On 14 Aug 2006 at 12:19, Miller, Raul D said:
> Fred Bone wrote:
> > Yes; perhaps I wasn't clear. I was talking not about
> > applying the verb with a given rank, but about defining
> > it. I eventually remembered the syntax, for example:
> > wossname =. 4 : 0 " 1 0
> > I still can't find where this is documented.
>
> J syntax is documented in appendix E of the dictionary
> ("Parsing and Execution").
>
> For example, wossname =. 4 : 0 " 1 0 has the syntax
> mark name asgn noun conj noun conj noun
>
> To parse this by the rules of appendix E, you would
> take the above sentence and use it to populate the
> parsing queue, and you'd start with an empty stack.
> Then, you'd shift values from the queue to the stack
> until the queue looks like
> mark name
> and the stack looks like
> asgn noun conj noun conj noun
>
> From appendix E, the first rule that triggers on
> this sentence is rule 4 (Conj), which executes
> 4 : 0, and produces a dyad. Once this has happened,
> the derived sentence has the syntax
> mark name asgn verb conj noun
>
> Again, rule 4 triggers, but this time the conjunction
> is " and the noun is 1 0, and the derived syntax is
> mark name asgn verb
>
> Finally, rule 7 (Is) triggers, and execution of the
> sentence is completed.
Well, yes. But on that argument almost all of the Dictionary is
redundant.
And I still don't see how you can tell from just the parsing rules that
the rank will become the default rank for the verb whose definition is in
the following lines. Let alone that this is the way to go about it.
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