python - 为什么 While 循环返回一个变量而不返回另一个变量？

Question

我正在尝试自学 Python 并正在为自己构建一个基本的刽子手游戏。我有一点办法，但现在我被卡住了。

我尝试将大部分用于猜测的“计数器”放入一个函数中，然后引用另一个函数来实际计算出猜测是否正确。

所以这是我的计数器函数:

def game(generated_word):
    hangman_lst = []
    guess_num = 0
    for i in range(0, len(generated_word)):
        hangman_lst.append("__")
    print(" ".join(hangman_lst))

    while "__" in hangman_lst:
        if guess_num == 0:
            l = input("This is your first guess. Guess a letter!")
            guess(l, random_word_stripped, hangman_lst, guess_num)
        elif guess_num == 9:
            print("Sorry - you lose!")
            print("The word was " + str(random_word_stripped))
            break
        else:
            l = input("This is guess #" + str(guess_num + 1) + ":")
            guess(l, random_word_stripped, hangman_lst, guess_num)

    if "__" not in hangman_lst:
        print("**Ta-da!** You're a winner!")

这是我用来判断猜测是否正确的函数:

def guess(ltr, word, lst, try_num):
    upper_ltr = ltr.upper()
    if len(upper_ltr) > 1:
        print("That's more than 1 letter, try again:")
        print(" ".join(lst))
    elif len(upper_ltr) < 1:
        print("You didn't enter a letter, try again:")
        print(" ".join(lst))
    elif upper_ltr in lst:
        print("You already guessed that letter, try again:")
        print(" ".join(lst))
    elif upper_ltr not in word:
        print("Sorry, that's incorrect. Try again.")
        print(" ".join(lst))
        try_num += 1
        return try_num
    else:
        for n in range(0, len(word)):
            if word[n] == upper_ltr:
                lst[n] = upper_ltr
        print("Nice job. That's one. You get a bonus guess.")
        print(" ".join(lst))
        return lst

所以它有点起作用，因为 hangman 单词会根据猜测的字母更新。但我无法让计数器工作 - guess_num 似乎总是保持在 0，所以它总是循环遍历这部分:

if guess_num == 0:
    l = input("This is your first guess. Guess a letter!")
    guess(l, random_word_stripped, hangman_lst, guess_num)

但是，guess 函数似乎返回了 lst，那么为什么 try_num 没有返回呢？

干杯!

Answer 1

您需要增加 guess_num。在 while 循环的末尾添加 guess_num = guess_num + 1 可以解决这个问题，但您只希望 guess_num 在猜错时递增，因此您可以这样做。 (我还没有测试过这个)

while "__" in hangman_lst:
    if guess_num == 0:
        l = input("This is your first guess. Guess a letter!")
        if guess(l, random_word_stripped, hangman_lst, guess_num):
            print("good guess")
        else:
            guess_num += 1
    elif guess_num == 9:
        print("Sorry - you lose!")
        print("The word was " + str(random_word_stripped))
        break
    else:
        l = input("This is guess #" + str(guess_num + 1) + ":")
        if guess(l, random_word_stripped, hangman_lst, guess_num):
            print("good guess")
        else:
            guess_num += 1

然后在您的猜测函数中，您需要添加行 return True 或 return False 以根据猜测是否有效返回适当的 True/False 值。

所以猜测函数的一部分

if len(upper_ltr) > 1:
    print("That's more than 1 letter, try again:")
    print(" ".join(lst))

需要改成

if len(upper_ltr) > 1:
    print("That's more than 1 letter, try again:")
    print(" ".join(lst))
    return False

然后您将为每个 if 条件执行此操作。一旦 python 看到 return 语句，它就会退出该函数，并返回您告诉它返回的任何内容，在我们的例子中，这将是 True 或 False。