thank you for your explanations dan, double =: 3 : 0 +: y : smoutput x +: y )
2 (double &+)&+: 3 4 12 2 double &(+&+:) 3 4 12 this I understand 4 (double &+) 6 4 12 and this (double &+) 6 12 so this feels like a bug: 2 double &(+&+:) 3 4 12 should be able to parenthesize to force monad. Though I get what is happening to a slight degree, its extremely unusual that x gets modified. ----- Original Message ----- From: Dan Bron <j...@bron.us> To: "programm...@jsoftware.com" <programm...@jsoftware.com> Cc: Sent: Sunday, March 2, 2014 1:59:14 PM Subject: Re: [Jprogramming] bug in & ? Looking at that table, it's clear that the monadic cases of @ and & are identical, so the difference lies in how they combine dyadic verbs. So maybe a simpler or more intuitive way to express the difference is: @ expresses that you want x and y to "meet" _first_, and & expresses you want x and y to meet last. I think this definition will allow us to understand expressions like yours more intuitively. Rather think of grammar and precedence and rules of substitution, we just note that +:&+&+: is composed solely with &, so the combination of x and y will be deferred to the very end, pushed to the leftmost verb; therefore that +: will be called dyadically, and since dyad +: has a Boolean domain (both left and right), it will raise a domain error when invoked with integers (4 and 6 in this case). Sent from my iPhone > On Mar 2, 2014, at 10:59 AM, Dan Bron <j@bron. That tablus> wrote: > > The sentence 2 +&+&+: 3 expands to (+ +: 2) + (+ +: 3) . This might be > clearer if we used 3 distinct verbs: x f&g&h y is (g h x) f (g h y) . > > In f&g, f retains the valence of the whole verb (is ambivalent), and g is > always called monadically. By contrast, in f@g, f is always called > monadically, while g retains the valence of the whole verb (is ambivalent). * > > expression | monad (exp y) | dyad (x exp y) > f&g | f g y | (g x) f (g y) > f@g | f g y | f (x g y) > > (All subject to the rank of g; the rank-independent analogs are &: and @: > respectively). > > -Dan > > * Note that The symmetry is not perfect: in f&g, g could be called at most > twice, if the whole verb is called dyadically (and g is applied monadically > the the right argument and again monadically to the left argument), whereas > in f@g, f will be called exactly once (monadically to the result of g, > whether g itself was called monadically or dyadically; in short, the > asymmetry mirrors the asymmetry of dyads, which take 2 inputs but produce 1 > output). > > Sent from my iPhone > >> On Mar 2, 2014, at 10:33 AM, Pascal Jasmin <godspiral2...@yahoo.ca> wrote: >> >> That is indeed the problem, but the code >> >> 2 +&+&+: 3 >> >> >> is not 2 + 2 +&+: 3 >> >> but rather >> 0 + 2 +&+: 3 >> 10 >> when it should be: >> + 2 +&+: 3 >> 10 >> >> There doesn't seem to be a good reason to insert 0 v in it. This is weird >> though: >> >> 2 ([: +: ])&(+&+:) 3 >> 12 >> >> I don't think it is: >> >> 2+ 2 +&+:3 ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
