Python Bool被正确返回，但仍然获胜'；行不通

我是Python新手

我想检查用户的输入，并检查输入是否被接受。该方法称为“check_input”，我在“running”方法第3行中调用了该方法。我给它传递一个字符串和bool变量

然后我想返回inputAccepted值，然后根据其值对其进行处理

我使用了断点，bool本身被正确地更改，但是当代码离开“check_input”方法时，bool“inputAccepted”被忘记了

我做错了什么

我的理论是，bool在方法之外是不可访问的

import random
import collections
import re

codeLength = 4
guessesRemaining = 10
inputAccepted = False

welcomeMessage = 'Welcome to Mastermind. Try and guess the code by using the following letters:\n\nA B C D E F \n\nThe code will NOT contain more than 1 instance of a letter.\n\nAfter you have entered your guess press the "ENTER" key.\nYou have a total of'
print(welcomeMessage, guessesRemaining, 'guesses.\n')


def gen_code(codeLength):
    symbols = ('ABCDEF')
    code = random.sample(symbols, k=codeLength)
    return str(code)

code = gen_code(codeLength)
print(code)
counted = collections.Counter(code)

def check_input(guess, inputAccepted):
    if not re.match("[A-F]", guess): #Only accepts the letters from A-F
        print("Please only use the letters 'ABCDEF.'")
        inputAccepted = False
        return (guess, inputAccepted)

    else:
        inputAccepted = True
        return (guess, inputAccepted)

def running():
    guess = input() #Sets the guess variable to what the user has inputted
    guess = guess.upper() #Converts the guess to uppercase
    check_input(guess, inputAccepted) #Checks if the letter the user put in is valid

    print(guess)
    print(inputAccepted)

    if inputAccepted == True:
        guessCount = collections.Counter(trueGuess)
        close = sum(min(counted[k], guessCount[k]) for k in counted)
        exact = sum(a == b for a,b in zip(code, guess))
        close -= exact
        print('\n','Exact: {}. Close: {}. '.format(exact,close))
        return exact != codeLength
    else:
        print("Input wasnt accepted")

for attempt in range(guessesRemaining):
    if not running():
        print('Done')
        break
    else:
        print('Guesses remaining:', guessesRemaining - 1 - attempt, '\n')

else:
        print('Game over. The code was {}'.format(''.join(code)))

---

非常感谢

您需要查看

检查输入的返回值

，而不是查看输入值

inputAccepted = check_input(guess)

您也没有理由返回最初的猜测，因此我建议重写函数

check\u input

：

def check_input(guess):
    if not re.match("[A-F]", guess): #Only accepts the letters from A-F
        print("Please only use the letters 'ABCDEF.'")
        return False

    else:
        return True

---

如果在函数检查输入中使用“inputAccepted”作为全局变量和形式参数，请在定义函数检查输入时更改参数名称，这可能会解决您的问题。

check\u输入有一个返回值，您不使用该值。您可能打算执行类似于guess，inputAccepted=check\u input（guess，inputAccepted）的操作。我认为您需要

inputAccepted=check\u input（guess，inputAccepted）

。另外，只返回InputAcceptedPossible duplicate的bool值，或者将函数重命名为inputAccepted（guess）并在if@KennyOstrom现在我得到了错误“UnboundLocalError….inputAccepted referenced before assignment”非常感谢！我知道我现在的问题是打印副作用不是很好的做法。一种更具python风格的方法会涉及一个异常，因此您可以稍后打印消息。但是对于一个新的程序员来说，这可能太多了。