above examples (I used ¨ instead of Elias' original ⍨ because ¨ is present in IBM APL2 while ⍨ is not). (+¨) is reduced by a pattern F M (function monadic-operator) to a derived function. In contrast  (/¨) is not reduced because there is no M M pattern (except in the cases where / was downgraded). The M M is shifted  rather than reduced and the first F in a sequence F M M ... causes the whole chain to be unrolled from left to right, This is the difference that Jay has observed between IBM APL2 and the others. Adding a M M rule forces the parser to reduce M M immediately rather than shifting it. After doing that, a few regression testcases did fail. Looking at the test results my impression was that the current behavior of GNU APL is the preferred one. 3. Conclusion -------------------- My conclusion so far is that we should leave things as they are. Putting things in parentheses is, in my opinion, not such a bad thing and it makes programs more explicit and more portable. When choosing between:       A (/⍨) B and       (A) /⍨ B I would recommend the former because that expresses better what is desired and the latter may change at some point in time. /// Jürgen On 11/25/2014 04:01 PM, Juergen Sauermann wrote: Hi Jay, yes, what I meant is that / is called like a dyadic function as in 1 1 1 / 1 2 3. But handling it always like an operator could be a better solution. Currently in GNU APL operators are distinguished from functions which works well except for / and friends which are parsed as function in some contexts and parsed as operator in others. I will look into changing this to making operators accept a non-function left argument. /// Jürgen On 11/25/2014 03:33 PM, Jay Foad wrote: On 25 November 2014 at 14:06, Jay Foad <jay.f...@gmail.com> wrote: On 25 November 2014 at 13:38, Juergen Sauermann <juergen.sauerm...@t-online.de> wrote: I have read the IBM binding rules a number of times but they seem not to help. The problem of these rules is that they give different results in the cases where / is an operator and where / is a function. In IBM APL2 / is always an operator. For example: 1 2/¨3 4 ⍝ GNU APL, NARS2000 and Dyalog: parse as 1 2(/¨)3 4 3 4 4 1 2/¨3 4 ⍝ APL2: parse as (1 2/)¨3 4 3 3 3 4 4 4 Jay.