> Also b&#^:_1 rather than b #^:_1 was mentioned, so that it does work without 
> the
> & seems an extra bonus, is it reliable assuming fit is not needed?

x f^:n y is documented (on the same dictionary page
you cited) as being equivalent to  x&f^:n y  which, 
with redundant parens inserted to make clear how it 
is parsed, is equivalent to  ((x&f)^:n) y .  
Therefore,  b #^:_1 y  is the same as   b&#^:_1 y .
This was previously mentioned on the J Forum in an 
exchange between Henry Rich and myself.

p.s. How is "extra bonus" different from "bonus"?
Is it like a "free gift"?



----- Original Message -----
From: bill lam <[EMAIL PROTECTED]>
Date: Friday, November 3, 2006 6:47 am
Subject: Re: [Jprogramming] Boxing according to a Pattern

> Thanks to Henry and Roger for help.
> 
> it is written in system/extras/help/dictionary/d202n.htm
> b&#^:_1 where b is a boolean list is Expand (whose fill atom f can 
> be specified
> by fit, b&#^:_1!.f)
> 
> I had read this page and this sentence before I asked, Why I 
> failed to notice
> the portion inside parenthesis? (concentration dysfunction?)
> 
> Also b&#^:_1 rather than b #^:_1 was mentioned, so that it does 
> work without the
> & seems an extra bonus, is it reliable assuming fit is not needed?


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