python - 我是一个初学者程序员，我需要查询return语句

Question

所以我在制作一个猜谜游戏功能时遇到了一些麻烦，该功能可以记录您正确的次数。我目前有两个问题:

1. 代码导致函数激活两次，有没有办法既打印函数的返回值并为其分配变量？
2. 变量“score”永远不会超过 1，因为如果您猜测的数字错误，该函数将不返回任何内容并返回到 0

这是我的代码，一团糟:

def GuessingGame(score):

    rng = random.Random()
    numbertoguess = rng.randrange(1,10)
    guess = int(input ("What is your guess for a number between 1 and 10?"))
    if numbertoguess == guess:
        print ("Great guess! You're correct.")
        int (score = score + 1)
        return score
    else:
        print ("Wrong, the number was "+str(numbertoguess)+".")

playagain = "yes"

score = 0

while playagain == "yes":

    print ("The score is now "+str(GuessingGame(score))+".")
    score = GuessingGame(score)
    playagain = input ("Play again? (yes or no)")
    if playagain != "yes":
        print ("Goodbye")
    else:
        pass

Answer 1

这行实际上是调用函数:

print ("The score is now "+str(GuessingGame(score))+".")

你应该只使用:

print ("The score is now "+ str(score) +".")

score 是一个变量，可以这样使用

要回答第二个问题，您不会在 else: 子句中return。

而不是这个:

if numbertoguess == guess:
    print ("Great guess! You're correct.")
    int (score = score + 1)
    return score
else:
    print ("Wrong, the number was "+str(numbertoguess)+".")

您可以在这两种情况下返回分数，如下所示:

if numbertoguess == guess:
    print ("Great guess! You're correct.")
    int (score = score + 1)
else:
    print ("Wrong, the number was "+str(numbertoguess)+".")
return score

此外，此行可能不会执行您想要的操作:

int (score = score + 1)

没有理由cast这个，只需使用这一行:

score = score + 1

或者:

score += 1

最后一点是 GuessingGame 的样式会更好，如下所示:

guessing_game

根据PEP8