Question

我和我的伴侣已经为此工作了几个小时，无法解决。方向在某些领域含糊不清，我们的教授没有很好地分解它以帮助我们。这是指南的链接。我相信他们不太清楚，但是如果我做错了，请改正我，请多加思考https://imgur.com/a/huHnwos

我相信我们最大的问题是unlock（combination）和set_new_combination（new_combination）方法。我可以弄清楚 str （）方法，因为它并不是很难做到的。我们已经尝试了教授告诉我们的尝试，但是没有成功。

class Lock:
    def __init__(self, combination = 0):
        self.combination = combination
        self.locked = False

    def lock(self):
        self.locked = True

    def unlock(self, combination):
        if combination == True or combination == 0:
             self.locked = False

    def set_new_combination(self, new_combination):
        if self.locked == False:
            self.combination = new_combination

   def is_locked(self):
        if self.locked == True or self.combination == True:
            return True
        else:
            return False

    def __eq__(self, other):
        if other is not None and type(other) == type(self):
            if self.combination == other.new_combination:
                return False

    def __str__(self):
        return self.combination, ',', self.locked

预期结果应该是可以正常工作的基本密码锁。

Answer 1

您的unlock方法正在尝试将布尔值与数字（组合）进行比较。更改为如下所示：

def unlock(self, combination):
    if combination == self.combination:
         self.locked = False

您还可以通过is_locked方法执行此操作，因此也应该更改此内容：

def is_locked(self):
    return self.locked

（每次发现自己写成if x return True else return False的东西时，只要条件简单，几乎总是可以用return x代替）。

set_new_combination工作正常；我不知道您看到了什么问题。

最后，您的__str__方法应该实际上返回一个字符串：

def __str__(self):
    return '[' + str(self.combination) + ', ' + 'locked' if self.locked else 'unlocked' + ']'

Answer 2

您的代码有几个问题。首先，只有在combination == 0 or combination == 1才执行解锁方法中的if语句，这与锁的组合（self.combination）没有关系。在is_locked方法中，您仅应返回self.locked，而无需返回if。 __eq__方法也可以简化。并且__str__方法实际上应该返回String。

class Lock:
    def __init__(self, combination = 0):
        self.combination = combination
        self.locked = False

    def lock(self):
        self.locked = True

    def unlock(self, combination):
        if self.combination == combination:
             self.locked = False

    def set_new_combination(self, new_combination):
        if not self.locked:
            self.combination = new_combination

    def is_locked(self):
        return self.locked

    def __eq__(self, other):
        return isinstance(other, Lock) and self.combination == other.combination

    def __str__(self):
        return f'{self.combination}, { "locked" if self.locked else "unlocked"}'

Answer 3

这是我基于提供的指令的实现，带有注释的地方与代码有出入。

class Lock:
    def __init__(self, combination = 0):  # No change here
        self.combination = combination
        self.locked = False

    def lock(self):
        # Although a test of self.locked is redundant, the instructions state
        # "...if invoked a second time this, method should do nothing."
        if not self.locked:
            self.locked = True

    def unlock(self, combination):
        # You were not testing the stored combination against the one passed to the method.
        # it does not matter if the stored combination is zero or a different number,
        # you still need to check for equality.
        # You also need a test as with lock() to satisfy the "if invoked a second time this, 
        # method should do nothing" requirement.
        if self.locked and self.combination == combination:
            self.locked = False

    def set_new_combination(self, new_combination):
        # You can simply the `if` condition, there's no need to test against False
        if not self.locked:
            self.combination = new_combination

    def is_locked(self):
        # I don't know why you are testing the self.combination value, you 
        # only need to return the state of the lock   
        return self.locked

    def __eq__(self, other):
        # You have the correct guard conditions but were returning False when 
        # the combinations matched. You can simply return the comparison result.
        if other is not None and type(other) == type(self):
            return self.combination == other.new_combination

    def __str__(self):
        # For some reason the output format specified for this appears to put it in a list
        # (the square brackets) but as it's only for display we'll "fake" the list.
        # The `if` statement prints the word 'locked' or 'unlocked' depending on the 
        # `self.locked` state.
        return '[{}, {}]'.format(self.combination, 'locked' if self.locked else 'unlocked')

密码锁程序

3 个答案: