While working on the Y combinator rosettacode entry ( https://rosettacode.org/wiki/Y_combinator#Explicit_alternate_implementation ), I ran into what looks like a bug.
If I write the combinator this way: Y=:2 :0(0 :0) f=. u (1 :n) (5!:1<'f') f y ) : g=. x&(x`:6) (5!:1<'g') u y ) I get a domain error when I try to use it, because that : line gets lost. It's present when I inspect the 0 :0 result, but it vanishes from the 5!:1 generated result. It looks like the code is trying to simplify for the 4 :0 case, even though it turns out that that is not relevant. Thanks, -- Raul ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
