Yes that is it.
>I think it's confusing because X (a0 a1) in this case is not the same as (X a0) a1 and not the same as (the implied parentheses created that usually work): X a0 a1 The workarounds are to use a linear representation of the adverb train and another adverb that combines it with its argument. (happy to share if any interest) ________________________________ From: Thomas Costigliola <[email protected]> To: [email protected] Sent: Tuesday, March 15, 2016 12:12 PM Subject: Re: [Jprogramming] Unbox request for requests I believe this is what Pascal is trying to demonstrate (correct me if I am wrong): 0 (1 : '/') \ /\ 0 ((1 : '/') \) |syntax error | 0((1 :'/')\) A sequence of adverb applications when the first application returns an adverb is okay (it produces the adverb train), however, applied as a train of adverbs, which one might expect to behave the same, it is a syntax error. I think it's confusing because X (a0 a1) in this case is not the same as (X a0) a1 On 03/13/2016 10:32 AM, Raul Miller wrote: > As you point out, this works: > + - (1 : '`u') `:6 > + - > > If we fully parenthesize that expression, it looks like this: > require 'trace' > paren '+ - (1 : ''`u'') `:6' > ((+ (- (1 : '`u'))) `: 6) > > So when you put parenthesis around (1 : '`u') `:6 you break the expression. > > And, as an aside, your original expression did not actually need any > parenthesis: > + - 1 : '`u' `:6 > + - > > Thanks, > ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
