Raul Miller wrote:
> On Fri, Mar 19, 2010 at 1:43 PM, David Ward Lambert
> <[email protected]> wrote:
>>   NB. j programmers MUST understand data flow.
>>   NB. x f^:3 y  differs from  (f&y)^:3 x
>
> To illustrate this we should find an operation which
> is not associative and which produces a plausible result
> for one of those two cases.

It is easy to show by example that these are different, for example

   x=:2
   y=:4
   f=:-
   x f^:3 y
_2
   (f&y)^:3 x
_10

Best wishes,

John


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