Python 猜谜游戏——if vs while

Question

我正在尝试用 Python 制作一个简单的猜谜游戏。我，用户，有 3 次猜测来猜出正确的数字。我尝试自己做，我使用了if语句，而在正确的解决方案中，应该使用while循环。我的解决方案运行良好，但是当我在第一次或第二次尝试中猜出正确的数字时，我收到一条错误消息，指出第二个/第三个数字的输入缺失，请参见下面的代码。我知道 'break' 只能在 while 循环中使用。有没有办法使用 if 语句来完成这项工作，还是只能通过使用 while 循环来解决？这里是初学者'coder'，请耐心等待。谢谢!

correct = 3

first_number = int(input('Guess: '))

if first_number == correct:
    print('You win!')
elif first_number != correct:
    second_number = int(input('Try again: '))

if second_number == correct:
    print('You win, second guess!')
elif second_number != correct:
    third_number = int(input('Last guess: '))

if third_number == correct:
    print('Finally!')
elif third_number != correct:
    print('You lose!')

Answer 1

python 3.8。我们在 while 循环中使用 walrus 运算符。

之前:

correct = 3

first_number = int(input('Guess: '))

if first_number == correct:
    print('You win!')
elif first_number != correct:
    second_number = int(input('Try again: '))

if second_number == correct:
    print('You win, second guess!')
elif second_number != correct:
    third_number = int(input('Last guess: '))

if third_number == correct:
    print('Finally!')
elif third_number != correct:
    print('You lose!')

之后:

correct = 3

while (first_number := int(input('Guess: ')) is not correct):
    print('Try again!')
else:
    print('You win!')