【问题标题】:Global variable is not working when trying to assign from the function尝试从函数分配时全局变量不起作用
【发布时间】:2020-08-13 19:43:01
【问题描述】:

我正在学习 Python,但我遇到了 gobal 变量的问题。我已经在全局范围内定义了变量,然后使用 global 关键字从函数中分配它们。但是,当它运行脚本时,它仍然无法分配变量。

看了这个解决方案,Using global variables in a function,我不知道为什么他的代码可以正常工作,而我的相同逻辑却失败了。几个小时都想不通并陷入小问题。

def player_input():
    player_1_name = input('Please enter your name Player 1.')
    player_2_name = input('Please enter your name Player 2.')
    player_1_symbol='None'
    player_2_symbol=''
    allowed_symbol=['X','O']
    symbol_accepted = False

def get_symbol():
    global player_1_symbol
    player_1_symbol = input('Please enter your preferred symbol {}. It can be either X or O.   '.format(player_1_name))
    print('value of symbol after assignment: '+ player_1_symbol)

def validate_symbol():
    global symbol_accepted
    print('value of symbol when process started: '+ player_1_symbol)
    if player_1_symbol.upper() in allowed_symbol:
        print('it passed')
        symbol_accepted = True
    else:
        print('it came here, because it did not find the value of player_1_symbol')
        get_symbol()

while symbol_accepted == False:
    validate_symbol()

if player_1_symbol.upper() == 'O':
    player_2_symbol = 'X'
else:
    player_2_symbol = 'O'

print("{}, unfortunately you don't have choice. You have been assigned the symbol {}.".format(player_2_name,player_2_symbol))

我想要达到的结果是我想 assign player_1_symbol 从 get_symbol 函数到全局变量。 symbol_accepted 也是如此。没有变量被分配。我正在打印它们,但我无法理解为什么它不起作用。我在 youtube 上看到了一些视频和一些文章。

我看到了一些关于输入验证的其他视频/文章,但是一旦我开始这个,我想了解它有什么问题,以便我将来可以使用 global

【问题讨论】:

  • 你是什么意思他们没有被分配?请提供minimal reproducible example。注意,这只是一个函数定义,它永远不会在任何地方调用
  • 注意,虽然player_1_symbolget_symbol 中是全局的,但在player_input 中是本地的,在validate_symbol 中是非本地的。在任何情况下,无论如何你都不应该依赖全局可变状态
  • 所以当我运行脚本时,变量player_1_symbol首先被分配为X,但是当它到达它运行validate_symbol函数的点时,它的值是None。我希望它根据用户输入将值保持为XO
  • 旁注:在player_input() 函数中定义player_1_symbolplayer_2_symbol 真的 没有用,因为您在运行时再次定义它们get_symbol()。第二条评论很好地解释了为什么不应该依赖单个 global 变量(太多的范围限制/堆栈中未使用的内存)。
  • 作为一名程序员,您应该非常、非常非常努力不使用全局变量。

标签: python input global-variables global variable-assignment


【解决方案1】:

我用 player_symbols 创建了一个数组(player_1 为 0,player_2 为 1),因此可以正确分配变量。我还修复了 validate_symbol 重复的错误,因为 symbol_accepted 有相同的错误。

def player_input():
    player_1_name = input('Please enter your name Player 1.')
    player_2_name = input('Please enter your name Player 2.')
    player_symbols = ["", ""]
    allowed_symbol = ['X', 'O']
    symbol_accepted = False

    def get_symbol():
        player_symbols[0] = input(
            'Please enter your preferred symbol {}. It can be either X or O.   '.format(player_1_name))
        print('value of symbol after assignment: ' + player_symbols[0])

    def validate_symbol(symbol_accepted):
        print('value of symbol when process started: ' + player_symbols[0])
        if player_symbols[0].upper() in allowed_symbol:
            print('it passed')
            symbol_accepted = True
        else:
            print('it came here, because it did not find the value of player_1_symbol')
            get_symbol()
        return symbol_accepted

    while symbol_accepted == False:
        symbol_accepted = validate_symbol(symbol_accepted)

    if player_symbols[0].upper() == 'O':
        player_symbols[1] = 'X'
    else:
        player_symbols[1] = 'O'

    print("{}, unfortunately you don't have choice. You have been assigned the symbol {}.".format(player_2_name,
                                                                                                  player_symbols[1]))
player_input()

【讨论】:

  • 所以您的解决方案有效,但为什么只有变量而不是列表无效?
  • 只有变量不起作用,因为当您在函数内部创建 player_1_symbol 时,python 会将其识别为新变量,而不是全局变量。正如其他评论者所说,在您的代码中使用“全局 x”不是最佳做法,因此我制作了一个包含您的变量的列表。
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