Brian Schott wrote: > I cannot parse ident and I cannot get trace to parse it either. > Is there a monadic part a:1 and a dyadic part (u b.1)? > Or is there an adverb 1 :(u b.1) within an adverb? > Neither of these seem plausible because the a: is not compatible.
Here's ident: ident=:1 :'a:1 :(u b.1)' The core of this statement is b. According to the dictionary: 'u b. y gives ... the linear representation of the identity function if y is 1'. For example: + b. 1 0 $~ [EMAIL PROTECTED] But to be useful I need to convert this linear representation to a verb. In ident, I'm using 1 :() to perform this conversion. A simpler expression than ident might be 1 :(+ b. 1) 1 : '0 $~ [EMAIL PROTECTED]' Except, I don't want an adverb, I want the verb, so I need to throw an argument at it (which will be ignored) to get the value the adverb produces ''1 :(+ b. 1) 0 $~ [EMAIL PROTECTED] Now I have a verb. But I need to use 'u' in place of + so that I can pass in the original verb I'm deriving from. Thus, the original 1 :'a:1 :()' formulation. Does that make sense? Thanks, -- Raul
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