the following attempts to code the one line pattern with one less if. : if. test1 y do. if. -. test2 x do. 'test2 failed' return. end. end.
The problem is that the verb sideff is called even if it shouldn't be. test=: 3 : 0 if. -. (2 = sideff y)"_ ^: (*./ y) 1 do. 'bad' return. end. ) sideff =: 3 : 'a=: +/ y' test 1 2 bad a 3 test 1 1 a 2 test 1 0 a 1 is this just due to nature of parentheses, and the interpreter needing to parse what is on other side of "_ This works as a one liner including shortcircuiting out the side effect when appropriate : 3 (2 = sideff)@:[ ^: (* 0) 1 1 3 (2 = sideff)@:[ ^: (* 1) 1 0 2 (2 = sideff)@:[ ^: (* 1) 1 1 but I don't understand why this fails: guard =: 2 : 0 : (u@:[)^: (v y) 1 ) 3 (2 = sideff) guard * 1 0 2 (2 = sideff) guard * 1 NB. should be 1 0 is there a way to define guard to properly short circuit and give the right answer? Is it impossible if the u argument evaluates to a noun as it was used in first example? ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
