another expr issue
Jan 05 2010 | 3:13 pm
Hello,
I want to manage 2 expr in one [jit.expr] object ... without success
Could anyone give help.
This:
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."
with
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 ...
thanks.
- Jan 05 2010 | 9:11 pmHere'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 pmaren't you looking for in[0] instead of cell[0]?
- Jan 12 2010 | 1:58 pmThx e.j---------Hello,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 pmjit.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 amwow! stupid boy!So the inverse would be @expr "abs(norm[0] *-1.)"Thanks again e.j
