Python猜数字游戏函数异常排查与优化请求

问题描述 #

我正在提升Python技能，编写了一个机器猜数字的游戏函数，但运行出现异常：首次输入1（表示猜中）后，程序重复显示输入提示语句再输出恭喜信息，而非直接提示成功。

错误输出：

We will play a guess game.
Is this 50 your number?
0 means too low, 1 this is the number and 2 means too high 1
0 means too low, 1 this is the number and 2 means too high 1
Congrats ! You guess after 1 tries

原函数代码：

def guess_game():
    m = 50
    count = 0
    print('We will play a guess game.')
    print('Is this ' + str(m) + ' your number?')
    myinput = input('0 means too low, 1 this is the number and 2 means too high')
    while myinput != 1:
        myinput = input('0 means too low, 1 this is the number and 2 means too high')
        count += 1
        if myinput == '0':
            print('Not the right one. Too low')
            for m in range(m,m+1):
                m += 1
            print('Is this ' + str(m) + ' your number?')
        elif myinput == '2':
            print('Not the right one. Too high.')
            for m in range(m,m+1):
                m -= 1
            print('Is this ' + str(m) + ' your number?')
        else: 
            print('Congrats ! You guess after ' + str(count) + ' tries')
            break

问题根源 #

1. 类型不匹配：input()返回的是字符串，而循环条件判断的是myinput != 1（整数），导致第一次输入'1'时，字符串'1'不等于整数1，循环会继续执行，所以会再次弹出输入提示。
2. 冗余代码：用for m in range(m,m+1)来修改m完全没必要，直接m +=1或m -=1即可。
3. 计数逻辑错误：首次猜中时，count被额外加了1，但这实际上是第一次尝试，计数时机不对。

修复后的代码 #

def guess_game():
    m = 50
    count = 1  # 第一次猜就算一次尝试
    print('We will play a guess game.')
    print(f'Is this {m} your number?')
    
    while True:
        try:
            myinput = int(input('0 means too low, 1 this is the number and 2 means too high: '))
        except ValueError:
            print('请输入0、1或2中的一个数字！')
            continue
            
        if myinput == 1:
            print(f'Congrats ! You guessed right after {count} tries')
            break
        elif myinput == 0:
            print('Not the right one. Too low')
            m += 1
            print(f'Is this {m} your number?')
            count += 1
        elif myinput == 2:
            print('Not the right one. Too high.')
            m -= 1
            print(f'Is this {m} your number?')
            count += 1
        else:
            print('请输入0、1或2中的一个数字！')

优化建议 #

• 改用二分法猜数：不要每次只±1，而是维护一个猜数范围（比如1-100），每次取中间值，大幅减少猜测次数。示例：

  def guess_game():
      low = 1
      high = 100
      count = 0
      print('请想一个1到100之间的数字，我来猜！')
      
      while low <= high:
          count += 1
          m = (low + high) // 2
          print(f'Is this {m} your number?')
          
          try:
            myinput = int(input('0 means too low, 1 this is the number and 2 means too high: '))
          except ValueError:
              print('请输入0、1或2！')
              count -=1  # 无效输入不计入次数
              continue
              
          if myinput == 1:
              print(f'恭喜！我用了{count}次猜中了！')
              return
          elif myinput == 0:
              low = m + 1
          elif myinput == 2:
              high = m - 1
          else:
              print('请输入0、1或2！')
              count -=1
      print('你是不是作弊了？我找不到这个数字！')
  

• 使用f-string格式化字符串：比str()拼接更简洁易读。
• 增加输入验证：捕获非数字输入，提示用户正确输入，避免程序崩溃。
• 明确猜数范围：给用户明确的数字范围，比如1-100，让游戏规则更清晰。

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