Python井字棋程序中while循环内in运算符判定失效问题

核心问题是类型不匹配：

• Python中input()函数的返回值默认是字符串（str）类型，你从输入拿到的num是类似"3"的字符串，而gone列表中存储的是已经落子位置的整数（int）类型值，比如3。不同类型的值在in判定时会被视为不相等，所以"3" in [3]的结果是False，直接导致判定失效。
• 你测试not num in gone得到相反效果也可以佐证这一点：因为num in gone本身就一直返回False，取反后变成True，自然会触发相反的判定逻辑。

根据你的存储习惯可以二选一处理：

1. 统一使用整数类型做判定和存储

gone = []
turn = 0
def play(XO,player):
 while True:
  num = input("\n"+player+" enter an available number where you want to put an '"+XO+"'.\n > ")
  # 合法数字转int后再做存在性判定
  while (not num.isdigit()) or (not (1 <= int(num) <= 9)) or (int(num) in gone):
   num = input("\nValue was not an available number.\n"+player+" enter an available number where you want to put an '"+XO+"'.\n > ")
  # 后续存入gone的也是int类型
  num = int(num)
  gone.append(num)
  # 剩余落子逻辑

2. 统一使用字符串类型做判定和存储

gone = []
turn = 0
def play(XO,player):
 while True:
  num = input("\n"+player+" enter an available number where you want to put an '"+XO+"'.\n > ")
  # 直接用字符串做存在性判定
  while (not num.isdigit()) or (not (1 <= int(num) <= 9)) or (num in gone):
   num = input("\nValue was not an available number.\n"+player+" enter an available number where you want to put an '"+XO+"'.\n > ")
  # 后续存入gone的是原始字符串类型
  gone.append(num)
  # 剩余落子逻辑

内容的提问来源于stack exchange，提问作者LeftHandedPsycho