Talk:Switchy

Interpreter
I have created a Switchy interpreter in ruby: stack = [] labels = {} flip = false counter = 0 print "#{Dir.pwd} filepath> " program = File.open(gets.delete("\n"),"r").read.split(//) while counter < program.count if program[counter] == "("		lbl = ""		counter+=1		while program[counter] != ")" lbl+=program[counter] counter += 1 end labels[lbl] = counter end counter+=1 end counter = 0 while counter < program.count case program[counter] when '#' stack << (flip ? stack.pop - stack.pop : stack.pop + stack.pop) when '*' (flip ? stack.pop : stack.push(1)) when "'" (flip ? (print stack[stack.count-1].chr) : stack << gets.delete("\n")[0]) when "[" lbl = "" counter+=1 while program[counter] != "]" lbl+=program[counter] counter += 1 end counter = labels[lbl] end ((program[counter]=~/[\#|\*|\']/)==0 ? begin flip=!flip;stack.reverse! end : flip=flip) counter+=1 end I think it works, but I have only tested the cat program with it. OriginalOldMan (talk) 17:42, 25 October 2013 (UTC)

Actually, I don't believe the cat program listed on the Switchy page is correct due to the fact that the stack is reversed every cycle. OriginalOldMan (talk) 22:19, 25 October 2013 (UTC)

Dang, you're right. I'll make a fixed cat program. Poolala (talk) 01:42, 26 October 2013 (UTC)

You could add a no-op to make a cat program possible, or there could be a 'Switchy++' with a no-op (of course the no-op would have to trigger a clock cycle). OriginalOldMan (talk) 02:53, 26 October 2013 (UTC)

The reason I made this language was so every program would be difficult to write, and a noop would make it too easy IMO. Poolala (talk) 03:58, 26 October 2013 (UTC)

True. OriginalOldMan (talk) 04:22, 26 October 2013 (UTC)

Does Switchy allow you to jump to labels further forward in the program than the current position? If so, I will need to change the interpreter. OriginalOldMan (talk) 05:56, 26 October 2013 (UTC)

Yes. Poolala (talk) 06:31, 26 October 2013 (UTC)

Compuational Class
It's not Turing complete. You can't get more than one item on the stack at a time. -A program can't start with an addition(# with flip=0) because the stack starts with 0 items. -Therefor it must start with a push 1(* with flip=0) or input(' with flip=0) (or a jump that leads to one of these two things). -The stack now must either have a single 1 or the first input, and flip=1. -The next action can't be a subtraction(# with flip=1) because the stack only has 1 item. -Therefor it must either pop(* with flip=1) or output(' with flip=1) (or a jump that leads to one of these two things) -Both of these pop from the stack, so our stack of either 1 or the first input is now empty again. -The cycle keeps repeating until the code ends JeffryThunderStrike (talk) 19:35, 24 October 2017 (UTC)

Be bold in editing pages --StellatedHexahedron (talk) 15:27, 25 October 2017 (UTC)