当'else'应该在python基本猜测操作中激活时，如何修复'if'语句激活'and'参数？

Question

这里是新手。我正在尝试应用我迄今为止在 Python 中学到的知识，以更好地理解迄今为止的工作方式。我正在将某人的数字猜谜游戏与 freecodecamp 的猜谜游戏结合起来。当达到“猜测限制”时，我输得正确。但是，当数字输入正确时，用户仍然“输”。

例如在下面的代码中，我用 x = 5 调用函数，当我“猜测”为 5 时，当我希望它是“else”操作时，我仍然会收到“if limit”操作。

我尝试了几种变体，看看是否能弄明白，但无济于事。任何支持将不胜感激。如果有可以指导我的链接或文章，那就太好了。

def numguess(x):
    number = x
    guess = 0
    guess_count = 1
    guess_limit = 5
    limit = False

    while guess != number and not(limit):
        guess = int(input("\nGuess the number: "))
        if guess < number and guess_count <= 4:
            print("too low")
            guess_count += 1
        elif guess > number and guess_count <= 4:
            print("too high")
            guess_count += 1
        else:
            limit = True
        
    if limit:
        print("you lose")
    else:
        print("you win")
numguess(5) 

Answer 1

您可以像这样简化代码：

def numguess(x):
    number = x
    guess = 0
    guess_count = 1
    guess_limit = 5

    while guess != number and guess_count <= guess_limit:
        guess = int(input("\nGuess the number: "))
        if guess < number:
            print("too low")
            guess_count += 1
        elif guess > number:
            print("too high")
            guess_count += 1
        
    if guess_count <= guess_limit:
        print("you win")
    else:
        print("you loose")
numguess(5)

或者，如果你真的想使用limit 变量：

def numguess(x):
    number = x
    guess = 0
    guess_count = 1
    guess_limit = 5
    limit = False

    while guess != number and not(limit):
        guess = int(input("\nGuess the number: "))
        if guess < number:
            print("too low")
            guess_count += 1
        elif guess > number:
            print("too high")
            guess_count += 1
        limit = guess_count > guess_limit
        
    if limit:
        print("you loose")
    else:
        print("you win")
numguess(5)