StackBBQ

StackBBQ is an Esolang designed by User:I am islptng.
BBQ stands for Binary-Based Quine(because you have to write quine to perform loops) It's a stack-based binary language, heavily inspired by bct and 1.
Yes, Binary. The program consists of 0s and 1s. And the stack is too.

Programs in StackBBQ consist of a mutable program string made of bits, and an unbounded stack of bits. Execution halts when the end of the program string is reached.

Program string instructions:

• 1: push a 1 (on/true) bit onto the stack
• 0: pop c, then b, then a, concatenate them into the number abc, then run a special command

Special commands:

abc
111 invert the top bit of the stack
110 if the stack has 2 or more values, pop the top two, OR them, and push result.
    if it has one, pop it.
    otherwise, do nothing.
101 move bottom of stack to the top
100 duplicate the top stack bit
011 move third element from the top (top of stack, second, third) to the top of the stack
010 swap top two bits of the stack
001 conditional I/O
000 conditional self modification

Conditional I/O pops the top bit from the stack, then:

• If the bit is 0, then one byte is read from the input stream, each bit of which is pushed to the stack starting from the least significant bit (such that the top of stack is the most significant bit at the end of the command).
• If the bit is instead 1, then pop a bit from the stack into a buffer. If the buffer has a full byte (eight outputs were issued) then output it. The buffer modification is equivalent to buffer <<= 1; buffer |= pop(), so the first of the eight outputs becomes the MSB.

Conditional self modification first pops a value, and checks if the top of the stack is 1, if so, it appends the firstly-popped value to the current program string.

Computational class

In order for looping to occur, the program must periodically reproduce itself using the self modification instruction. This means that a given program of n bits must be able to invoke the self modification instruction at least n times. Since there are no other looping constructs, there must be at least n many 0 instructions in a quine in order to invoke the n self modifications. The self modification instruction consumes 4 stack values (the instruction number + the bit to append), meaning that at least 4n pushes must occur for the quine to work. This supposed quine would need to fit on the order of 4n + n operations in n instructions to reproduce itself. Since the quine only has at most n instructions (and therefore n operations because of a lack of primitive loops) to reproduce itself, a quine is impossible. Due to this, StackBBQ is total.

Interpreter

Quite buggy Python one.

prog = input("Enter program > ")
stack = []

inbuffer = []
outbuffer = []

i = -1
while i < len(prog):
	i = i + 1
	try: prog[i]
	except IndexError: break
	if prog[i] == "1":
		stack.append(1)
	elif prog[i] == "0":
		c = stack.pop()
		b = 2*stack.pop() + c
		a = 4*stack.pop() + b
		if a == 7:
			stack.append(not(stack.pop()))
		if a == 6:
			if len(stack) >= 2:
				stack.append(stack.pop() or stack.pop())
			else:
				stack.pop()
		if a == 5:
			stack.append(stack.pop(0))
		if a == 4:
			stack.append(stack[-1])
		if a == 3:
			stack.append(stack.pop(-3))
		if a == 2:
			stack.append(stack.pop(-2))
		if a == 1:
			ind = stack.pop()
			if ind == 0:
				if len(inbuffer) == 0:
					newbuffer = input() + '\n'
					for j in newbuffer:
						x = ord(i)
						temp = []
						for k in range(8):
							temp.append(x % 2)
							x = int(x/2)
						for k in range(8): inbuffer.append(temp.pop())
				stack.append(inbuffer.pop(0))
			else:
				outbuffer.append(stack.pop())
				if len(outbuffer) >= 8:
					outchar = 0
					for k in range(8):
						outchar *= 2
						outchar += outbuffer.pop(0)
					print(chr(outchar),end="")
		if a == 0:
			if stack[-2] == 1:
				prog += stack.pop()
			else:
				stack.pop()
	else:
		pass
	# print("Instruction No.",i+1,"i.e.",prog[i],"\nProgram:",prog,"\nStack: ", end = "")
	for j in range(len(stack)):
		stack[j] = int(stack[j])
		# print(stack[j], end = "")
	# print("\n")
print("---------- Halts.")
while True: print(end="")