如何在Python中输入"如果函数被多次调用"？

Question

所以我正在使用Python和Kivy设计一个刽子手游戏,我想添加一个赢/输选项.

我定义的其中一个函数是Button_pressed,如果它被按下则会隐藏按钮,但是我希望函数man_is_hung()能够显示"如果按钮被按下了6次,则显示"游戏结束"."

有人请帮帮我吗？

 def button_pressed(button):
        for (letter, label) in CurrentWord:
            if (letter.upper() == button.text): label.text=letter 
        button.text=" " # hide the letter to indicate it's been tried

def man_is_hung():
    if button_pressed(button)

Answer 1

使用装饰者:

例:

class count_calls(object):
    def __init__(self, func):
        self.count = 0
        self.func = func
    def __call__(self, *args, **kwargs):
        # if self.count == 6 : do something
        self.count += 1
        return self.func(*args, **kwargs)

@count_calls
def func(x, y):
    return x + y

演示:

>>> for _ in range(4): func(0, 0)
>>> func.count
4
>>> func(0, 0)
0
>>> func.count
5

在py3.x中,您可以使用nonlocal函数而不是类来实现相同的功能:

def count_calls(func):
    count = 0
    def wrapper(*args, **kwargs):
        nonlocal count
        if count == 6:
            raise TypeError('Enough button pressing')
        count += 1
        return func(*args, **kwargs)
    return wrapper

@count_calls
def func(x, y):
    return x + y

演示:

>>> for _ in range(6):func(1,1)
>>> func(1, 1)
    ...
    raise TypeError('Enough button pressing')
TypeError: Enough button pressing