Roger Hui wrote: > 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 .
Thank you for taking the time to explain. It became obvious. -- regards, bill ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
