p j wrote:
> a=: 3 4
> (;{.(t=.>5!:1<'a'))
> 0
3{.(;{.(t=.>5!:1<'a'))
0
You're looking at a character argument
> (Can't seem to transform result of first expression
> into $ of 0 (blank result).. so that an expression
> like the following returns a 1 where it matches:
> (;{.(t=.>5!:1<'a')) (="0 1) 0 2 3 4 7
> 0 0 0 0 0
Here's a quote from the message you were responding to:
> > test=: -:&(((#~[:+./\1&~:)@$ ($,) ])@:":)
> > Or, use test"0 if you prefer.
Here's how that might apply to this case you've
presented:
(;{.(t=.>5!:1<'a')) (test"0"0 1) 0 2 3 4 7
1 0 0 0 0
Remember: the result of " is a derived verb. You're
not setting values on an existing verb -- you're
creating a new verb.
--
Raul
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