bug-apl
[Top][All Lists]

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

 From: Elias Mårtenson Subject: Re: [Bug-apl] Bug in the parser? Date: Wed, 26 Nov 2014 23:38:22 +0800

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 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 <address@hidden> wrote:
```
```On 25 November 2014 at 13:38, Juergen Sauermann
```
```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.

```

reply via email to