another expr issue

    Jan 05 2010 | 3:13 pm
    I want to manage 2 expr in one [jit.expr] object ... without success
    Could anyone give help.
    jit.expr @expr "((cell[0]*0.1)*2.)-1." "((cell[0]*0.1)*1.5)-0.75"
    does not work
    but this:
    jit.expr @expr "((cell[0]*0.1)*2.)-1."
    jit.expr @expr "((cell[0]*0.1)*1.5)-0.75"
    packed inside [jit.pack 2] , work nice.
    In Max5 I tryed:
    jit.expr @expr (((cell[0]*0.1)*2.)-1.) (((cell[0]*0.1)*1.5)-0.75)
    jit.expr @expr "((cell[0]*0.1)*2.)-1." "((cell[0]*0.1)*1.5)-0.75"
    jit.expr @expr ((cell[0]*0.1)*2.)-1. ((cell[0]*0.1)*1.5)-0.75
    but it is a hard way ...
    max v2;
    max v2;

    • Jan 05 2010 | 9:11 pm
      Here's a similar problem.
      the side with [expr] is good but my [jit.expr] gives bas result.
      Any help would be greatly appreciated.
    • Jan 05 2010 | 10:09 pm
      aren't you looking for in[0] instead of cell[0]?
    • Jan 12 2010 | 1:58 pm
      Thx e.j
      I want to multiply the expr by its inverse, no success within one [jit.expr],
      and I don't want the [jit.op @op * @val -1.]
      Any idea?
    • Jan 12 2010 | 3:42 pm
      jit.expr @expr "(in[0] * -1.)" just works fine. jit.pwindow represent negative values by black... which is the reason why it gives the impression of not working properly.
    • Jan 14 2010 | 8:32 am
      wow! stupid boy!
      So the inverse would be @expr "abs(norm[0] *-1.)"
      Thanks again e.j