Bundled cable troubles.

col = {
[1] = 0,	 -- Inactive
[2] = 1,	 -- White
[3] = 2,	 -- Orange
[4] = 4,	 -- Magenta
[5] = 8,	 -- Light Blue
[6] = 16,	-- Yellow
[7] = 32,	-- Lime
[8] = 64,	-- Pink
[9] = 128,   -- Gray
[10] = 256,   -- Light Gray
[11] = 512,   -- Cyan
[12] = 1024,  -- Purple
[13] = 2048,  -- Blue
[14] = 4096,  -- Brown
[15] = 8192,  -- Green
[16] = 16384, -- Red
[17] = 32768, -- Black
}
on = {}
function getOn()
  for i = 1,#col do
	if rs.getBundledOutput("right") == col[i] then
	  table.insert(on, col[i])
	end
  end
end

getOn()
for i = 1,#on do
print(on[i])
end
input = read()
if input then  
   if input == "white" then
	 if not on == col[1] then
   for i = 1,#on do
			rs.setbundledOutput("right", col[2]+on[i])
	  end
	 else
		 rs.setBundledOutput("right", col[2])
  end
	end
	
	if input == "green" then
   if not on == col[1] then
   for i = 1,#on do
			rs.setbundledOutput("right", col[2]+on[i])
	  end
  else
		 rs.setBundledOutput("right", col[15])  
  end  
end
	
	if input == "red" then
	 if not on == col[1] then
   for i = 1,#on do
		  rs.setbundledOutput("right", col[2]+on[i])
	 end
  else
   rs.setBundledOutput("right", col[16])
  end
end
end

Dragon53535, on 23 August 2016 - 03:25 PM said:

getBundledOutput doesn't work like you think it does. It gives back a 16bit number in which each of the individual bits corresponds to a different wire being on or off, 1 or 0.

There is a method called colors.test() in which it tests if a number, usually 16 bit color code, contains another number, 16 bit color code.

if(colors.test(rs.getBundledOutput("right"),colors.green)) then
--#If the input from the right has the color green on.
end

You can use this to setup an if-else statement in which it tests for colors that are on/off.
Or you could of course use this to test for if the color is in the output and then save it.

Another way to do so, would be to grab the Output number and work in order backwards and modulus each number, starting at bit 16, and down to bit 1, if the modulus is 0, then the output contains that color.