# StackBBQ

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StackBBQ is an Esolang designed by User:ZCX 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.

Each bit is an instruction. Yes, no loops.

A 1 pushes a True to the stack, and what about 0?

A 0 first pops 3 values, first c, second b, third a.

abc 111: Push = not Pop 110: Push = Pop or Pop 101: Move bottom->top 100: Duplicate 011: Move top 3 to top 010: Swap top 2 001: Pop 1, if 0, input 8, else pop and output 000: Pop a value, if it is 1, Pop and push to the end of the program, then push it back

## 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[-1] == 1: prog += stack.pop(-2) 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="")